k largest(or smallest) elements in an array
Question: Write an efficient program for printing k largest elements in an array. Elements in an array can be in any order.
For example, if the given array is [1, 23, 12, 9, 30, 2, 50] and you are asked for the largest 3 elements i.e., k = 3 then your program should print 50, 30, and 23.
Method 1 (Use Bubble k times)
Thanks to Shailendra for suggesting this approach.
1) Modify Bubble Sort to run the outer loop at most k times.
2) Print the last k elements of the array obtained in step 1.
Time Complexity: O(n*k)
Like Bubble sort, other sorting algorithms like Selection Sort can also be modified to get the k largest elements.
Method 2 (Use temporary array)
K largest elements from arr[0..n-1]
1) Store the first k elements in a temporary array temp[0..k-1].
2) Find the smallest element in temp[], let the smallest element be min.
3-a) For each element x in arr[k] to arr[n-1]. O(n-k)
If x is greater than the min then remove min from temp[] and insert x.
3-b)Then, determine the new min from temp[]. O(k)
4) Print final k elements of temp[]
Time Complexity: O((n-k)*k). If we want the output sorted then O((n-k)*k + k*log(k))
Thanks to nesamani1822 for suggesting this method.
Method 3(Use Sorting)
1) Sort the elements in descending order in O(n*log(n))
2) Print the first k numbers of the sorted array O(k).
Following is the implementation of the above.
C++
// C++ code for k largest elements in an array #include <bits/stdc++.h> using namespace std; void kLargest( int arr[], int n, int k) { // Sort the given array arr in reverse order. sort(arr, arr + n, greater< int >()); // Print the first kth largest elements for ( int i = 0; i < k; i++) cout << arr[i] << " " ; } // driver program int main() { int arr[] = { 1, 23, 12, 9, 30, 2, 50 }; int n = sizeof (arr) / sizeof (arr[0]); int k = 3; kLargest(arr, n, k); } // This code is contributed by Aditya Kumar (adityakumar129) |
C
// C code for k largest elements in an array #include <stdio.h> #include <stdlib.h> // Compare function for qsort int cmpfunc( const void * a, const void * b) { return (*( int *)b - *( int *)a); } void kLargest( int arr[], int n, int k) { // Sort the given array arr in reverse order. qsort (arr, n, sizeof ( int ), cmpfunc); // Print the first kth largest elements for ( int i = 0; i < k; i++) printf ( "%d " , arr[i]); } // driver program int main() { int arr[] = { 1, 23, 12, 9, 30, 2, 50 }; int n = sizeof (arr) / sizeof (arr[0]); int k = 3; kLargest(arr, n, k); } // This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java code for k largest elements in an array import java.util.Arrays; import java.util.Collections; import java.util.ArrayList; class GFG { public static void kLargest(Integer[] arr, int k) { // Sort the given array arr in reverse order // This method doesn't work with primitive data // types. So, instead of int, Integer type // array will be used Arrays.sort(arr, Collections.reverseOrder()); // Print the first kth largest elements for ( int i = 0 ; i < k; i++) System.out.print(arr[i] + " " ); } //This code is contributed by Niraj Dubey public static ArrayList<Integer> kLargest( int [] arr, int k) { //Convert using stream Integer[] obj_array = Arrays.stream( arr ).boxed().toArray( Integer[] :: new ); Arrays.sort(obj_array, Collections.reverseOrder()); ArrayList<Integer> list = new ArrayList<>(k); for ( int i = 0 ; i < k; i++) list.add(obj_array[i]); return list; } public static void main(String[] args) { Integer arr[] = new Integer[] { 1 , 23 , 12 , 9 , 30 , 2 , 50 }; int k = 3 ; kLargest(arr, k); //This code is contributed by Niraj Dubey //What if primitive datatype array is passed and wanted to return in ArrayList<Integer> int [] prim_array = { 1 , 23 , 12 , 9 , 30 , 2 , 50 }; System.out.print(kLargest(prim_array, k)); } } // This code is contributed by Kamal Rawal |
