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;Â
// Function to find the nearest even and odd// perfect squares for even and odd array elementsvoid nearestPerfectSquare(int arr[], int N){Â
    // Traverse the array    for (int i = 0; i < N; i++) {Â
        // Calculate square root of        // current array element        int sr = sqrt(arr[i]);Â
        // If both are of same parity        if ((sr & 1) == (arr[i] & 1))            cout << sr * sr << " ";Â
      // Otherwise        else {            sr++;            cout << sr * sr << " ";        }    }}Â
// Driver Codeint main(){Â Â Â Â int arr[] = { 6, 3, 2, 15 };Â Â Â Â int N = sizeof(arr) / sizeof(arr[0]);Â Â Â Â nearestPerfectSquare(arr, N);Â Â Â Â return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;Â
class GFG{Â
// Function to find the nearest even and odd// perfect squares for even and odd array elementsstatic void nearestPerfectSquare(int arr[], int N){Â
    // Traverse the array    for (int i = 0; i < N; i++) {Â
        // Calculate square root of        // current array element        int sr = (int)Math.sqrt(arr[i]);Â
        // If both are of same parity        if ((sr & 1) == (arr[i] & 1))            System.out.print((sr * sr) + " ");Â
      // Otherwise        else {            sr++;            System.out.print((sr * sr) + " ");        }    }}Â
// Driver Codepublic static void main(String[] args){Â Â Â Â int arr[] = { 6, 3, 2, 15 };Â Â Â Â int N = arr.length;Â Â Â Â nearestPerfectSquare(arr, N);}}Â
// This code is contributed by souravghosh0416. |
Python3
# Python3 program for the above approachimport mathÂ
# Function to find the nearest even and odd# perfect squares for even and odd array elementsdef nearestPerfectSquare(arr, N) :Â
    # Traverse the array    for i in range(N):Â
        # Calculate square root of        # current array element        sr = int(math.sqrt(arr[i]))Â
        # If both are of same parity        if ((sr & 1) == (arr[i] & 1)) :            print(sr * sr, end = " ")Â
      # Otherwise        else :            sr += 1            print(sr * sr, end = " ")         # Driver Codearr = [ 6, 3, 2, 15 ]N = len(arr)nearestPerfectSquare(arr, N)Â
# This code is contributed by sanjoy_62. |
C#
// C# program for the above approachusing System;Â
class GFG{Â
  // Function to find the nearest even and odd  // perfect squares for even and odd array elements  static void nearestPerfectSquare(int[] arr, int N)  {Â
    // Traverse the array    for (int i = 0; i < N; i++) {Â
      // Calculate square root of      // current array element      int sr = (int)Math.Sqrt(arr[i]);Â
      // If both are of same parity      if ((sr & 1) == (arr[i] & 1))        Console.Write((sr * sr) + " ");Â
      // 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);  }}Â
// This code is contributed by splevel62. |
Javascript
<script>Â
// Javascript program to implement// the above approachÂ
// 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]));Â
        // If both are of same parity        if ((sr & 1) == (arr[i] & 1))            document.write((sr * sr) + " ");Â
        // Otherwise        else        {            sr++;            document.write((sr * sr) + " ");        }    }}Â
// Driver code     // Given arraylet arr = [ 6, 3, 2, 15 ];let N = arr.length;Â
nearestPerfectSquare(arr, N);Â
// This code is contributed by target_2Â
</script> |
Output:Â
4 1 4 9
Â
Time Complexity: O(N*logN) where N is size of given array
Auxiliary Space: O(1)Â
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