Given an array arr[ ] of size N, the task for each array element is to print the nearest perfect square having same parity.
Examples:
Input: arr[ ] = {6, 3, 2, 15}
Output: 4 1 4 9
Explanation:
The nearest even perfect square of arr[0] (= 6) is 4.
The nearest odd perfect square of arr[1] (= 3) is 1.
The nearest even perfect square of arr[2] (= 2) is 4
The nearest odd perfect square of arr[3] (= 15) is 9.Input: arr[ ] = {31, 18, 64}
Output: 25 16 64
Approach: Follow the steps below to solve the problem:
- Traverse the array and perform the following operations:
- Find the square root of the current array element and store it in a variable, say sr.
- If sr is of same parity as arr[i], then sr*sr is the nearest perfect square.
- Otherwise, (sr + 1)2 is the nearest perfect square.
- Print the nearest perfect square obtained in the above step.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the nearest even and odd// perfect squares for even and odd array elementsvoid nearestPerfectSquare(int arr[], int N){ // Traverse the array for (int i = 0; i < N; i++) { // Calculate square root of // current array element int sr = sqrt(arr[i]); // If both are of same parity if ((sr & 1) == (arr[i] & 1)) cout << sr * sr << " "; // Otherwise else { sr++; cout << sr * sr << " "; } }}// Driver Codeint main(){ int arr[] = { 6, 3, 2, 15 }; int N = sizeof(arr) / sizeof(arr[0]); nearestPerfectSquare(arr, N); return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to find the nearest even and odd// perfect squares for even and odd array elementsstatic void nearestPerfectSquare(int arr[], int N){ // Traverse the array for (int i = 0; i < N; i++) { // Calculate square root of // current array element int sr = (int)Math.sqrt(arr[i]); // If both are of same parity if ((sr & 1) == (arr[i] & 1)) System.out.print((sr * sr) + " "); // Otherwise else { sr++; System.out.print((sr * sr) + " "); } }}// Driver Codepublic static void main(String[] args){ int arr[] = { 6, 3, 2, 15 }; int N = arr.length; nearestPerfectSquare(arr, N);}}// This code is contributed by souravghosh0416. |
Python3
# Python3 program for the above approachimport math# Function to find the nearest even and odd# perfect squares for even and odd array elementsdef nearestPerfectSquare(arr, N) : # Traverse the array for i in range(N): # Calculate square root of # current array element sr = int(math.sqrt(arr[i])) # If both are of same parity if ((sr & 1) == (arr[i] & 1)) : print(sr * sr, end = " ") # Otherwise else : sr += 1 print(sr * sr, end = " ") # Driver Codearr = [ 6, 3, 2, 15 ]N = len(arr)nearestPerfectSquare(arr, N)# This code is contributed by sanjoy_62. |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the nearest even and odd // perfect squares for even and odd array elements static void nearestPerfectSquare(int[] arr, int N) { // Traverse the array for (int i = 0; i < N; i++) { // Calculate square root of // current array element int sr = (int)Math.Sqrt(arr[i]); // If both are of same parity if ((sr & 1) == (arr[i] & 1)) Console.Write((sr * sr) + " "); // Otherwise else { sr++; Console.Write((sr * sr) + " "); } } } // Driver Code static public void Main() { int[] arr = { 6, 3, 2, 15 }; int N = arr.Length; nearestPerfectSquare(arr, N); }}// This code is contributed by splevel62. |
Javascript
<script>// Javascript program to implement// the above approach// Function to find the nearest even and odd// perfect squares for even and odd array elementsfunction nearestPerfectSquare(arr, N){ // Traverse the array for(let i = 0; i < N; i++) { // Calculate square root of // current array element let sr = Math.floor(Math.sqrt(arr[i])); // If both are of same parity if ((sr & 1) == (arr[i] & 1)) document.write((sr * sr) + " "); // Otherwise else { sr++; document.write((sr * sr) + " "); } }}// Driver code // Given arraylet arr = [ 6, 3, 2, 15 ];let N = arr.length;nearestPerfectSquare(arr, N);// This code is contributed by target_2</script> |
Output:
4 1 4 9
Time Complexity: O(N*logN) where N is size of given array
Auxiliary Space: O(1)
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