Python
''' Python3 code for k largest elements in an array''' def kLargest(arr, k): # Sort the given array arr in reverse # order. arr.sort(reverse = True ) # Print the first kth largest elements for i in range (k): print (arr[i], end = " " ) # Driver program arr = [ 1 , 23 , 12 , 9 , 30 , 2 , 50 ] # n = len(arr) k = 3 kLargest(arr, k) # This code is contributed by shreyanshi_arun. |
C#
// C# code for k largest elements in an array using System; class GFG { public static void kLargest( int [] arr, int k) { // Sort the given array arr in reverse order // This method doesn't work with primitive data // types. So, instead of int, Integer type // array will be used Array.Sort(arr); Array.Reverse(arr); // Print the first kth largest elements for ( int i = 0; i < k; i++) Console.Write(arr[i] + " " ); } // Driver code public static void Main(String[] args) { int [] arr = new int [] { 1, 23, 12, 9, 30, 2, 50 }; int k = 3; kLargest(arr, k); } } // This code contributed by Rajput-Ji |
PHP
<?php // PHP code for k largest // elements in an array function kLargest(& $arr , $n , $k ) { // Sort the given array arr // in reverse order. rsort( $arr ); // Print the first kth // largest elements for ( $i = 0; $i < $k ; $i ++) echo $arr [ $i ] . " " ; } // Driver Code $arr = array (1, 23, 12, 9, 30, 2, 50); $n = sizeof( $arr ); $k = 3; kLargest( $arr , $n , $k ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // JavaScript code for k largest // elements in an array function kLargest(arr, n, k) { // Sort the given array arr in reverse // order. arr.sort((a, b) => b - a); // Print the first kth largest elements for (let i = 0; i < k; i++) document.write(arr[i] + " " ); } // driver program let arr = [ 1, 23, 12, 9, 30, 2, 50 ]; let n = arr.length; let k = 3; kLargest(arr, n, k); // This code is contributed by Manoj. </script> |
50 30 23
Time complexity: O(n*log(n))
Auxiliary Space: O(1)
Method 4 (Use Max Heap)
1) Build a Max Heap tree in O(n)
2) Use Extract Max k times to get k maximum elements from the Max Heap O(k*log(n))
Time complexity: O(n + k*log(n))
Method 5(Use Order Statistics)
1) Use an order statistic algorithm to find the kth largest element. Please see the topic selection in worst-case linear time O(n)
2) Use QuickSort Partition algorithm to partition around the kth largest number O(n).
3) Sort the k-1 elements (elements greater than the kth largest element) O(k*log(k)). This step is needed only if the sorted output is required.
Time complexity: O(n) if we don’t need the sorted output, otherwise O(n+k*log(k))
Thanks to Shilpi for suggesting the first two approaches.
Method 6 (Use Min Heap)
This method is mainly an optimization of method 1. Instead of using temp[] array, use Min Heap.
1) Build a Min Heap MH of the first k elements (arr[0] to arr[k-1]) of the given array. O(k*log(k))
2) For each element, after the kth element (arr[k] to arr[n-1]), compare it with root of MH.
……a) If the element is greater than the root then make it root and call heapify for MH
……b) Else ignore it.
// The step 2 is O((n-k)*log(k))
3) Finally, MH has k largest elements, and the root of the MH is the kth largest element.
Time Complexity: O(k*log(k) + (n-k)*log(k)) without sorted output. If sorted output is needed then O(k*log(k) + (n-k)*log(k) + k*log(k)) so overall it is O(k*log(k) + (n-k)*log(k))
All of the above methods can also be used to find the kth largest (or smallest) element.
