K’th Smallest/Largest Element in Unsorted Array
Given an array and a number K where K is smaller than the size of the array. Find the K’th smallest element in the given array. Given that all array elements are distinct.
Examples:
Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 3
Output: 7Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 4
Output: 10
We have discussed a similar problem to print k largest elements.
K’th smallest element in an unsorted array using sorting:
Sort the given array and return the element at index K-1 in the sorted array.
Follow the given steps to solve the problem:
- Sort the input array in the increasing order
- Return the element at the K-1 index (0 – Based indexing) in the sorted array
Below is the Implementation of the above approach:
C
// C program to find K'th smallest element #include <stdio.h> #include <stdlib.h> // Compare function for qsort int cmpfunc( const void * a, const void * b) { return (*( int *)a - *( int *)b); } // Function to return K'th smallest // element in a given array int kthSmallest( int arr[], int N, int K) { // Sort the given array qsort (arr, N, sizeof ( int ), cmpfunc); // Return k'th element in the sorted array return arr[K - 1]; } // Driver's code int main() { int arr[] = { 12, 3, 5, 7, 19 }; int N = sizeof (arr) / sizeof (arr[0]), K = 2; // Function call printf ( "K'th smallest element is %d" , kthSmallest(arr, N, K)); return 0; } // This code is contributed by Sania Kumari Gupta // (kriSania804) |
C++
// C++ program to find K'th smallest element #include <bits/stdc++.h> using namespace std; // Function to return K'th smallest element in a given array int kthSmallest( int arr[], int N, int K) { // Sort the given array sort(arr, arr + N); // Return k'th element in the sorted array return arr[K - 1]; } // Driver's code int main() { int arr[] = { 12, 3, 5, 7, 19 }; int N = sizeof (arr) / sizeof (arr[0]), K = 2; // Function call cout << "K'th smallest element is " << kthSmallest(arr, N, K); return 0; } // This code is contributed by Sania Kumari Gupta // (kriSania804) |
Java
// Java code for Kth smallest element // in an array import java.util.Arrays; import java.util.Collections; class GFG { // Function to return K'th smallest // element in a given array public static int kthSmallest(Integer[] arr, int K) { // Sort the given array Arrays.sort(arr); // Return K'th element in // the sorted array return arr[K - 1 ]; } // driver's code public static void main(String[] args) { Integer arr[] = new Integer[] { 12 , 3 , 5 , 7 , 19 }; int K = 2 ; // Function call System.out.print( "K'th smallest element is " + kthSmallest(arr, K)); } } // This code is contributed by Chhavi |
Python3
# Python3 program to find K'th smallest # element # Function to return K'th smallest # element in a given array def kthSmallest(arr, N, K): # Sort the given array arr.sort() # Return k'th element in the # sorted array return arr[K - 1 ] # Driver code if __name__ = = '__main__' : arr = [ 12 , 3 , 5 , 7 , 19 ] N = len (arr) K = 2 # Function call print ( "K'th smallest element is" , kthSmallest(arr, N, K)) # This code is contributed by # Shrikant13 |
C#
// C# code for Kth smallest element // in an array using System; class GFG { // Function to return K'th smallest // element in a given array public static int kthSmallest( int [] arr, int K) { // Sort the given array Array.Sort(arr); // Return k'th element in // the sorted array return arr[K - 1]; } // driver's program public static void Main() { int [] arr = new int [] { 12, 3, 5, 7, 19 }; int K = 2; // Function call Console.Write( "K'th smallest element" + " is " + kthSmallest(arr, K)); } } // This code is contributed by nitin mittal. |
PHP
<?php // Simple PHP program to find // K'th smallest element // Function to return K'th smallest // element in a given array function kthSmallest( $arr , $N , $K ) { // Sort the given array sort( $arr ); // Return k'th element // in the sorted array return $arr [ $K - 1]; } // Driver's Code $arr = array (12, 3, 5, 7, 19); $N = count ( $arr ); $K = 2; // Function call echo "K'th smallest element is " , kthSmallest( $arr , $N , $K ); // This code is contributed by anuj_67. ?> |
Javascript
// Simple Javascript program to find K'th smallest element // Function to return K'th smallest element in a given array function kthSmallest(arr, N, K) { // Sort the given array arr.sort((a,b) => a-b); // Return k'th element in the sorted array return arr[K - 1]; } // Driver program to test above methods let arr = [12, 3, 5, 7, 19]; let N = arr.length, K = 2; document.write( "K'th smallest element is " + kthSmallest(arr, N, K)); //This code is contributed by Mayank Tyagi |
K'th smallest element is 5
Time Complexity: O(N log N)
Auxiliary Space: O(1)
K’th smallest element in an unsorted array using set data structure:
Set data structure can be used to find the kth smallest element as it stores the distinct elements in sorted order. Set can be used because it is mentioned in the question that all the elements in the array are distinct.
Follow the given steps to solve the problem:
- Insert all array elements into the set
- Advance the iterator to the Kth element in the set
- Return the value of the element at which the iterator is pointing
Below is the Implementation of the above approach:
C++
// C++ code for the above approach #include <bits/stdc++.h> using namespace std; int main() { int arr[] = { 12, 3, 5, 7, 19 }; int N = sizeof (arr) / sizeof (arr[0]); int K = 4; set< int > s(arr, arr + N); // s.begin() returns a pointer to first // element in the set set< int >::iterator itr = s.begin(); advance(itr, K - 1); // itr points to kth element in set cout << *itr << "\n" ; return 0; } |
Java
// Java code for the above approach import java.util.*; class GFG { public static void main(String[] args) { int [] arr = { 12 , 3 , 5 , 7 , 19 }; int N = arr.length; int K = 4 ; // since counting starts from 0 so to find kth // element we need to reduce K by 1 K--; // for storing elements in sorted form // in set we will use TreeSet Set<Integer> s = new TreeSet<Integer>(); // Adding elements to set for ( int i = 0 ; i < N; i++) s.add(arr[i]); // Use iterator method of Iterator // for the traversal Iterator<Integer> itr = s.iterator(); while (K > 0 ) { itr.next(); K--; } // itr points to the Kth element in the set System.out.println(itr.next()); } } // This code is contributed by Abhijeet Kumar(abhijeet19403) |
Python3
# Python3 code for the above approach if __name__ = = '__main__' : arr = [ 12 , 3 , 5 , 7 , 19 ] N = len (arr) K = 4 s = set (arr) for itr in s: if K = = 1 : print (itr) # itr is the Kth element in the set break K - = 1 # This code is contributed by Abhijeet Kumar(abhijeet19403) |
C#
// C# code for the above approach using System; using System.Collections.Generic; class GFG { // Driver code public static void Main() { int [] arr = { 12, 3, 5, 7, 19 }; int N = arr.Length; int K = 4; SortedSet< int > s = new SortedSet< int >(); foreach ( int i in arr) s.Add(i); foreach ( int itr in s) { if (K == 1) { Console.WriteLine(itr); // itr is the kth element in the set break ; } K--; } } } // This code is contributed by Abhijeet Kumar(abhijeet19403) |
12
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
K’th smallest element in an unsorted array using heap data structure:
K’th smallest element in an unsorted array using Min-Heap
Min-Heap can be used to find the kth smallest element, by inserting all the elements into Min-Heap and then and call extractMin() function K times.
