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 functionint 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 approachimport 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 approachdef PointInKSquares(n, a, k) : a.sort() return a[n - k]# Driver Codeif __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 Codepublic 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 approachfunction 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.
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