C++
#include <iostream> using namespace std; // Swap function to interchange // the value of variables x and y int swap( int & x, int & y) { int temp = x; x = y; y = temp; } // Min Heap Class // arr holds reference to an integer // array size indicate the number of // elements in Min Heap class MinHeap { int size; int * arr; public : // Constructor to initialize the size and arr MinHeap( int size, int input[]); // Min Heapify function, that assumes that // 2*i+1 and 2*i+2 are min heap and fix the // heap property for i. void heapify( int i); // Build the min heap, by calling heapify // for all non-leaf nodes. void buildHeap(); }; // Constructor to initialize data // members and creating mean heap MinHeap::MinHeap( int size, int input[]) { // Initializing arr and size this ->size = size; this ->arr = input; // Building the Min Heap buildHeap(); } // Min Heapify function, that assumes // 2*i+1 and 2*i+2 are min heap and // fix min heap property for i void MinHeap::heapify( int i) { // If Leaf Node, Simply return if (i >= size / 2) return ; // variable to store the smallest element // index out of i, 2*i+1 and 2*i+2 int smallest; // Index of left node int left = 2 * i + 1; // Index of right node int right = 2 * i + 2; // Select minimum from left node and // current node i, and store the minimum // index in smallest variable smallest = arr[left] < arr[i] ? left : i; // If right child exist, compare and // update the smallest variable if (right < size) smallest = arr[right] < arr[smallest] ? right : smallest; // If Node i violates the min heap // property, swap current node i with // smallest to fix the min-heap property // and recursively call heapify for node smallest. if (smallest != i) { swap(arr[i], arr[smallest]); heapify(smallest); } } // Build Min Heap void MinHeap::buildHeap() { // Calling Heapify for all non leaf nodes for ( int i = size / 2 - 1; i >= 0; i--) { heapify(i); } } void FirstKelements( int arr[], int size, int k){ // Creating Min Heap for given // array with only k elements MinHeap* m = new MinHeap(k, arr); // Loop For each element in array // after the kth element for ( int i = k; i < size; i++) { // if current element is smaller // than minimum element, do nothing // and continue to next element if (arr[0] > arr[i]) continue ; // Otherwise Change minimum element to // current element, and call heapify to // restore the heap property else { arr[0] = arr[i]; m->heapify(0); } } // Now min heap contains k maximum // elements, Iterate and print for ( int i = 0; i < k; i++) { cout << arr[i] << " " ; } } // Driver Program int main() { int arr[] = { 11, 3, 2, 1, 15, 5, 4, 45, 88, 96, 50, 45 }; int size = sizeof (arr) / sizeof (arr[0]); // Size of Min Heap int k = 3; FirstKelements(arr,size,k); return 0; } // This code is contributed by Ankur Goel |
Java
import java.io.*; import java.util.*; class GFG{ public static void FirstKelements( int arr[], int size, int k) { // Creating Min Heap for given // array with only k elements // Create min heap with priority queue PriorityQueue<Integer> minHeap = new PriorityQueue<>(); for ( int i = 0 ; i < k; i++) { minHeap.add(arr[i]); } // Loop For each element in array // after the kth element for ( int i = k; i < size; i++) { // If current element is smaller // than minimum ((top element of // the minHeap) element, do nothing // and continue to next element if (minHeap.peek() > arr[i]) continue ; // Otherwise Change minimum element // (top element of the minHeap) to // current element by polling out // the top element of the minHeap else { minHeap.poll(); minHeap.add(arr[i]); } } // Now min heap contains k maximum // elements, Iterate and print Iterator iterator = minHeap.iterator(); while (iterator.hasNext()) { System.out.print(iterator.next() + " " ); } } // Driver code public static void main (String[] args) { int arr[] = { 11 , 3 , 2 , 1 , 15 , 5 , 4 , 45 , 88 , 96 , 50 , 45 }; int size = arr.length; // Size of Min Heap int k = 3 ; FirstKelements(arr, size, k); } } // This code is contributed by Vansh Sethi |