Follow the given steps to solve the problem:
- Insert all the array elements into the Min-Heap
- Call extractMin() function K times
- Return the value obtained at the last call of extractMin() function
Below is the Implementation of the above approach:
C++
// C++ program to find K'th smallest element using min heap #include <climits> #include <iostream> using namespace std; // Prototype of a utility function to swap two integers void swap( int * x, int * y); // A class for Min Heap class MinHeap { int * harr; // pointer to array of elements in heap int capacity; // maximum possible size of min heap int heap_size; // Current number of elements in min heap public : MinHeap( int a[], int size); // Constructor // To minheapify subtree rooted with index i void MinHeapify( int i); int parent( int i) { return (i - 1) / 2; } int left( int i) { return (2 * i + 1); } int right( int i) { return (2 * i + 2); } int extractMin(); // extracts root (minimum) element int getMin() { return harr[0]; } // Returns minimum }; MinHeap::MinHeap( int a[], int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1) / 2; while (i >= 0) { MinHeapify(i); i--; } } // Method to remove minimum element (or root) from min heap int MinHeap::extractMin() { if (heap_size == 0) return INT_MAX; // Store the minimum value. int root = harr[0]; // If there are more than 1 items, move the last item to // root and call heapify. if (heap_size > 1) { harr[0] = harr[heap_size - 1]; MinHeapify(0); } heap_size--; return root; } // A recursive method to heapify a subtree with root at // given index This method assumes that the subtrees are // already heapified void MinHeap::MinHeapify( int i) { int l = left(i); int r = right(i); int smallest = i; if (l < heap_size && harr[l] < harr[i]) smallest = l; if (r < heap_size && harr[r] < harr[smallest]) smallest = r; if (smallest != i) { swap(&harr[i], &harr[smallest]); MinHeapify(smallest); } } // A utility function to swap two elements void swap( int * x, int * y) { int temp = *x; *x = *y; *y = temp; } // Function to return k'th smallest element in a given array int kthSmallest( int arr[], int N, int K) { // Build a heap of n elements: O(n) time MinHeap mh(arr, N); // Do extract min (K-1) times for ( int i = 0; i < K - 1; i++) mh.extractMin(); // Return root return mh.getMin(); } // Driver's code int main() { int arr[] = { 12, 3, 5, 7, 19 }; int N = sizeof (arr) / sizeof (arr[0]), K = 2; // Function call cout << "K'th smallest element is " << kthSmallest(arr, N, K); return 0; } |
Java
// Java program to find K'th smallest element using min heap import java.util.*; public class GFG { // A class for Max Heap class MinHeap { int [] harr; // pointer to array of elements in heap int capacity; // maximum possible size of min heap int heap_size; // Current number of elements in min // heap int parent( int i) { return (i - 1 ) / 2 ; } int left( int i) { return (( 2 * i) + 1 ); } int right( int i) { return (( 2 * i) + 2 ); } int getMin() { return harr[ 0 ]; } // Returns minimum // to replace root with new node x and heapify() new // root void replaceMax( int x) { this .harr[ 0 ] = x; minHeapify( 0 ); } MinHeap( int a[], int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1 ) / 2 ; while (i >= 0 ) { minHeapify(i); i--; } } // Method to remove maximum element (or root) from // min heap int extractMin() { if (heap_size == 0 ) return Integer.MAX_VALUE; // Store the maximum value. int root = harr[ 0 ]; // If there are more than 1 items, move the last // item to root and call heapify. if (heap_size > 1 ) { harr[ 0 ] = harr[heap_size - 1 ]; minHeapify( 0 ); } heap_size--; return root; } // A recursive method to heapify a subtree with root // at given index This method assumes that the // subtrees are already heapified void minHeapify( int i) { int l = left(i); int r = right(i); int smallest = i; if (l < heap_size && harr[l] < harr[i]) smallest = l; if (r < heap_size && harr[r] < harr[smallest]) smallest = r; if (smallest != i) { int t = harr[i]; harr[i] = harr[smallest]; harr[smallest] = t; minHeapify(smallest); } } }; // Function to return k'th largest element in a given // array int kthSmallest( int arr[], int N, int K) { // Build a heap of first k elements: O(k) time MinHeap mh = new MinHeap(arr, N); // Process remaining n-k elements. If current // element is smaller than root, replace root with // current element for ( int i = 0 ; i < K - 1 ; i++) mh.extractMin(); // Return root return mh.getMin(); } // Driver's code public static void main(String[] args) { int arr[] = { 12 , 3 , 5 , 7 , 19 }; int N = arr.length, K = 2 ; GFG gfg = new GFG(); // Function call System.out.print( "K'th smallest element is " + gfg.kthSmallest(arr, N, K)); } } // This code is contributed by avanitrachhadiya2155 |
Python3
# Python3 program to find K'th smallest element # using min heap # Class for Min Heap class MinHeap: # Constructor def __init__( self , a, size): # list of elements in the heap self .harr = a # maximum possible size of min heap self .capacity = None # current number of elements in min heap self .heap_size = size i = int (( self .heap_size - 1 ) / 2 ) while i > = 0 : self .minHeapify(i) i - = 1 def parent( self , i): return (i - 1 ) / 2 def left( self , i): return 2 * i + 1 def right( self , i): return 2 * i + 2 # Returns minimum def getMin( self ): return self .harr[ 0 ] # Method to remove minimum element (or root) # from min heap def extractMin( self ): if self .heap_size = = 0 : return float ( "inf" ) # Store the minimum value root = self .harr[ 0 ] # If there are more than 1 items, move the last item # to root and call heapify if self .heap_size > 1 : self .harr[ 0 ] = self .harr[ self .heap_size - 1 ] self .minHeapify( 0 ) self .heap_size - = 1 return root # A recursive method to heapify a subtree with root at # given index. This method assumes that the subtrees # are already heapified def minHeapify( self , i): l = self .left(i) r = self .right(i) smallest = i if ((l < self .heap_size) and ( self .harr[l] < self .harr[i])): smallest = l if ((r < self .heap_size) and ( self .harr[r] < self .harr[smallest])): smallest = r if smallest ! = i: self .harr[i], self .harr[smallest] = ( self .harr[smallest], self .harr[i]) self .minHeapify(smallest) # Function to return k'th smallest element in a given array def kthSmallest(arr, N, K): # Build a heap of n elements in O(n) time mh = MinHeap(arr, N) # Do extract min (k-1) times for i in range (K - 1 ): mh.extractMin() # Return root return mh.getMin() # Driver's code if __name__ = = '__main__' : arr = [ 12 , 3 , 5 , 7 , 19 ] N = len (arr) K = 2 # Function call print ( "K'th smallest element is" , kthSmallest(arr, N, K)) # This Code is contributed by Kevin Joshi |
C#
