在不改变平均值的情况下可以从 Array 中移除的对数

给定一个大小为N的数组arr[] ，任务是找出数组的对数，删除后数组的均值（即所有元素的平均值）保持不变。

例子：

Input: N = 5, arr[] = {1, 4, 7, 3, 5}
Output:  2
Explanation:  The required pairs of positions are – {0, 2} and {3, 4}.
Mean of original array = (1 + 4 + 7 + 3 + 5) / 5 = 4.
On deletion of elements at positions 0 and 2, array becomes {4, 3, 5}. 
Mean of the new array = (4 + 3 + 5) / 3 = 4. which is same as the mean of original array.
On deletion of elements at positions 3 and 4, array becomes {1, 4, 7}. 
Mean of the new array = (1 + 4 + 7) / 3 = 4. which is same as the mean of original array.

Input: N = 3, A = {50, 20, 10}
Output: 0
Explanation: No such pair is possible.

方法：考虑以下观察：

Observation:

If a sum (S = 2*mean) is subtracted from the original array (with initial mean = M), the mean of the new updated array will still remains as M.

Derivation:

• Sum of array (of size N) = N * mean = N * M
• new mean = (sum of array – 2 * M) / (N – 2)
                    =(N*M – 2*M)/(N – 2)
                    = M.

可以按照以下步骤来解决问题 -

• 计算给定数组的平均值。
• 计算所需的总和S即2*mean 。
• 如果S不是整数，则返回0 ，因为不会有任何对具有非整数和。
• 否则，通过散列找到许多所需的对。

下面是上述方法的实现：

C++

// C++ code to implement the above approach
#include 
using namespace std;
 
// Function to calculate the number
// of pairs of positions [i, j] (i 0) {
        return 0;
    }
    else {
 
        // Initialising count variable to
        // store total number of valid pairs
        int count = 0;
 
        // Declaring a map to calculate
        // number of pairs with required sum
        map mp;
 
        // Standard procedure to calculate
        // total number of pairs
        // of required sum
        for (int i = 0; i < n; i++) {
            if (i > 0) {
                count += mp[required_sum
                            - A[i]];
            }
            mp[A[i]]++;
        }
 
        // Returning count
        return count;
    }
 
    // If no pairs with required sum
    return 0;
}
 
// Driver code
int main()
{
    int N = 5;
    int arr[] = { 1, 4, 7, 3, 5 };
    int numberOfPairs = countPairs(N, arr);
    cout << numberOfPairs;
    return 0;
}

Python

# Python code to implement the above approach
 
# Function to calculate the number
# of pairs of positions [i, j] (i 0):
        return 0
         
    else:
 
        # Initialising count variable to
        # store total number of valid pairs
        count = 0
 
        # Declaring a map to calculate
        # number of pairs with required sum
        mp = {}
 
        # Standard procedure to calculate
        # total number of pairs
        # of required sum
        for i in range(0, n):
            if (i > 0 and required_sum - A[i] in mp):
                count += mp[required_sum - A[i]]
             
            if arr[i] in mp:
                mp[arr[i]] += 1
            else:
                mp[arr[i]] = 1
 
        # Returning count
        return count
 
    # If no pairs with required sum
    return 0
 
# Driver code
 
N = 5
arr = [ 1, 4, 7, 3, 5 ]
numberOfPairs = countPairs(N, arr)
print(numberOfPairs)
 
# This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code to implement the above approach
 
    // Function to calculate the number
    // of pairs of positions [i, j] (i<j),
    // such that on deletion of elements
    // at these positions, the mean of
    // remaining (n-2) elements does not change
    const countPairs = (n, A) => {
        // Calculating sum of array
        let sum = 0;
        for (let i = 0; i < n; i++) {
            sum += A[i];
        }
 
        // Calculating mean of array
        let mean = (sum * 1.0) / n;
 
        // Required sum = twice of mean
        let required_sum = 2.0 * mean;
 
        // Checking if required_sum is integral
        let check = required_sum;
        let temp = required_sum - check;
 
        if (temp > 0) {
            return 0;
        }
        else {
 
            // Initialising count variable to
            // store total number of valid pairs
            let count = 0;
 
            // Declaring a map to calculate
            // number of pairs with required sum
            let mp = {};
 
            // Standard procedure to calculate
            // total number of pairs
            // of required sum
            for (let i = 0; i < n; i++) {
                if (i > 0 && (required_sum - A[i] in mp)) {
                    count += mp[required_sum
                        - A[i]];
                }
                if (A[i] in mp) mp[A[i]] += 1;
                else mp[A[i]] = 1;
            }
 
            // Returning count
            return count;
        }
 
        // If no pairs with required sum
        return 0;
    }
 
    // Driver code
 
    let N = 5;
    let arr = [1, 4, 7, 3, 5];
    let numberOfPairs = countPairs(N, arr);
    document.write(numberOfPairs);
 
    // This code is contributed by rakeshsahni
 

输出

2

时间复杂度： O(N)
辅助空间： O(N)