Python3
def FirstKelements(arr,size,k): # Creating Min Heap for given # array with only k elements # Create min heap with priority queue minHeap = [] for i in range (k): minHeap.append(arr[i]) # Loop For each element in array # after the kth element for i in range (k, size): minHeap.sort() # If current element is smaller # than minimum ((top element of # the minHeap) element, do nothing # and continue to next element if (minHeap[ 0 ] > arr[i]): continue # Otherwise Change minimum element # (top element of the minHeap) to # current element by polling out # the top element of the minHeap else : minHeap.pop( 0 ) minHeap.append(arr[i]) # Now min heap contains k maximum # elements, Iterate and print for i in minHeap: print (i, end = " " ) # Driver code arr = [ 11 , 3 , 2 , 1 , 15 , 5 , 4 , 45 , 88 , 96 , 50 , 45 ] size = len (arr) # Size of Min Heap k = 3 FirstKelements(arr, size, k) # This code is contributed by avanitrachhadiya2155 |
C#
using System; using System.Collections.Generic; public class GFG { public static void FirstKelements( int []arr, int size, int k) { // Creating Min Heap for given // array with only k elements // Create min heap with priority queue List< int > minHeap = new List< int >(); for ( int i = 0; i < k; i++) { minHeap.Add(arr[i]); } // Loop For each element in array // after the kth element for ( int i = k; i < size; i++) { minHeap.Sort(); // If current element is smaller // than minimum ((top element of // the minHeap) element, do nothing // and continue to next element if (minHeap[0] > arr[i]) continue ; // Otherwise Change minimum element // (top element of the minHeap) to // current element by polling out // the top element of the minHeap else { minHeap.RemoveAt(0); minHeap.Add(arr[i]); } } // Now min heap contains k maximum // elements, Iterate and print foreach ( int i in minHeap) { Console.Write(i + " " ); } } // Driver code public static void Main(String[] args) { int []arr = { 11, 3, 2, 1, 15, 5, 4, 45, 88, 96, 50, 45 }; int size = arr.Length; // Size of Min Heap int k = 3; FirstKelements(arr, size, k); } } // This code is contributed by aashish1995. |
Javascript
<script> function FirstKelements(arr , size , k) { // Creating Min Heap for given // array with only k elements // Create min heap with priority queue var minHeap = []; for ( var i = 0; i < k; i++) { minHeap.push(arr[i]); } // Loop For each element in array // after the kth element for ( var i = k; i < size; i++) { minHeap.sort((a,b)=>a-b); // If current element is smaller // than minimum ((top element of // the minHeap) element, do nothing // and continue to next element if (minHeap[minHeap.length-3] > arr[i]) continue ; // Otherwise Change minimum element // (top element of the minHeap) to // current element by polling out // the top element of the minHeap else { minHeap.reverse(); minHeap.pop(); minHeap.reverse(); minHeap.push(arr[i]); } } // Now min heap contains k maximum // elements, Iterate and print for ( var iterator of minHeap) { document.write(iterator + " " ); } } // Driver code var arr = [11, 3, 2, 1, 15, 5, 4, 45, 88, 96, 50, 45]; var size = arr.length; // Size of Min Heap var k = 3; FirstKelements(arr, size, k); // This code is contributed by gauravrajput1 </script> |
50 88 96
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
Method 7(Using Quick Sort partitioning algorithm):
- Choose a pivot number.
- if K is lesser than the pivot_Index then repeat the step.
- if K == pivot_Index : Print the array (low to pivot to get K-smallest elements and (n-pivot_Index) to n for K-largest elements)
- if K > pivot_Index : Repeat the steps for right part.
We can improve on the standard quicksort algorithm by using the random() function. Instead of using the pivot element as the last element, we can randomly choose the pivot element. The worst-case time complexity of this version is O(n2) and the average time complexity is O(n).