using System; public class GFG { public class MinHeap { int [] harr; // Pointer to array of elements in heap // int capacity; // maximum possible size of min // heap int heap_size; // Current number of elements in min // heap int parent( int i) { return (i - 1) / 2; } int left( int i) { return ((2 * i) + 1); } int right( int i) { return ((2 * i) + 2); } public int getMin() { return harr[0]; } // Returns minimum // To replace root with new node x and heapify() new // root public void replaceMax( int x) { this .harr[0] = x; minHeapify(0); } public MinHeap( int [] a, int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1) / 2; while (i >= 0) { minHeapify(i); i--; } } // Method to remove maximum element (or root) from // min heap public int extractMin() { if (heap_size == 0) return Int32.MaxValue; // Store the maximum value. int root = harr[0]; // If there are more than 1 items, move the last // item to root and call heapify. if (heap_size > 1) { harr[0] = harr[heap_size - 1]; minHeapify(0); } heap_size--; return root; } // A recursive method to heapify a subtree with root // at given index This method assumes that the // subtrees are already heapified public void minHeapify( int i) { int l = left(i); int r = right(i); int smallest = i; if (l < heap_size && harr[l] < harr[i]) smallest = l; if (r < heap_size && harr[r] < harr[smallest]) smallest = r; if (smallest != i) { int t = harr[i]; harr[i] = harr[smallest]; harr[smallest] = t; minHeapify(smallest); } } }; // Function to return k'th largest element in a given // array int kthSmallest( int [] arr, int N, int K) { // Build a heap of first k elements: O(k) time MinHeap mh = new MinHeap(arr, N); // Process remaining n-k elements. If current // element is smaller than root, replace root with // current element for ( int i = 0; i < K - 1; i++) mh.extractMin(); // Return root return mh.getMin(); } // Driver's code static public void Main() { int [] arr = { 12, 3, 5, 7, 19 }; int N = arr.Length, K = 2; GFG gfg = new GFG(); // Function call Console.Write( "K'th smallest element is " + gfg.kthSmallest(arr, N, K)); } } // This code is contributed by rag2127 |
Javascript
// Javascript program to find K'th smallest element using min heap // class for Max Heap class MinHeap { parent(i) { return (i - 1) / 2; }; left(i) { return ((2 * i) + 1); }; right(i) { return ((2 * i) + 2); } getMin() { return this .harr[0]; } // Returns minimum // to replace root with new node x and heapify() new root replaceMax(x) { this .harr[0] = x; minHeapify(0); } constructor(a, size) { this .heap_size = size; this .harr = a; // store address of array let i = ( this .heap_size - 1) / 2; while (i >= 0) { this .minHeapify(i); i--; } } // Method to remove maximum element (or root) from min heap extractMin() { if ( this .heap_size == 0) return Number.MAX_SAFE_INTEGER; // Store the maximum value. let root = this .harr[0]; // If there are more than 1 items, move the last item to root // and call heapify. if ( this .heap_size > 1) { this .harr[0] = this .harr[ this .heap_size - 1]; this .minHeapify(0); } this .heap_size--; return root; } // A recursive method to heapify a subtree with root at given index // This method assumes that the subtrees are already heapified minHeapify(i) { let l = this .left(i); let r = this .right(i); let smallest = i; if (l < this .heap_size && this .harr[l] < this .harr[i]) smallest = l; if (r < this .heap_size && this .harr[r] < this .harr[smallest]) smallest = r; if (smallest != i) { let t = this .harr[i]; this .harr[i] = this .harr[smallest]; this .harr[smallest] = t; this .minHeapify(smallest); } } }; // Function to return k'th largest element in a given array function kthSmallest(arr, N, K) { // Build a heap of first k elements: O(k) time let mh = new MinHeap(arr, N); // Process remaining N-K elements. If current element is // smaller than root, replace root with current element for (let i = 0; i < K - 1; i++) mh.extractMin(); // Return root return mh.getMin(); } // Driver program to test above methods let arr = [12, 3, 5, 7, 19]; let N = arr.length, K = 2; document.write( "K'th smallest element is " + kthSmallest(arr, N, K)); // This code is contributed by gfgking. |
K'th smallest element is 5
Time complexity: O(N + K Log N).
Auxiliary Space: O(N)
K’th smallest element in an unsorted array using Max-Heap
Max-Heap can be used to find the kth smallest element, by inserting first K elements into Max-Heap and then compare remaining elements with the root of the Max-Heap and if the element is less than the root then remove the root and insert this element into the heap and finally return root of the Max-Heap
Follow the given steps to solve the problem:
- Build a Max-Heap MH of the first K elements (arr[0] to arr[K-1]) of the given array.
- For each element, after the Kth element (arr[K] to arr[n-1]), compare it with the root of MH.
- If the element is less than the root then make it the root and call heapify for Max-Heap MH
- b) Else ignore it.
- Finally, the root of the MH is the Kth smallest element.
