K-th Smallest Element in an Unsorted Array using Priority Queue
Given an array arr[] consisting of N integers and an integer K, the task is to find the Kth smallest element in the array using Priority Queue.
Examples:
Input: arr[] = {5, 20, 10, 7, 1}, N = 5, K = 2
Output: 5
Explanation: In the given array, the 2nd smallest element is 5. Therefore, the required output is 5.
Input: arr[] = {5, 20, 10, 7, 1}, N = 5, K = 5
Output: 20
Explanation: In the given array, the 5th smallest element is 20. Therefore, the required output is 20.
Approach: The idea is to use PriorityQueue Collection in Java or priority_queue STL library to implement Max_Heap to find the Kth smallest array element. Follow the steps below to solve the problem:
- Implement Max Heap using a priority_queue.
- Push first K array elements into the priority_queue.
- From there on, after every insertion of an array element, pop the element at the top of the priority_queue.
- After complete traversal of the array, print the element at the top of the priority queue as the required answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void kthSmallest(vector< int >& v, int N, int K)
{
priority_queue< int > heap1;
for ( int i = 0; i < N; ++i) {
heap1.push(v[i]);
if (heap1.size() > K) {
heap1.pop();
}
}
cout << heap1.top() << endl;
}
int main()
{
vector< int > vec = { 5, 20, 10, 7, 1 };
int N = vec.size();
int K = 2;
kthSmallest(vec, N, K % (N+1));
return 0;
}
|
Java
import java.util.*;
class CustomComparator implements Comparator<Integer> {
@Override
public int compare(Integer number1, Integer number2) {
int value = number1.compareTo(number2);
if (value > 0 ) {
return - 1 ;
}
else if (value < 0 ) {
return 1 ;
}
else {
return 0 ;
}
}
}
class GFG{
static void kthSmallest( int []v, int N, int K)
{
PriorityQueue<Integer> heap1 = new PriorityQueue<Integer>( new CustomComparator());
for ( int i = 0 ; i < N; ++i) {
heap1.add(v[i]);
if (heap1.size() > K) {
heap1.remove();
}
}
System.out.print(heap1.peek() + "\n" );
}
public static void main(String[] args)
{
int []vec = { 5 , 20 , 10 , 7 , 1 };
int N = vec.length;
int K = 2 ;
kthSmallest(vec, N, K % N);
}
}
|
Python3
def kthSmallest(v, N, K):
heap1 = []
for i in range (N):
heap1.append(v[i])
if ( len (heap1) > K):
heap1.sort()
heap1.reverse()
del heap1[ 0 ]
heap1.sort()
heap1.reverse()
print (heap1[ 0 ])
vec = [ 5 , 20 , 10 , 7 , 1 ]
N = len (vec)
K = 2
kthSmallest(vec, N, K % N)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void kthSmallest( int []v, int N, int K)
{
List< int > heap1 = new List< int >();
for ( int i = 0; i < N; ++i) {
heap1.Add(v[i]);
if (heap1.Count > K) {
heap1.Sort();
heap1.Reverse();
heap1.RemoveAt(0);
}
}
heap1.Sort();
heap1.Reverse();
Console.WriteLine(heap1[0]);
}
public static void Main(String[] args)
{
int []vec = { 5, 20, 10, 7, 1 };
int N = vec.Length;
int K = 2;
kthSmallest(vec, N, K % N);
}
}
|
Javascript
<script>
function kthSmallest(v,N,K)
{
let heap1 = [];
for (let i = 0; i < N; ++i) {
heap1.push(v[i]);
if (heap1.length > K)
{
heap1.sort( function (a,b){
return a-b;
});
heap1.reverse();
heap1.shift();
}
}
heap1.sort( function (a,b){
return a-b;
});
heap1.reverse();
document.write(heap1[0] + "<br>" );
}
let vec=[5, 20, 10, 7, 1 ];
let N = vec.length;
let K = 2;
kthSmallest(vec, N, K % N);
</script>
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Time Complexity: O(N LogK)
Auxiliary Space: O(K), since the priority queue at any time holds at max k elements.
Last Updated :
16 Mar, 2023
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