Following is the implementation of the above algorithm:
C++
#include <bits/stdc++.h> using namespace std; //picks up last element between start and end int findPivot( int a[], int start, int end) { // Selecting the pivot element int pivot = a[end]; // Initially partition-index will be // at starting int pIndex = start; for ( int i = start; i < end; i++) { // If an element is lesser than pivot, swap it. if (a[i] <= pivot) { swap(a[i], a[pIndex]); // Incrementing pIndex for further // swapping. pIndex++; } } // Lastly swapping or the // correct position of pivot swap(a[pIndex], a[end]); return pIndex; } //THIS PART OF CODE IS CONTRIBUTED BY - rjrachit //Picks up random pivot element between start and end int findRandomPivot( int arr[], int start, int end) { int n = end - start + 1; // Selecting the random pivot index int pivotInd = random()%n; swap(arr[end],arr[start+pivotInd]); int pivot = arr[end]; //initialising pivoting point to start index pivotInd = start; for ( int i = start; i < end; i++) { // If an element is lesser than pivot, swap it. if (arr[i] <= pivot) { swap(arr[i], arr[pivotInd]); // Incrementing pivotIndex for further // swapping. pivotInd++; } } // Lastly swapping or the // correct position of pivot swap(arr[pivotInd], arr[end]); return pivotInd; } //THIS PART OF CODE IS CONTRIBUTED BY - rjrachit void SmallestLargest( int a[], int low, int high, int k, int n) { if (low == high) return ; else { int pivotIndex = findRandomPivot(a, low, high); if (k == pivotIndex) { cout << k << " smallest elements are : " ; for ( int i = 0; i < pivotIndex; i++) cout << a[i] << " " ; cout << endl; cout << k << " largest elements are : " ; for ( int i = (n - pivotIndex); i < n; i++) cout << a[i] << " " ; } else if (k < pivotIndex) SmallestLargest(a, low, pivotIndex - 1, k, n); else if (k > pivotIndex) SmallestLargest(a, pivotIndex + 1, high, k, n); } } // Driver Code int main() { int a[] = { 11, 3, 2, 1, 15, 5, 4, 45, 88, 96, 50, 45 }; int n = sizeof (a) / sizeof (a[0]); int low = 0; int high = n - 1; // Lets assume k is 3 int k = 3; // Function Call SmallestLargest(a, low, high, k, n); return 0; } |
Java
import java.util.*; class GFG{ //picks up last element between start and end static int findPivot( int a[], int start, int end) { // Selecting the pivot element int pivot = a[end]; // Initially partition-index will be // at starting int pIndex = start; for ( int i = start; i < end; i++) { // If an element is lesser than pivot, swap it. if (a[i] <= pivot) { int temp =a[i]; a[i]= a[pIndex]; a[pIndex] = temp; // Incrementing pIndex for further // swapping. pIndex++; } } // Lastly swapping or the // correct position of pivot int temp = a[pIndex]; a[pIndex] = a[end]; a[end] = temp; return pIndex; } //THIS PART OF CODE IS CONTRIBUTED BY - rjrachit //Picks up random pivot element between start and end static int findRandomPivot( int arr[], int start, int end) { int n = end - start + 1 ; // Selecting the random pivot index int pivotInd = ( int ) ((Math.random()* 1000000 )%n); int temp = arr[end]; arr[end] = arr[start+pivotInd]; arr[start+pivotInd] = temp; int pivot = arr[end]; //initialising pivoting point to start index pivotInd = start; for ( int i = start; i < end; i++) { // If an element is lesser than pivot, swap it. if (arr[i] <= pivot) { int temp1 = arr[i]; arr[i]= arr[pivotInd]; arr[pivotInd] = temp1; // Incrementing pivotIndex for further // swapping. pivotInd++; } } // Lastly swapping or the // correct position of pivot int tep = arr[pivotInd]; arr[pivotInd] = arr[end]; arr[end] = tep; return pivotInd; } //THIS PART OF CODE IS CONTRIBUTED BY - rjrachit static void SmallestLargest( int a[], int low, int high, int k, int n) { if (low == high) return ; else { int pivotIndex = findRandomPivot(a, low, high); if (k == pivotIndex) { System.out.print(k+ " smallest elements are : " ); for ( int i = 0 ; i < pivotIndex; i++) System.out.print(a[i]+ " " ); System.out.println(); System.out.print(k+ " largest elements are : " ); for ( int i = (n - pivotIndex); i < n; i++) System.out.print(a[i]+ " " ); } else if (k < pivotIndex) SmallestLargest(a, low, pivotIndex - 1 , k, n); else if (k > pivotIndex) SmallestLargest(a, pivotIndex + 1 , high, k, n); } } // Driver Code public static void main(String[] args) { int a[] = { 11 , 3 , 2 , 1 , 15 , 5 , 4 , 45 , 88 , 96 , 50 , 45 }; int n = a.length; int low = 0 ; int high = n - 1 ; // Lets assume k is 3 int k = 3 ; // Function Call SmallestLargest(a, low, high, k, n); } } // This code is contributed by Rajput-Ji |
