Find a point that lies inside exactly K given squares

Given an integer K and an array arr each of whose element x represents a square with two of its vertices as (0, 0) and (x, x). The task is to find a point which lies in exactly K squares.

Examples:

Input: arr[] = {1, 2, 3, 4}, K = 2 
Output: (3, 3) 
The point (3, 3) lies inside 3rd and 4th square only.

Input: arr[] = {8, 1, 55, 90}, K = 3 
Output: (8, 8) 

Approach: Since all squares have a common corner point (0, 0), any point which lies in any square would also lie in any larger square. Hence, we can simply print the other corner of the Kth largest square.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

int PointInKSquares(int n, int a[], int k)
{
    sort(a, a + n);
    return a[n - k];
}

// Driver Program to test above function
int main()
{
    int k = 2;
    int a[] = { 1, 2, 3, 4 };
    int n = sizeof(a) / sizeof(a[0]);

    int x = PointInKSquares(n, a, k);
    cout << "(" << x << ", " << x << ")";
}

Java

// Java implementation of the approach

import java.io.*;
import java.util.*;
class GFG {


static int PointInKSquares(int n, int a[], int k)
{
    Arrays.sort(a);
    return a[n - k];
}

// Driver Program to test above function

    public static void main (String[] args) {
            int k = 2;
    int []a = { 1, 2, 3, 4 };
    int n = a.length;

    int x = PointInKSquares(n, a, k);
    System.out.println( "(" + x + ", " + x +")");


    }
}
// This code is contributed by anuj_67..

Python3

# Python 3 implementation of the
# above approach
def PointInKSquares(n, a, k) :

    a.sort()
    return a[n - k]

# Driver Code
if __name__ == "__main__" :

    k = 2
    a = [1, 2, 3, 4]
    n = len(a)

    x = PointInKSquares(n, a, k)
    print("(", x, ",", x, ")")

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
class GFG
{

static int PointInKSquares(int n,
                           int []a, int k)
{
    Array.Sort(a);
    return a[n - k];
}

// Driver Code
public static void Main (String[] args)
{
    int k = 2;
    int []a = { 1, 2, 3, 4 };
    int n = a.Length;

    int x = PointInKSquares(n, a, k);
    Console.WriteLine("(" + x + ", " + x +")");
}
}

// This code is contributed
// by Arnab Kundu

PHP

<?php
// PHP implementation of the approach

function PointInKSquares($n, $a, $k)
{
    sort($a);
    return $a[$n - $k];
}

// Driver Code
$k = 2;
$a = array(1, 2, 3, 4);
$n = sizeof($a);

$x = PointInKSquares($n, $a, $k);
echo "(" . $x . ", " . $x . ")";

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// Javascript implementation of the approach


function PointInKSquares(n, a, k)
{
    a.sort();
    return a[n - k];
}

// Driver Program to test above function

    let k = 2;
    let a = [ 1, 2, 3, 4 ];
    let n = a.length;

    let x = PointInKSquares(n, a, k);
    document.write("(" + x + ", " + x + ")");

// This code is contributed by Mayank Tyagi

</script>

Output

(3, 3)

Complexity Analysis:

• Time Complexity: O(nlog(n)), since to sort the elements of the array, the best algorithm takes (n * log n) time.
• Auxiliary Space: O(1), since no extra space has been taken.