We recommend to read following post as a prerequisite of this post.
K’th Smallest/Largest Element in Unsorted Array | Set 1
Given an array and a number k where k is smaller than size of array, we need to find the k’th largest element in the given array. It is given that all array elements are distinct.
Examples:
Input: arr[] = {7, 10, 4, 3, 20, 15} k = 3 Output: 7 Input: arr[] = {7, 10, 4, 3, 20, 15} k = 4 Output: 10
We have discussed three different solutions here.
In this post method 4 is discussed which is mainly an extension of method 3 (QuickSelect) discussed in the previous post.
C++
// C++ implementation of above implementation #include<iostream> #include<climits> #include<cstdlib> using namespace std; int randomPartition( 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 = randomPartition(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; } int randomPartition( int arr[], int l, int r) { int n = r-l+1; int pivot = rand () % n; swap(&arr[l + pivot], &arr[r]); return partition(arr, l, r); } // Driver program to test above methods int main() { int arr[] = {12, 3, 5, 7, 4, 19, 26}; int n = sizeof (arr)/ sizeof (arr[0]), k = 3; cout << "K'th smallest element is " << kthSmallest(arr, 0, n-1, k); return 0; } |
Java
// Java program of above implementation import java.util.Random; public class GFG { // This function returns k'th smallest element in arr[l..r] using // QuickSort based method. ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT 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 = randomPartition(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 Integer.MAX_VALUE; } static void swap( int [] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = 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 static 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, j); i++; } } swap(arr, i, r); return i; } static int randomPartition( int arr[], int l, int r) { int n = r - l + 1 ; int pivot = new Random().nextInt( 1 ); swap(arr, l + pivot, r); return partition(arr, l, r); } // Driver program to test above methods public static void main(String args[]) { int arr[] = { 12 , 3 , 5 , 7 , 4 , 19 , 26 }; int n = arr.length, k = 3 ; System.out.println( "K'th smallest element is " + kthSmallest(arr, 0 , n - 1 , k)); } } /*This code is contributed by 29AjayKumar*/ |
Python3
# Python3 implementation of above implementation # This function returns k'th smallest element # in arr[l..r] using QuickSort based method. # ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT from random import randint def randomPartition(arr, l, r): n = r - l + 1 pivot = randint( 1 , 100 ) % n arr[l + pivot], arr[r] = arr[l + pivot], arr[r] return partition(arr, l, r) 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 = randomPartition(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 10 * * 9 # 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 Code 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 Mohit Kumar |
C#
// C# program of above implementation using System; class GFG { // This function returns k'th smallest // element in arr[l..r] using // QuickSort based method. ASSUMPTION: // ALL ELEMENTS IN ARR[] ARE DISTINCT 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 = randomPartition(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; } static void swap( int [] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } // Standard partition process of QuickSort(). // It considers the last element as pivot and // oves all smaller element to left of it // and greater elements to right static 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, j); i++; } } swap(arr, i, r); return i; } static int randomPartition( int []arr, int l, int r) { int n = r - l + 1; int pivot = new Random().Next(1); swap(arr, l + pivot, r); return partition(arr, l, r); } // Driver Code public static void Main() { int []arr = {12, 3, 5, 7, 4, 19, 26}; int n = arr.Length, k = 3; Console.WriteLine( "K'th smallest element is " + kthSmallest(arr, 0, n - 1, k)); } } // his code is contributed by 29AjayKumar |
Output:
K'th smallest element is 5
References:
https://www.geeksforgeeks.org/kth-smallestlargest-element-unsorted-array/
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.