Python3
# Python program to implement above approach # picks up last element between start and end import random def findPivot(a, start, end): # Selecting the pivot element pivot = a[end] # Initially partition-index will be # at starting pIndex = start for i in range (start,end): # If an element is lesser than pivot, swap it. if (a[i] < = pivot): a[i],a[pIndex] = a[pIndex],a[i] # Incrementing pIndex for further # swapping. pIndex + = 1 # Lastly swapping or the # correct position of pivot a[end],a[pIndex] = a[pIndex],a[end] return pIndex #THIS PART OF CODE IS CONTRIBUTED BY - rjrachit #Picks up random pivot element between start and end def findRandomPivot(arr, start, end): n = end - start + 1 # Selecting the random pivot index pivotInd = ( int ((random.random() * 1000000 )) % n) arr[end],arr[start + pivotInd] = arr[start + pivotInd],arr[end] pivot = arr[end] #initialising pivoting poto start index pivotInd = start for i in range (start,end): # If an element is lesser than pivot, swap it. if (arr[i] < = pivot): arr[i],arr[pivotInd] = arr[pivotInd],arr[i] # Incrementing pivotIndex for further # swapping. pivotInd + = 1 # Lastly swapping or the # correct position of pivot arr[pivotInd],arr[end] = arr[end],arr[pivotInd] return pivotInd def SmallestLargest(a, low, high, k, n): if (low = = high): return else : pivotIndex = findRandomPivot(a, low, high) if (k = = pivotIndex): print ( str (k) + " smallest elements are :" ,end = " " ) for i in range (pivotIndex): print (a[i],end = " " ) print () print ( str (k) + " largest elements are :" ,end = " " ) for i in range (n - pivotIndex,n): print (a[i],end = " " ) elif (k < pivotIndex): SmallestLargest(a, low, pivotIndex - 1 , k, n) elif (k > pivotIndex): SmallestLargest(a, pivotIndex + 1 , high, k, n) # Driver code a = [ 11 , 3 , 2 , 1 , 15 , 5 , 4 , 45 , 88 , 96 , 50 , 45 ] n = len (a) low = 0 high = n - 1 # assume k is 3 k = 3 # Function Call SmallestLargest(a, low, high, k, n) # This code is contributed by shinjanpatra |
C#
using System; using System.Text; public class GFG { // picks up last element between start and end static int findPivot( int []a, int start, int end) { // Selecting the pivot element int pivot = a[end]; // Initially partition-index will be // at starting int pIndex = start; for ( int i = start; i < end; i++) { // If an element is lesser than pivot, swap it. if (a[i] <= pivot) { int temp6 = a[i]; a[i] = a[pIndex]; a[pIndex] = temp6; // Incrementing pIndex for further // swapping. pIndex++; } } // Lastly swapping or the // correct position of pivot int temp = a[pIndex]; a[pIndex] = a[end]; a[end] = temp; return pIndex; } // THIS PART OF CODE IS CONTRIBUTED BY - rjrachit // Picks up random pivot element between start and end static int findRandomPivot( int []arr, int start, int end) { int n = end - start + 1; // Selecting the random pivot index Random _random = new Random(); var randomNumber = _random.Next(0, n); int pivotInd = randomNumber; int temp = arr[end]; arr[end] = arr[start + pivotInd]; arr[start + pivotInd] = temp; int pivot = arr[end]; // initialising pivoting point to start index pivotInd = start; for ( int i = start; i < end; i++) { // If an element is lesser than pivot, swap it. if (arr[i] <= pivot) { int temp1 = arr[i]; arr[i] = arr[pivotInd]; arr[pivotInd] = temp1; // Incrementing pivotIndex for further // swapping. pivotInd++; } } // Lastly swapping or the // correct position of pivot int tep = arr[pivotInd]; arr[pivotInd] = arr[end]; arr[end] = tep; return pivotInd; } static void SmallestLargest( int []a, int