Below is the Implementation of the above approach:
C++
// C++ program to find k'th smallest element using max heap #include <bits/stdc++.h> using namespace std; // Prototype of a utility function to swap two integers void swap( int * x, int * y); // A class for Max Heap class MaxHeap { int * harr; // pointer to array of elements in heap int capacity; // maximum possible size of max heap int heap_size; // Current number of elements in max heap public : MaxHeap( int a[], int size); // Constructor void maxHeapify( int i); // To maxHeapify subtree rooted with index i int parent( int i) { return (i - 1) / 2; } int left( int i) { return (2 * i + 1); } int right( int i) { return (2 * i + 2); } int extractMax(); // extracts root (maximum) element int getMax() { return harr[0]; } // Returns maximum // to replace root with new node x and heapify() new // root void replaceMax( int x) { harr[0] = x; maxHeapify(0); } }; MaxHeap::MaxHeap( int a[], int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1) / 2; while (i >= 0) { maxHeapify(i); i--; } } // Method to remove maximum element (or root) from max heap int MaxHeap::extractMax() { if (heap_size == 0) return INT_MAX; // Store the maximum value. int root = harr[0]; // If there are more than 1 items, move the last item to // root and call heapify. if (heap_size > 1) { harr[0] = harr[heap_size - 1]; maxHeapify(0); } heap_size--; return root; } // A recursive method to heapify a subtree with root at // given index This method assumes that the subtrees are // already heapified void MaxHeap::maxHeapify( int i) { int l = left(i); int r = right(i); int largest = i; if (l < heap_size && harr[l] > harr[i]) largest = l; if (r < heap_size && harr[r] > harr[largest]) largest = r; if (largest != i) { swap(&harr[i], &harr[largest]); maxHeapify(largest); } } // A utility function to swap two elements void swap( int * x, int * y) { int temp = *x; *x = *y; *y = temp; } // Function to return k'th largest element in a given array int kthSmallest( int arr[], int N, int K) { // Build a heap of first k elements: O(k) time MaxHeap mh(arr, K); // Process remaining n-k elements. If current element // is smaller than root, replace root with current // element for ( int i = K; i < N; i++) if (arr[i] < mh.getMax()) mh.replaceMax(arr[i]); // Return root return mh.getMax(); } // Driver's code int main() { int arr[] = { 12, 3, 5, 7, 19 }; int N = sizeof (arr) / sizeof (arr[0]), K = 4; // Function call cout << "K'th smallest element is " << kthSmallest(arr, N, K); return 0; } |
Java
// A Java program to find k'th smallest element using max // heap import java.util.*; class GFG { // A class for Max Heap class MaxHeap { int [] harr; // pointer to array of elements in heap int capacity; // maximum possible size of max heap int heap_size; // Current number of elements in max // heap int parent( int i) { return (i - 1 ) / 2 ; } int left( int i) { return ( 2 * i + 1 ); } int right( int i) { return ( 2 * i + 2 ); } int getMax() { return harr[ 0 ]; } // Returns maximum // to replace root with new node x and heapify() new // root void replaceMax( int x) { this .harr[ 0 ] = x; maxHeapify( 0 ); } MaxHeap( int a[], int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1 ) / 2 ; while (i >= 0 ) { maxHeapify(i); i--; } } // Method to remove maximum element (or root) from // max heap int extractMax() { if (heap_size == 0 ) return Integer.MAX_VALUE; // Store the maximum value. int root = harr[ 0 ]; // If there are more than 1 items, move the last // item to root and call heapify. if (heap_size > 1 ) { harr[ 0 ] = harr[heap_size - 1 ]; maxHeapify( 0 ); } heap_size--; return root; } // A recursive method to heapify a subtree with root // at given index This method assumes that the // subtrees are already heapified void maxHeapify( int i) { int l = left(i); int r = right(i); int largest = i; if (l < heap_size && harr[l] > harr[i]) largest = l; if (r < heap_size && harr[r] > harr[largest]) largest = r; if (largest != i) { int t = harr[i]; harr[i] = harr[largest]; harr[largest] = t; maxHeapify(largest); } } }; // Function to return k'th largest element in a given // array int kthSmallest( int arr[], int N, int K) { // Build a heap of first k elements: O(k) time MaxHeap mh = new MaxHeap(arr, K); // Process remaining n-k elements. If current // element is smaller than root, replace root with // current element for ( int i = K; i < N; i++) if (arr[i] < mh.getMax()) mh.replaceMax(arr[i]); // Return root return mh.getMax(); } // Driver's code public static void main(String[] args) { int arr[] = { 12 , 3 , 5 , 7 , 19 }; int N = arr.length, K = 4 ; GFG gfg = new GFG(); // Function call System.out.print( "K'th smallest element is " + gfg.kthSmallest(arr, N, K)); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 program to find K'th smallest element # using max heap # Class for Max Heap class MaxHeap: # Constructor def __init__( self , a, size): # list of elements in the heap self .harr = a # maximum possible size of max heap self .capacity = None # current number of elements in max heap self .heap_size = size i = int (( self .heap_size - 1 ) / 2 ) while i > = 0 : self .maxHeapify(i) i - = 1 def parent( self , i): return (i - 1 ) / 2 def left( self , i): return 2 * i + 1 def right( self , i): return 2 * i + 2 # Returns maximum def getMax( self ): return self .harr[ 0 ] # to replace root with new node x and heapify() new root def replaceMax( self , x): self .harr[ 0 ] = x self .maxHeapify( 0 ) # Method to remove maximum element (or root) # from max heap def extractMin( self ): if self .heap_size = = 0 : return float ( "inf" ) # Store the maximum value. root = self .harr[ 0 ] # If there are more than 1 items, move the # last item to root and call heapify if self .heap_size > 1 : self .harr[ 0 ] = self .harr[ self .heap_size - 1 ] self .maxHeapify( 0 ) self .heap_size - = 1 return root # A recursive method to heapify a subtree with root at # given index. This method assumes that the subtrees # are already heapified def maxHeapify( self , i): l = self .left(i) r = self .right(i) largest = i if ((l < self .heap_size) and ( self .harr[l] > self .harr[i])): largest = l if ((r < self .heap_size) and ( self .harr[r] > self .harr[largest])): largest = r if largest ! = i: self .harr[i], self .harr[largest] = ( self .harr[largest], self .harr[i]) self .maxHeapify(largest) # Function to return k'th smallest element in a given array def kthSmallest(arr, N, K): # Build a heap of first k elements in O(k) time mh = MaxHeap(arr, K) # Process remaining n-k elements. If current element is # smaller than root, replace root with current element for i in range (K, N): if arr[i] < mh.getMax(): mh.replaceMax(arr[i]) # Return root return mh.getMax() # Driver's code if __name__ = = '__main__' : arr = [ 12 , 3 , 5 , 7 , 19 ] N = len (arr) K = 4 # Function call print ( "K'th smallest element is" , kthSmallest(arr, N, K)) # Code contributed by Kevin Joshi |
C#
// C# program to find k'th smallest element using max heap using System; public class GFG { // A class for Max Heap public class MaxHeap { public int [] harr; // pointer to array of elements in // heap public int capacity; // maximum possible size of max // heap public int heap_size; // Current number of elements in // max heap public int parent( int i) { return (i - 1) / 2; } public int left( int i) { return (2 * i + 1); } public int right( int i) { return (2 * i + 2); } public int getMax() { return harr[0]; } // Returns maximum // to replace root with new node x and heapify() new // root public void replaceMax( int x) { this .harr[0] = x; maxHeapify(0); } public MaxHeap( int [] a, int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1) / 2; while (i >= 0) { maxHeapify(i); i--; } } // Method to remove maximum element (or root) from // max heap public int extractMax() { if (heap_size == 0) return int .MaxValue; // Store the maximum value. int root = harr[0]; // If there are more than 1 items, move the last // item to root and call heapify. if (heap_size > 1) { harr[0] = harr[heap_size - 1]; maxHeapify(0); } heap_size--; return root; } // A recursive method to heapify a subtree with root // at given index This method assumes that the // subtrees are already heapified public void maxHeapify( int i) { int l = left(i); int r = right(i); int largest = i; if (l < heap_size && harr[l] > harr[i]) largest = l; if (r < heap_size && harr[r] > harr[largest]) largest = r; if (largest != i) { int t = harr[i]; harr[i] = harr[largest]; harr[largest] = t; maxHeapify(largest); } } }; // Function to return k'th largest element in a given // array int kthSmallest( int [] arr, int N, int K) { // Build a heap of first k elements: O(k) time MaxHeap mh = new MaxHeap(arr, K); // Process remaining n-k elements. If current // element is smaller than root, replace root with // current element for ( int i = K; i < N; i++) if (arr[i] < mh.getMax()) mh.replaceMax(arr[i]); // Return root return mh.getMax(); } // Driver's code public static void Main(String[] args) { int [] arr = { 12, 3, 5, 7, 19 }; int N = arr.Length, K = 4; GFG gfg = new GFG(); // Function call Console.Write( "K'th smallest element is " + gfg.kthSmallest(arr, N, K)); } } // This code is contributed by gauravrajput1 |
K'th smallest element is 12
Time Complexity: O(K + (N-K) * Log K)
Auxiliary Space: O(K)
K’th smallest element in an unsorted array using QuickSelect:
This is an optimization over method 1, if QuickSort is used as a sorting algorithm in first step. In QuickSort, pick a pivot element, then move the pivot element to its correct position and partition the surrounding array. The idea is, not to do complete quicksort, but stop at the point where pivot itself is k’th smallest element. Also, not to recur for both left and right sides of pivot, but recur for one of them according to the position of pivot.
