Given a circular array arr[] of size N, the task is to check if it is possible to make all array elements of the circular array equal by increasing pairs of adjacent elements by 1.
Examples:
Input: N = 4, arr[] = {2, 1, 3, 4}
Output:Yes
Explanation:
Step 1: {2, 1, 3, 4} -> {3, 2, 3, 4}
Step 2: {3, 2, 3, 4} -> {4, 3, 3, 4}
Step 3: {4, 3, 3, 4} -> {4, 4, 4, 4}
Input: N = 6, arr[]={1, 5, 9, 6, 1, 1}
Output: No
Approach: To solve the problem, it can be observed that the two indices consisting of elements to be incremented, one is even and the other one is odd. Therefore, if we increase the value of an even-indexed element, consequently an odd-indexed element will also be increased. Therefore, all the array elements can be made equal only if the sum of odd-indexed elements and even-indexed elements are equal. Follow the steps below to solve the problem:
- Calculate the sum of all even-indexed numbers, i.e sumEven.
- Calculate the sum of all odd-indexed numbers, i.e sumOdd.
- If sumEven and sumOdd are found to be equal, then print “Yes” else “No”.
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 check if all array// elements can be made equalbool checkEquall(int arr[], int N){ // Stores the sum of even and // odd array elements int sumEven = 0, sumOdd = 0; for (int i = 0; i < N; i++) { // If index is odd if (i & 1) sumOdd += arr[i]; else sumEven += arr[i]; } if (sumEven == sumOdd) return true; else return false;}// Driver Codeint main(){ int arr[] = { 2, 7, 3, 5, 7 }; int N = sizeof(arr) / sizeof(arr[0]); if (checkEquall(arr, N)) cout << "YES" << endl; else cout << "NO" << endl; return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to check if all array// elements can be made equalstatic boolean checkEquall(int arr[], int N){ // Stores the sum of even and // odd array elements int sumEven = 0, sumOdd = 0; for(int i = 0; i < N; i++) { // If index is odd if (i % 2 == 1) sumOdd += arr[i]; else sumEven += arr[i]; } if (sumEven == sumOdd) return true; else return false;}// Driver Codepublic static void main(String[] args){ int arr[] = { 2, 7, 3, 5, 7 }; int N = arr.length; if (checkEquall(arr, N)) System.out.print("YES" + "\n"); else System.out.print("NO" + "\n");}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to implement# the above approach# Function to check if all array# elements can be made equaldef checkEquall(arr, N): # Stores the sum of even and # odd array elements sumEven, sumOdd = 0, 0 for i in range(N): # If index is odd if (i & 1): sumOdd += arr[i] else: sumEven += arr[i] if (sumEven == sumOdd): return True else: return False# Driver Codeif __name__ == "__main__": arr = [ 2, 7, 3, 5, 7 ] N = len(arr) if (checkEquall(arr, N)): print("YES") else: print("NO")# This code is contributed by chitranayal |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to check if all array// elements can be made equalstatic bool checkEquall(int []arr, int N){ // Stores the sum of even and // odd array elements int sumEven = 0, sumOdd = 0; for(int i = 0; i < N; i++) { // If index is odd if (i % 2 == 1) sumOdd += arr[i]; else sumEven += arr[i]; } if (sumEven == sumOdd) return true; else return false;}// Driver Codepublic static void Main(String[] args){ int []arr = { 2, 7, 3, 5, 7 }; int N = arr.Length; if (checkEquall(arr, N)) Console.Write("YES" + "\n"); else Console.Write("NO" + "\n");}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript program to implement// the above approach// Function to check if all array// elements can be made equalfunction checkEquall(arr, N){ // Stores the sum of even and // odd array elements let sumEven = 0, sumOdd = 0; for(let i = 0; i < N; i++) { // If index is odd if (i % 2 == 1) sumOdd += arr[i]; else sumEven += arr[i]; } if (sumEven == sumOdd) return true; else return false;}// Driver Code let arr = [ 2, 7, 3, 5, 7 ]; let N = arr.length; if (checkEquall(arr, N)) document.write("YES" ); else document.write("NO"); </script> |
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
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