low, int high, int k, int n) { if (low == high) return ; else { int pivotIndex = findRandomPivot(a, low, high); if (k == pivotIndex) { Console.Write(k + " smallest elements are : " ); for ( int i = 0; i < pivotIndex; i++) Console.Write(a[i] + " " ); Console.WriteLine(); Console.Write(k + " largest elements are : " ); for ( int i = (n - pivotIndex); i < n; i++) Console.Write(a[i] + " " ); } else if (k < pivotIndex) SmallestLargest(a, low, pivotIndex - 1, k, n); else if (k > pivotIndex) SmallestLargest(a, pivotIndex + 1, high, k, n); } } // Driver Code public static void Main(String[] args) { int []a = { 11, 3, 2, 1, 15, 5, 4, 45, 88, 96, 50, 45 }; int n = a.Length; int low = 0; int high = n - 1; // Lets assume k is 3 int k = 3; // Function Call SmallestLargest(a, low, high, k, n); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript code to implement the approach //picks up last element between start and end function findPivot( a, start, end) { // Selecting the pivot element let pivot = a[end]; // Initially partition-index will be // at starting let pIndex = start; for (let i = start; i < end; i++) { // If an element is lesser than pivot, swap it. if (a[i] <= pivot) { let temp =a[i]; a[i]= a[pIndex]; a[pIndex] = temp; // Incrementing pIndex for further // swapping. pIndex++; } } // Lastly swapping or the // correct position of pivot let temp = a[pIndex]; a[pIndex] = a[end]; a[end] = temp; return pIndex; } //THIS PART OF CODE IS CONTRIBUTED BY - rjrachit //Picks up random pivot element between start and end function findRandomPivot(arr, start, end) { let n = end - start + 1; // Selecting the random pivot index let pivotInd = (parseInt((Math.random()*1000000))%n); let temp = arr[end]; arr[end] = arr[start+pivotInd]; arr[start+pivotInd] = temp; let pivot = arr[end]; //initialising pivoting point to start index pivotInd = start; for (let i = start; i < end; i++) { // If an element is lesser than pivot, swap it. if (arr[i] <= pivot) { let temp1 = arr[i]; arr[i]= arr[pivotInd]; arr[pivotInd] = temp1; // Incrementing pivotIndex for further // swapping. pivotInd++; } } // Lastly swapping or the // correct position of pivot let tep = arr[pivotInd]; arr[pivotInd] = arr[end]; arr[end] = tep; return pivotInd; } //THIS PART OF CODE IS CONTRIBUTED BY - rjrachit function SmallestLargest( a, low, high, k, n) { if (low == high) return ; else { let pivotIndex = findRandomPivot(a, low, high); if (k == pivotIndex) { document.write(k+ " smallest elements are : " ); for (let i = 0; i < pivotIndex; i++) document.write(a[i]+ " " ); document.write( "<br/>" ); document.write(k+ " largest elements are : " ); for (let i = (n - pivotIndex); i < n; i++) document.write(a[i]+ " " ); } else if (k < pivotIndex) SmallestLargest(a, low, pivotIndex - 1, k, n); else if (k > pivotIndex) SmallestLargest(a, pivotIndex + 1, high, k, n); } } // Driver code let a = [ 11, 3, 2, 1, 15, 5, 4, 45, 88, 96, 50, 45 ]; let n = a.length; let low = 0; let high = n - 1; // Lets assume k is 3 let k = 3; // Function Call SmallestLargest(a, low, high, k, n); // This code is contributed by sanjoy_62. </script> |
3 smallest elements are : 3 2 1 3 largest elements are : 96 50 88
Time Complexity: O(nlogn)
Auxiliary Space: O(1)
Method 8(Using priority queue STL library):
In this approach, we can efficiently print the k largest/smallest elements of an array using a priority queue in O(n*log(k)) time complexity. First, we push k elements into the priority queue from the array. From there on, after every insertion of an array element, we will pop the element at the top of priority_queue. In the case of the k largest element, the priority_queue will be in increasing order, and thus top most element will be the smallest so we are removing it. Similarly, in the case of the k smallest element, the priority_queue is in decreasing order and hence the top most element is the largest one so we will remove it. In this fashion whole array is traversed and the priority queue of size k is printed which contains k largest/smallest elements.