Follow the given steps to solve the problem:
- Run quick sort algorithm on the input array
- In this algorithm pick a pivot element and move it to it’s correct position
- Now, if index of pivot is equal to K then return the value, else if the index of pivot is greater than K, then recur for the left subarray, else recur for the right subarray
- Repeat this process until the element at index K is not found
Below is the Implementation of the above approach:
C
// C code for the above approach #include <limits.h> #include <stdio.h> int partition( int arr[], int l, int r); // This function returns K'th smallest element in arr[l..r] // using QuickSort based method. ASSUMPTION: ALL ELEMENTS IN // ARR[] ARE DISTINCT int kthSmallest( int arr[], int l, int r, int K) { // If k is smaller than number of elements in array if (K > 0 && K <= r - l + 1) { // Partition the array around last element and get // position of pivot element in sorted array int pos = partition(arr, l, r); // If position is same as k if (pos - l == K - 1) return arr[pos]; if (pos - l > K - 1) // If position is more, recur // for left subarray return kthSmallest(arr, l, pos - 1, K); // Else recur for right subarray return kthSmallest(arr, pos + 1, r, K - pos + l - 1); } // If k is more than number of elements in array return INT_MAX; } void swap( int * a, int * b) { int temp = *a; *a = *b; *b = temp; } // Standard partition process of QuickSort(). It considers // the last element as pivot and moves all smaller element // to left of it and greater elements to right int partition( int arr[], int l, int r) { int x = arr[r], i = l; for ( int j = l; j <= r - 1; j++) { if (arr[j] <= x) { swap(&arr[i], &arr[j]); i++; } } swap(&arr[i], &arr[r]); return i; } // Driver's code int main() { int arr[] = { 12, 3, 5, 7, 4, 19, 26 }; int N = sizeof (arr) / sizeof (arr[0]), K = 3; // Function call printf ( "K'th smallest element is %d" , kthSmallest(arr, 0, N - 1, K)); return 0; } |
C++
// C++ code for the above approach #include <bits/stdc++.h> using namespace std; int partition( int arr[], int l, int r); // This function returns K'th smallest element in arr[l..r] // using QuickSort based method. ASSUMPTION: ALL ELEMENTS IN // ARR[] ARE DISTINCT int kthSmallest( int arr[], int l, int r, int K) { // If k is smaller than number of elements in array if (K > 0 && K <= r - l + 1) { // Partition the array around last element and get // position of pivot element in sorted array int pos = partition(arr, l, r); // If position is same as k if (pos - l == K - 1) return arr[pos]; if (pos - l > K - 1) // If position is more, recur // for left subarray return kthSmallest(arr, l, pos - 1, K); // Else recur for right subarray return kthSmallest(arr, pos + 1, r, K - pos + l - 1); } // If k is more than number of elements in array return INT_MAX; } void swap( int * a, int * b) { int temp = *a; *a = *b; *b = temp; } // Standard partition process of QuickSort(). It considers // the last element as pivot and moves all smaller element // to left of it and greater elements to right int partition( int arr[], int l, int r) { int x = arr[r], i = l; for ( int j = l; j <= r - 1; j++) { if (arr[j] <= x) { swap(&arr[i], &arr[j]); i++; } } swap(&arr[i], &arr[r]); return i; } // Driver's code int main() { int arr[] = { 12, 3, 5, 7, 4, 19, 26 }; int N = sizeof (arr) / sizeof (arr[0]), K = 3; // Function call cout << "K'th smallest element is " << kthSmallest(arr, 0, N - 1, K); return 0; } |
Java
// Java code for kth smallest element in an array import java.util.Arrays; import java.util.Collections; class GFG { // Standard partition process of QuickSort. // It considers the last element as pivot // and moves all smaller element to left of // it and greater elements to right public static int partition(Integer[] arr, int l, int r) { int x = arr[r], i = l; for ( int j = l; j <= r - 1 ; j++) { if (arr[j] <= x) { // Swapping arr[i] and arr[j] int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i++; } } // Swapping arr[i] and arr[r] int temp = arr[i]; arr[i] = arr[r]; arr[r] = temp; return i; } // This function returns k'th smallest element // in arr[l..r] using QuickSort based method. // ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT public static int kthSmallest(Integer[] arr, int l, int r, int K) { // If k is smaller than number of elements // in array if (K > 0 && K <= r - l + 1 ) { // Partition the array around last // element and get position of pivot // element in sorted array int pos = partition(arr, l, r); // If position is same as k if (pos - l == K - 1 ) return arr[pos]; // If position is more, recur for // left subarray if (pos - l > K - 1 ) return kthSmallest(arr, l, pos - 1 , K); // Else recur for right subarray return kthSmallest(arr, pos + 1 , r, K - pos + l - 1 ); } // If k is more than number of elements // in array return Integer.MAX_VALUE; } // Driver's code public static void main(String[] args) { Integer arr[] = new Integer[] { 12 , 3 , 5 , 7 , 4 , 19 , 26 }; int K = 3 ; // Function call System.out.print( "K'th smallest element is " + kthSmallest(arr, 0 , arr.length - 1 , K)); } } // This code is contributed by Chhavi |
Python3
# Python3 code for the above approach # This function returns k'th smallest element # in arr[l..r] using QuickSort based method. # ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT import sys def kthSmallest(arr, l, r, K): # If k is smaller than number of # elements in array if (K > 0 and K < = r - l + 1 ): # Partition the array around last # element and get position of pivot # element in sorted array pos = partition(arr, l, r) # If position is same as k if (pos - l = = K - 1 ): return arr[pos] if (pos - l > K - 1 ): # If position is more, # recur for left subarray return kthSmallest(arr, l, pos - 1 , K) # Else recur for right subarray return kthSmallest(arr, pos + 1 , r, K - pos + l - 1 ) # If k is more than number of # elements in array return sys.maxsize # Standard partition process of QuickSort(). # It considers the last element as pivot and # moves all smaller element to left of it # and greater elements to right def partition(arr, l, r): x = arr[r] i = l for j in range (l, r): if (arr[j] < = x): arr[i], arr[j] = arr[j], arr[i] i + = 1 arr[i], arr[r] = arr[r], arr[i] return i # Driver's Code if __name__ = = "__main__" : arr = [ 12 , 3 , 5 , 7 , 4 , 19 , 26 ] N = len (arr) K = 3 print ( "K'th smallest element is" , kthSmallest(arr, 0 , N - 1 , K)) # This code is contributed by ita_c |
C#
