Check if all elements of a Circular Array can be made equal by increments of adjacent pairs

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:

  1. Calculate the sum of all even-indexed numbers, i.e sumEven.
  2. Calculate the sum of all odd-indexed numbers, i.e sumOdd.
  3. If sumEven and sumOdd are found to be equal, then print “Yes” else “No”.

Below is the implementation of the above approach:

C++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if all array
// elements can be made equal
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 & 1)
            sumOdd += arr[i];
        else
            sumEven += arr[i];
    }
 
    if (sumEven == sumOdd)
        return true;
    else
        return false;
}
 
// Driver Code
int 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;
}

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Java

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// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to check if all array
// elements can be made equal
static 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 Code
public 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

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Python3

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# Python3 program to implement
# the above approach
 
# Function to check if all array
# elements can be made equal
def 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 Code
if __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

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C#

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// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to check if all array
// elements can be made equal
static 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 Code
public 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

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Output: 

YES


 

Time Complexity: O(N)
Auxiliary Space: O(1)

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Improved By : chitranayal, princiraj1992