查找 Array 的所有索引，其值与其他元素的平均值相同

给定一个包含N个整数的数组arr[] ，任务是找到数组中的所有索引，使得对于每个索引i ，除arr[i]之外的所有元素的算术平均值等于该索引处元素的值。

例子：

Input: N = 5, arr[] = {1, 2, 3, 4, 5}
Output : {2}
Explanation : Upon removing array element at index 2 (i.e. 3),
arithmetic mean of rest of the array equals (1 + 2 + 4 + 5) / 4 = 3, which is equal to the element at index 2.
It can be seen that 2 is the only such index.

Input : N = 6, arr[] = {5, 5, 5, 5, 5, 5}
Output : {0, 1, 2, 3, 4, 5}

编程需要懂一点英语

方法：问题可以通过以下思路解决：

Calculate the total sum of the array and for each element (arr[i]) check if the average of all the other elements is same as arr[i].

编程需要懂一点英语

请按照以下步骤解决问题：

求数组所有元素的总和，并存储在变量中，比如sum。
遍历数组，并且，
在每次迭代中，通过从总和中减去当前元素值来找到没有当前元素的数组的总和。说， current_sum
通过将其除以N-1找到current_sum的平均值
如果均值等于当前索引处的值，则将索引推入答案向量中，否则继续。
返回答案向量。

以下是基于上述方法的代码：

C++

// C++ code for above approach
 
#include 
using namespace std;
 
// Function to find indices such that
// elements at those indices equals to the
// mean of the array except those indices
vector findIndices(int N, int A[])
{
 
    // Vector to store answer (i.e. indices)
    vector answer;
 
    // Calculation of sum of all elements
    int sum = 0;
    for (int i = 0; i < N; i++) {
        sum += A[i];
    }
 
    // For each element checking if its
    // value is equal to the mean of the
    // array except the current element
    for (int i = 0; i < N; i++) {
        int curr_sum = sum - A[i];
 
        if (curr_sum % (N - 1) == 0
            && curr_sum / (N - 1) == A[i]) {
            answer.push_back(i);
        }
    }
 
    // returning answer
    return answer;
}
 
// Driver Code
int main()
{
    int N = 5;
    int A[] = { 5, 5, 5, 5, 5 };
 
    vector ans = findIndices(N, A);
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
    }
}

Java

// Java code for above approach
import java.util.*;
 
class GFG {
 
  // Function to find indices such that
  // elements at those indices equals to the
  // mean of the array except those indices
  static  Vector findIndices(int N, int[] A)
  {
 
    // Vector to store answer (i.e. indices)
    Vector answer = new Vector();
 
    // Calculation of sum of all elements
    int sum = 0;
    for (int i = 0; i < N; i++) {
      sum += A[i];
    }
 
    // For each element checking if its
    // value is equal to the mean of the
    // array except the current element
    for (int i = 0; i < N; i++) {
      int curr_sum = sum - A[i];
 
      if (curr_sum % (N - 1) == 0
          && curr_sum / (N - 1) == A[i]) {
        answer.add(i);
      }
    }
 
    // returning answer
    return answer;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int N = 5;
    int A[] = { 5, 5, 5, 5, 5 };
 
    Vector ans = findIndices(N, A);
 
    for (int i = 0; i < ans.size(); i++) {
      System.out.print(ans.get(i) + " ");
    }
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code for above approach
 
# Function to find indices such that
# elements at those indices equals to the
# mean of the array except those indices
def findIndices(N, A):
 
    # Vector to store answer (i.e. indices)
    answer = []
 
    # Calculation of sum of all elements
    sum = 0
    for i in range(0, N):
        sum += A[i]
 
    # For each element checking if its
    # value is equal to the mean of the
    # array except the current element
    for i in range(0, N):
        curr_sum = sum - A[i]
 
        if (curr_sum % (N - 1) == 0 and curr_sum // (N - 1) == A[i]):
            answer.append(i)
 
    # returning answer
    return answer
 
# Driver Code
N = 5
A = [5, 5, 5, 5, 5]
 
ans = findIndices(N, A)
print(*ans)
 
# This code is contributed by Samim Hossain Mondal.

C#

// C# code for above approach
using System;
using System.Collections;
 
class GFG {
 
  // Function to find indices such that
  // elements at those indices equals to the
  // mean of the array except those indices
  static ArrayList findIndices(int N, int[] A)
  {
 
    // Vector to store answer (i.e. indices)
    ArrayList answer = new ArrayList();
 
    // Calculation of sum of all elements
    int sum = 0;
    for (int i = 0; i < N; i++) {
      sum += A[i];
    }
 
    // For each element checking if its
    // value is equal to the mean of the
    // array except the current element
    for (int i = 0; i < N; i++) {
      int curr_sum = sum - A[i];
 
      if (curr_sum % (N - 1) == 0
          && curr_sum / (N - 1) == A[i]) {
        answer.Add(i);
      }
    }
 
    // returning answer
    return answer;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 5;
    int[] A = { 5, 5, 5, 5, 5 };
 
    ArrayList ans = findIndices(N, A);
    for (int i = 0; i < ans.Count; i++) {
      Console.Write(ans[i] + " ");
    }
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

// JavaScript code for the above approach
 
        // Function to find indices such that
        // elements at those indices equals to the
        // mean of the array except those indices
        function findIndices(N, A) {
 
            // Vector to store answer (i.e. indices)
            let answer = [];
 
            // Calculation of sum of all elements
            let sum = 0;
            for (let i = 0; i < N; i++) {
                sum += A[i];
            }
 
            // For each element checking if its
            // value is equal to the mean of the
            // array except the current element
            for (let i = 0; i < N; i++) {
                let curr_sum = sum - A[i];
 
                if (curr_sum % (N - 1) == 0
                    && curr_sum / (N - 1) == A[i]) {
                    answer.push(i);
                }
            }
 
            // returning answer
            return answer;
        }
 
        // Driver Code
        let N = 5;
        let A = [5, 5, 5, 5, 5];
 
        let ans = findIndices(N, A);
        for (let i = 0; i < ans.length; i++) {
            document.write(ans[i] + " ")
        }
 
      // This code is contributed by Potta Lokesh

输出

0 1 2 3 4

时间复杂度： O(N)
辅助空间： O(N)