// C# code for kth smallest element // in an array using System; class GFG { // Standard partition process of QuickSort. // It considers the last element as pivot // and moves all smaller element to left of // it and greater elements to right public static int partition( int [] arr, int l, int r) { int x = arr[r], i = l; int temp = 0; for ( int j = l; j <= r - 1; j++) { if (arr[j] <= x) { // Swapping arr[i] and arr[j] temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i++; } } // Swapping arr[i] and arr[r] temp = arr[i]; arr[i] = arr[r]; arr[r] = temp; return i; } // This function returns k'th smallest // element in arr[l..r] using QuickSort // based method. ASSUMPTION: ALL ELEMENTS // IN ARR[] ARE DISTINCT public static int kthSmallest( int [] arr, int l, int r, int K) { // If k is smaller than number // of elements in array if (K > 0 && K <= r - l + 1) { // Partition the array around last // element and get position of pivot // element in sorted array int pos = partition(arr, l, r); // If position is same as k if (pos - l == K - 1) return arr[pos]; // If position is more, recur for // left subarray if (pos - l > K - 1) return kthSmallest(arr, l, pos - 1, K); // Else recur for right subarray return kthSmallest(arr, pos + 1, r, K - pos + l - 1); } // If k is more than number // of elements in array return int .MaxValue; } // Driver's Code public static void Main() { int [] arr = { 12, 3, 5, 7, 4, 19, 26 }; int K = 3; // Function call Console.Write( "K'th smallest element is " + kthSmallest(arr, 0, arr.Length - 1, K)); } } // This code is contributed // by 29AjayKumar |
Javascript
// JavaScript code for kth smallest // element in an array // Standard partition process of QuickSort. // It considers the last element as pivot // and moves all smaller element to left of // it and greater elements to right function partition( arr , l , r) { var x = arr[r], i = l; for (j = l; j <= r - 1; j++) { if (arr[j] <= x) { // Swapping arr[i] and arr[j] var temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i++; } } // Swapping arr[i] and arr[r] var temp = arr[i]; arr[i] = arr[r]; arr[r] = temp; return i; } // This function returns k'th smallest element // in arr[l..r] using QuickSort based method. // ASSUMPTION: ALL ELEMENTS IN ARR ARE DISTINCT function kthSmallest( arr , l , r , k) { // If k is smaller than number of elements // in array if (k > 0 && k <= r - l + 1) { // Partition the array around last // element and get position of pivot // element in sorted array var pos = partition(arr, l, r); // If position is same as k if (pos - l == k - 1) return arr[pos]; // If position is more, recur for // left subarray if (pos - l > k - 1) return kthSmallest(arr, l, pos - 1, k); // Else recur for right subarray return kthSmallest(arr, pos + 1, r, k - pos + l - 1); } // If k is more than number of elements // in array return Number.MAX_VALUE; } // Driver program to test above methods var arr = [ 12, 3, 5, 7, 4, 19, 26 ]; var k = 3; document.write( "K'th smallest element is " + kthSmallest(arr, 0, arr.length - 1, k)); // This code contributed by Rajput-Ji |
K'th smallest element is 5
Time Complexity: O(N2) in worst case and O(N) on average
Auxiliary Space: O(1)
K’th smallest element in an unsorted array using Map:
This approach is very much similar to the QuickSelect and counting sort algorithm but much easier to implement. Use a map and then map each element with its frequency. And as an ordered map would store the data in a sorted manner, so keep on adding the frequency of each element till it does not become greater than or equal to k so that the k’th element from the start can be reached i.e. the k’th smallest element.
Example: A[] = {7, 0, 25, 6, 16, 17, 0}, K = 3
Follow the given steps to solve the problem:
- Store frequency of every element in a Map mp
- Now traverse over sorted elements in the Map mp and add their frequencies in a variable freq
- If at any point the value of freq is greater than or equal to K, then return the value of iterator of Map mp
Below is the Implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; int Kth_smallest(map< int , int > mp, int K) { int freq = 0; for ( auto it = mp.begin(); it != mp.end(); it++) { freq += (it->second); // adding the frequencies of // each element if (freq >= K) // if at any point frequency becomes // greater than or equal to k then // return that element { return it->first; } } return -1; // returning -1 if k>size of the array which // is an impossible scenario } // Driver's code int main() { int N = 5; int K = 2; vector< int > arr = { 12, 3, 5, 7, 19 }; map< int , int > mp; for ( int i = 0; i < N; i++) { mp[arr[i]] += 1; // mapping every element with it's // frequency } // Function call int ans = Kth_smallest(mp, K); cout << "The " << K << "th smallest element is " << ans << endl; return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG { static int Kth_smallest(TreeMap<Integer, Integer> mp, int K) { int freq = 0 ; for (Map.Entry it : mp.entrySet()) { // adding the frequencies of each element freq += ( int )it.getValue(); // if at any point frequency becomes // greater than or equal to k then // return that element if (freq >= K) { return ( int )it.getKey(); } } return - 1 ; // returning -1 if k>size of the array // which is an impossible scenario } // Driver's code public static void main(String[] args) { int N = 5 ; int K = 2 ; int [] arr = { 12 , 3 , 5 , 7 , 19 }; TreeMap<Integer, Integer> mp = new TreeMap<>(); for ( int i = 0 ; i < N; i++) { // mapping every element with // it's // frequency mp.put(arr[i], mp.getOrDefault(arr[i], 0 ) + 1 ); } // Function call int ans = Kth_smallest(mp, K); System.out.println( "The " + K + "th smallest element is " + ans); } } // This code is contributed by harshit17. |
Python3
# Python3 program for the above approach def Kth_smallest(mp, K): freq = 0 for it in sorted (mp.keys()): freq + = mp[it] # adding the frequencies of # each element if freq > = K: # if at any point frequency becomes return it # greater than or equal to k then # return that element return - 1 # returning -1 if k>size of the array which # is an impossible scenario # driver's code if __name__ = = "__main__" : N = 5 K = 2 arr = [ 12 , 3 , 5 , 7 , 19 ] mp = {} for i in range (N): if arr[i] in mp: # mapping every element with it's mp[arr[i]] = mp[arr[i]] + 1 # frequency else : mp[arr[i]] = 1 # Function call ans = Kth_smallest(mp, K) print ( "The " , K, "th smallest element is " , ans) # This code is contributed by Abhijeet Kumar(abhijeet19403) |
C#
// C# program for the above approach using System; using System.Collections; using System.Collections.Generic; public class GFG { static int Kth_smallest(SortedDictionary< int , int > mp, int K) { int freq = 0; foreach ( var it in mp) { // Adding the frequencies of each element freq += ( int )it.Value; // If at any point frequency becomes // greater than or equal to k then // return that element if (freq >= K) { return ( int )it.Key; } } return -1; // returning -1 if K > size of the array // which is an impossible scenario } // Driver's code public static void Main(String[] args) { int N = 5; int K = 2; int [] arr = { 12, 3, 5, 7, 19 }; SortedDictionary< int , int > mp = new SortedDictionary< int , int >(); for ( int i = 0; i < N; i++) { // mapping every element with // it's frequency mp.Add(arr[i], mp.ContainsKey(arr[i]) ? mp[arr[i]] + 1 : 1); } // Function call int ans = Kth_smallest(mp, K); Console.WriteLine( "The " + K + "th smallest element is " + ans); } } // This code is contributed by Abhijeet Kumar(abhijeet19403) |
Javascript