Below is the implementation of the above approach:
C++
// C++ code for k largest/ smallest elements in an array #include <bits/stdc++.h> using namespace std; // Function to find k largest array element void kLargest(vector< int >& v, int N, int K) { // Implementation using // a Priority Queue priority_queue< int , vector< int >, greater< int > >pq; for ( int i = 0; i < N; ++i) { // Insert elements into // the priority queue pq.push(v[i]); // If size of the priority // queue exceeds k if (pq.size() > K) { pq.pop(); } } // Print the k largest element while (!pq.empty()) { cout << pq.top() << " " ; pq.pop(); } cout<<endl; } // Function to find k smallest array element void kSmalest(vector< int >& v, int N, int K) { // Implementation using // a Priority Queue priority_queue< int > pq; for ( int i = 0; i < N; ++i) { // Insert elements into // the priority queue pq.push(v[i]); // If size of the priority // queue exceeds k if (pq.size() > K) { pq.pop(); } } // Print the k smallest element while (!pq.empty()) { cout << pq.top() << " " ; pq.pop(); } } // driver program int main() { // Given array vector< int > arr = { 11, 3, 2, 1, 15, 5, 4, 45, 88, 96, 50, 45 }; // Size of array int n = arr.size(); int k = 3; cout<<k<< " largest elements are : " ; kLargest(arr, n, k); cout<<k<< " smallest elements are : " ; kSmalest(arr, n, k); } // This code is contributed by Pushpesh Raj. |
Java
// Java code for k largest/ smallest elements in an array import java.util.*; class GFG { // Function to find k largest array element static void kLargest( int a[], int n, int k) { // Implementation using // a Priority Queue PriorityQueue<Integer> pq = new PriorityQueue<Integer>(); for ( int i = 0 ; i < n; ++i) { // Insert elements into // the priority queue pq.add(a[i]); // If size of the priority // queue exceeds k if (pq.size() > k) { pq.poll(); } } // Print the k largest element while (!pq.isEmpty()) { System.out.print(pq.peek() + " " ); pq.poll(); } System.out.println(); } // Function to find k smallest array element static void kSmallest( int a[], int n, int k) { // Implementation using // a Priority Queue PriorityQueue<Integer> pq = new PriorityQueue<Integer>( Collections.reverseOrder()); for ( int i = 0 ; i < n; ++i) { // Insert elements into // the priority queue pq.add(a[i]); // If size of the priority // queue exceeds k if (pq.size() > k) { pq.poll(); } } // Print the k largest element while (!pq.isEmpty()) { System.out.print(pq.peek() + " " ); pq.poll(); } } // Driver Code public static void main(String[] args) { int a[] = { 11 , 3 , 2 , 1 , 15 , 5 , 4 , 45 , 88 , 96 , 50 , 45 }; int n = a.length; int k = 3 ; System.out.print(k + " largest elements are : " ); // Function Call kLargest(a, n, k); System.out.print(k + " smallest elements are : " ); // Function Call kSmallest(a, n, k); } } // This code is contributed by Aarti_Rathi |
3 largest elements are : 50 88 96 3 smallest elements are : 3 2 1
Time Complexity: O(n*log(k))
Auxiliary Space: O(k)
Please write comments if you find any of the above explanations/algorithms incorrect, or find better ways to solve the same problem.
References:
http://en.wikipedia.org/wiki/Selection_algorithm