通过将对递减和递增 1 来最小化所有数组元素对之间的绝对差之和

给定一个由N个整数组成的数组arr[] （从 1 开始的索引），任务是通过将任意一对元素递减和递增1任意次数来找到所有数组元素对之间的绝对差的最小和。

例子：

Input: arr[] = {1, 2, 3}
Output: 0
Explanation:
Modify the array elements by performing the following operations:

Choose the pairs of element (arr[1], arr[3]) and incrementing and decrementing the pairs modifies the array to {2, 2, 2}.

After the above operations, the sum of the absolute differences is |2 – 2| + |2 – 2| + |2 – 2| = 0. Therefore, print 0.

Input: arr[] = {0, 1, 0, 1}
Output: 4

编程需要懂一点英语

方法：给定的问题可以通过使用贪婪方法来解决。可以看出，为了最小化每对数组元素arr[]之间的绝对差之和，使每个数组元素彼此接近的想法。请按照以下步骤解决问题：

找到数组元素arr[]的总和并将其存储在一个变量中，比如sum 。
现在，如果sum % N的值为0 ，则打印0 ，因为可以使所有数组元素相等，并且表达式的结果值始终为 0 。否则，找到sum % N的值并将其存储在变量中，例如R 。
现在，如果所有数组元素都是sum/N ，那么我们可以将数组元素的R个数设为1 ，将其余数组元素设为0 ，以最小化结果值。
在上述步骤之后，绝对差的最小和由R*(N – R)给出。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum value
// of the sum of absolute difference
// between all pairs of arrays
int minSumDifference(int ar[], int n)
{
    // Stores the sum of array elements
    int sum = 0;
 
    // Find the sum of array element
    for (int i = 0; i < n; i++)
        sum += ar[i];
 
    // Store the value of sum%N
    int rem = sum % n;
 
    // Return the resultant value
    return rem * (n - rem);
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 6, 8, 5, 2,
                  1, 11, 7, 10, 4 };
    int N = sizeof(arr) / sizeof(int);
    cout << minSumDifference(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG {
 
    // Function to find the minimum value
    // of the sum of absolute difference
    // between all pairs of arrays
    public static int minSumDifference(int ar[], int n) {
        // Stores the sum of array elements
        int sum = 0;
 
        // Find the sum of array element
        for (int i = 0; i < n; i++)
            sum += ar[i];
 
        // Store the value of sum%N
        int rem = sum % n;
 
        // Return the resultant value
        return rem * (n - rem);
    }
 
    // Driver Code
    public static void main(String args[]) {
        int[] arr = { 3, 6, 8, 5, 2, 1, 11, 7, 10, 4 };
        int N = arr.length;
        System.out.println(minSumDifference(arr, N));
 
    }
}
 
// This code is contributed by gfgking.

Python3

# Python 3 program for the above approach
 
# Function to find the minimum value
# of the sum of absolute difference
# between all pairs of arrays
def minSumDifference(ar, n):
    # Stores the sum of array elements
    sum = 0
 
    # Find the sum of array element
    for i in range(n):
        sum += ar[i]
 
    # Store the value of sum%N
    rem = sum % n
 
    # Return the resultant value
    return rem * (n - rem)
 
# Driver Code
if __name__ == '__main__':
    arr = [3, 6, 8, 5, 2, 1, 11, 7, 10, 4]
    N = len(arr)
    print(minSumDifference(arr, N))
     
    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the minimum value
// of the sum of absolute difference
// between all pairs of arrays
public static int minSumDifference(int[] ar, int n)
{
     
    // Stores the sum of array elements
    int sum = 0;
 
    // Find the sum of array element
    for(int i = 0; i < n; i++)
        sum += ar[i];
 
    // Store the value of sum%N
    int rem = sum % n;
 
    // Return the resultant value
    return rem * (n - rem);
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 3, 6, 8, 5, 2,
                  1, 11, 7, 10, 4 };
    int N = arr.Length;
     
    Console.Write(minSumDifference(arr, N));
}
}
 
// This code is contributed by sanjoy_62

Javascript

输出：

时间复杂度： O(N)
辅助空间： O(1)