// Javascript program to implement the approach //Function for creating Map function crateMap(arr, N) { var brr = arr.sort( function (a, b) { return a - b; }); let mp = {}; for ( var i = 0; i < N; i++) { mp[brr[i]] = 0; } for ( var i = 0; i < N; i++) { mp[brr[i]] += 1; } return mp; } //Function for finding kth Smallest function Kth_smallest(mp, K) { var freq = 0; for (const key in mp) { freq += mp[key]; // adding the frequencies of each element if (freq >= K) { // if at any point frequency becomes // greater than or equal to k then // return that element return key; } } return -1; // returning -1 if k>size of the array which // is an impossible scenario } var N = 5; var K = 2; var arr = [12, 3, 5, 7, 19]; let mpp = crateMap(arr, N); // Function call var ans = Kth_smallest(mpp, K); console.log( "The " + K + "th smallest element is " + ans); |
The 2th smallest element is 5
Time Complexity: O(N log N)
Auxiliary Space: O(N)
K’th smallest element in an unsorted array using Priority Queue:
To find the Kth minimum element in an array, insert the elements into the priority queue until the size of it is less than K, and then compare remaining elements with the root of the priority queue and if the element is less than the root then remove the root and insert this element into the priority queue and finally return root of the priority queue
Follow the given steps to solve the problem:
- Build a priority queue of the first K elements (arr[0] to arr[K-1]) of the given array.
- For each element, after the Kth element (arr[K] to arr[n-1]), compare it with the root of priority queue.
- If the element is less than the root then remove the root and insert this element into the priority queue
- b) Else ignore it.
- Finally, the root of the priority queue is the Kth smallest element.
Below is the Implementation of the above approach:
C++
// C++ code to implement the approach #include <bits/stdc++.h> using namespace std; // Function to find the kth smallest array element int kthSmallest( int arr[], int N, int K) { // For finding min element we need (Max heap)priority // queue priority_queue< int > pq; for ( int i = 0; i < K; i++) { // First push first K elements into heap pq.push(arr[i]); } // Now check from k to last element for ( int i = K; i < N; i++) { // If current element is < top that means // there are other k-1 lesser elements // are present at bottom thus, pop that element // and add kth largest element into the heap till // curr at last all the greater element than kth // element will get pop off and at the top of heap // there will be kth smallest element if (arr[i] < pq.top()) { pq.pop(); // Push curr element pq.push(arr[i]); } } // Return top of element return pq.top(); } // Driver's code: int main() { int N = 10; int arr[N] = { 10, 5, 4, 3, 48, 6, 2, 33, 53, 10 }; int K = 4; // Function call cout << "Kth Smallest Element is: " << kthSmallest(arr, N, K); } |
Java
// Java code to implement the approach import java.util.*; // Custom comparator class to form the Max heap class MinHeapComparator implements Comparator<Integer> { @Override public int compare(Integer number1, Integer number2) { int value = number1.compareTo(number2); // Elements are sorted in reverse order if (value > 0 ) { return - 1 ; } else if (value < 0 ) { return 1 ; } else { return 0 ; } } } class GFG { // Function to find kth smallest array element static int kthSmallest( int [] v, int N, int K) { // For finding min element we need (Max // heap)priority queue PriorityQueue<Integer> heap1 = new PriorityQueue<Integer>( new MinHeapComparator()); for ( int i = 0 ; i < N; ++i) { // Insert elements into // the priority queue heap1.add(v[i]); // If current element is less than top, that // means there are other k-1 lesser elements are // present at bottom // thus pop that element and add kth largest // element into the heap till curr at last all // the greater element than kth element will get // pop off and at the top of heap there will be // kth smallest element if (heap1.size() > K) { heap1.remove(); } } // Return the top of the heap as kth smallest // element return heap1.peek(); } // Driver's code public static void main(String[] args) { // Given array int [] vec = { 10 , 5 , 4 , 3 , 48 , 15 , 6 , 2 , 33 , 53 , 10 }; // Size of array int N = vec.length; // Given K int K = 4 ; // Function Call System.out.println( "Kth Smallest Element: " + kthSmallest(vec, N, K)); } } |
Python3
# Python3 code to implement the approach import heapq # Function to find the kth smallest array element def kthSmallest(arr, N, K): # For finding min element we need (Max heap)priority queue pq = [] for i in range (K): # First push first K elements into heap heapq.heappush(pq, arr[i]) heapq._heapify_max(pq) # Now check from k to last element for i in range (K, N): # If current element is < first that means # there are other k-1 lesser elements # are present at bottom thus, pop that element # and add kth largest element into the heap till curr # at last all the greater element than kth element will get pop off # and at the top of heap there will be kth smallest element if arr[i] < pq[ 0 ]: heapq.heappop(pq) # Push curr element heapq.heappush(pq, arr[i]) heapq._heapify_max(pq) # Return first of element return pq[ 0 ] # Driver's code: if __name__ = = "__main__" : N = 10 arr = [ 10 , 5 , 4 , 3 , 48 , 6 , 2 , 33 , 53 , 10 ] K = 4 # Function call print ( "Kth Smallest Element is:" , kthSmallest(arr, N, K)) # This code is contributed by Tapesh(tapeshdua420) |
C#
// C# program to implement // the above approach using System; using System.Collections.Generic; class GFG { // Function to find the kth smallest array element static int kthSmallest( int [] arr, int N, int K) { // For finding min element we need (Max // heap)priority queue List< int > pq = new List< int >(); for ( int i = 0; i < K; i++) { // First push first K elements into heap pq.Add(arr[i]); } // Now check from k to last element for ( int i = K; i < N; i++) { // If current element is < top that means // there are other k-1 lesser elements // are present at bottom thus, pop that element // and add kth largest element into the heap // till curr at last all the greater element // than kth element will get pop off and at the // top of heap there will be kth smallest // element if (arr[i] < pq[0]) { pq.Sort(); pq.Reverse(); pq.RemoveAt(0); // Push curr element pq.Add(arr[i]); } } // Return top of element return pq[0]; } // Driver's Code public static void Main() { // Given array int [] vec = { 10, 5, 4, 3, 48, 15, 6, 2, 33, 53, 10 }; // Size of array int N = vec.Length; // Given K int K = 4; // Function Call Console.WriteLine( "Kth Smallest Element: " + kthSmallest(vec, N, K)); } } // This code is contributed by sanjoy_62. |
Kth Smallest Element is: 5
Time complexity: O(K log K + (N – K) log K)
Auxiliary Space: O(K)
K’th smallest element in an unsorted array using Binary Search:
The idea to solve this problem is that the Kth smallest element would be the element at the kth position if the array was sorted in increasing order. Using this logic, binary search can be used to predict the index of an element as if the array was sorted but without actually sorting the array.
