检查是否可以通过相邻对的增量使圆形阵列的所有元素相等

给定一个大小为N的圆形数组arr[] ，任务是检查是否可以通过将相邻元素对增加1来使圆形数组的所有数组元素相等。

例子：

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

处理方法：解决这个问题，可以观察到由要递增的元素组成的两个索引，一个是偶数，另一个是奇数。因此，如果我们增加偶数索引元素的值，那么奇数索引元素也会增加。因此，只有奇数索引元素和偶数索引元素之和相等，才能使所有数组元素相等。请按照以下步骤解决问题：

1. 计算所有偶数索引数的总和，即sumEven 。
2. 计算所有奇数索引数的总和，即sumOdd 。
3. 如果发现sumEven和sumOdd相等，则打印“是”，否则打印“否”。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to check if all array
// elements can be made equal
function 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");
                      

输出：

YES

时间复杂度： O(N)
辅助空间： O(1)

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