Follow the given steps to solve the problem:
- Find low and high that is the range where our answer can lie.
- Apply Binary Search on this range.
- If the selected element which would be mid has less than K elements lesser to it then increase the number that is low = mid + 1.
- Otherwise, Decrement the high pointer, i.e high = mid
- The Binary Search will end when only one element remains in the answer space which would be the answer.
Below is the implementation of the above approach:
C++
// C++ code for the above approach #include <bits/stdc++.h> #include <iostream> using namespace std; int count(vector< int >& nums, int & mid) { // function to calculate number of elements less than // equal to mid int cnt = 0; for ( int i = 0; i < nums.size(); i++) if (nums[i] <= mid) cnt++; return cnt; } int kthSmallest(vector< int > nums, int & k) { int low = INT_MAX; int high = INT_MIN; // calculate minimum and maximum the array. for ( int i = 0; i < nums.size(); i++) { low = min(low, nums[i]); high = max(high, nums[i]); } // Our answer range lies between minimum and maximum // element of the array on which Binary Search is // Applied while (low < high) { int mid = low + (high - low) / 2; /*if the count of number of elements in the array less than equal to mid is less than k then increase the number. Otherwise decrement the number and try to find a better answer. */ if (count(nums, mid) < k) low = mid + 1; else high = mid; } return low; } // Driver's code int main() { vector< int > nums{ 1, 4, 5, 3, 19, 3 }; int k = 3; // Function call cout << "K'th smallest element is " << kthSmallest(nums, k); return 0; } // This code is contributed by garvjuneja98 |
Java
// Java code for kth smallest element in an array import java.util.Arrays; import java.util.Collections; class GFG { static int count( int [] nums, int mid) { // function to calculate number of elements less // than equal to mid int cnt = 0 ; for ( int i = 0 ; i < nums.length; i++) if (nums[i] <= mid) cnt++; return cnt; } static int kthSmallest( int [] nums, int k) { int low = Integer.MAX_VALUE; int high = Integer.MIN_VALUE; // calculate minimum and maximum the array. for ( int i = 0 ; i < nums.length; i++) { low = Math.min(low, nums[i]); high = Math.max(high, nums[i]); } // Our answer range lies between minimum and maximum // element of the array on which Binary Search is // Applied while (low < high) { int mid = low + (high - low) / 2 ; /*if the count of number of elements in the array less than equal to mid is less than k then increase the number. Otherwise decrement the number and try to find a better answer. */ if (count(nums, mid) < k) low = mid + 1 ; else high = mid; } return low; } // Driver's code public static void main(String[] args) { int arr[] = { 1 , 4 , 5 , 3 , 19 , 3 }; int k = 3 ; // Function call System.out.print( "Kth smallest element is " + kthSmallest(arr, k)); } } // This code is contributed by CodeWithMini |
Python3
# Python3 code for kth smallest element in an array import sys # function to calculate number of elements # less than equal to mid def count(nums, mid): cnt = 0 for i in range ( len (nums)): if nums[i] < = mid: cnt + = 1 return cnt def kthSmallest(nums, k): low = sys.maxsize high = - sys.maxsize - 1 # calculate minimum and maximum the array. for i in range ( len (nums)): low = min (low, nums[i]) high = max (high, nums[i]) # Our answer range lies between minimum and maximum element # of the array on which Binary Search is Applied while low < high: mid = low + (high - low) / / 2 # if the count of number of elements in the array less than equal # to mid is less than k then increase the number. Otherwise decrement # the number and try to find a better answer. if count(nums, mid) < k: low = mid + 1 else : high = mid return low # Driver's code if __name__ = = "__main__" : nums = [ 1 , 4 , 5 , 3 , 19 , 3 ] k = 3 # Function call print ( "K'th smallest element is" , kthSmallest(nums, k)) # This code is contributed by Tapesh(tapeshdua420) |
C#
// C# program to implement // the above approach using System; using System.Collections.Generic; class GFG { static int count( int [] nums, int mid) { // function to calculate number of elements less // than equal to mid int cnt = 0; for ( int i = 0; i < nums.Length; i++) if (nums[i] <= mid) cnt++; return cnt; } static int kthSmallest( int [] nums, int k) { int low = Int32.MaxValue; int high = Int32.MinValue; // calculate minimum and maximum the array. for ( int i = 0; i < nums.Length; i++) { low = Math.Min(low, nums[i]); high = Math.Max(high, nums[i]); } // Our answer range lies between minimum // and maximum element of the array on which Binary // Search is Applied while (low < high) { int mid = low + (high - low) / 2; /*if the count of number of elements in the array less than equal to mid is less than k then increase the number. Otherwise decrement the number and try to find a better answer. */ if (count(nums, mid) < k) low = mid + 1; else high = mid; } return low; } // Driver's Code public static void Main() { // Given array int [] vec = { 1, 4, 5, 3, 19, 3 }; // Given K int K = 3; // Function Call Console.WriteLine( "Kth Smallest Element: " + kthSmallest(vec, K)); } } // This code is contributed by CodeWithMini |
Javascript
// Javascript program to find the K’th // Smallest/Largest Element in Unsorted Array function count(nums, mid) { // function to calculate number of elements less than equal to mid var cnt = 0; for ( var i = 0; i < nums.length; i++) if (nums[i] <= mid) cnt++; return cnt; } function kthSmallest(nums,k){ var low = Number. MAX_VALUE; var high = Number. MIN_VALUE; // calculate minimum and maximum the array. for ( var i = 0; i < nums.length; i++) { low = Math.min(low, nums[i]); high = Math.max(high, nums[i]); } // Our answer range lies between minimum and // maximum element of the array on which Binary Search is Applied while (low < high) { var mid = Math.floor(low + ((high - low) / 2)); /*if the count of number of elements in the array less than equal to mid is less than k then increase the number. Otherwise decrement the number and try to find a better answer. */ if (count(nums, mid) < k) low = mid + 1; else high = mid; } return low; } var k = 3; var nums = [1, 4, 5, 3, 19, 3]; document.write( "K'th smallest element is " + kthSmallest(nums, k)); // This code is contributed by shruti456rawal |
K'th smallest element is 3
Time complexity: O(n * log (mx-mn)), where mn be minimum and mx be maximum.
Auxiliary Space: O(1)
Next:
- K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)
- K’th Smallest/Largest Element in Unsorted Array | Set 3 (Worst-Case Linear Time)
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