两个数组对的绝对差的最小和

给定两个数组a []和b [] ，它们的长度为n 。任务是将数组a的每个元素与数组b中的元素配对，以使所有对的绝对差之和S最小。
假设，两个元素A [1]和a [j]的(I！= j)的一个与元件B [P]和B B的[Q]分别配对，
那么p不应该等于q 。
例子：

Input :  a[] = {3, 2, 1}
         b[] = {2, 1, 3}
Output : 0
Explaination :
 1st pairing: |3 - 2| + |2 - 1| + |1 - 3|
         = 1 + 1 + 2 = 4
 2nd pairing: |3 - 2| + |1 - 1| + |2 - 3|
         = 1 + 0 + 1 = 2
 3rd pairing: |2 - 2| + |3 - 1| + |1 - 3|
         = 0 + 2 + 2 = 4
 4th pairing: |1 - 2| + |2 - 1| + |3 - 3|
         = 1 + 1 + 0 = 2
 5th pairing: |2 - 2| + |1 - 1| + |3 - 3|
         = 0 + 0 + 0 = 0
 6th pairing: |1 - 2| + |3 - 1| + |2 - 3|
         = 1 + 2 + 1 = 4
 Therefore, 5th pairing has minimum sum of
 absolute difference.

Input :  n = 4
         a[] = {4, 1, 8, 7}
         b[] = {2, 3, 6, 5}
Output : 6

解决问题的方法是一种简单的贪婪方法。它包括两个步骤。
步骤1 ：以O(n log n)时间对两个数组进行排序。
步骤2 ：找到两个数组的每对对应元素(元素在相同索引处)的绝对差，并将结果加到和S上。此步骤的时间复杂度为O(n) 。
因此，该程序的总时间复杂度为O(n log n) 。

C++

// C++ program to find minimum sum of absolute
// differences of two arrays.
#include 
using namespace std;
 
// Returns minimum possible pairwise absolute
// difference of two arrays.
long long int findMinSum(int a[], int b[], int n)
{
    // Sort both arrays
    sort(a, a+n);
    sort(b, b+n);
 
    // Find sum of absolute differences
    long long int sum= 0 ;
    for (int i=0; i

---

Java

// Java program to find minimum sum of
// absolute differences of two arrays.
import java.util.Arrays;
 
class MinSum
{
    // Returns minimum possible pairwise
    // absolute difference of two arrays.
    static long findMinSum(long a[], long b[], long n)
    {
        // Sort both arrays
        Arrays.sort(a);
        Arrays.sort(b);
      
        // Find sum of absolute differences
        long sum = 0 ;
        for (int i = 0; i < n; i++)
            sum = sum + Math.abs(a[i] - b[i]);
      
        return sum;
    }
      
    // Driver code
    public static void main(String[] args)
    {
        // Both a[] and b[] must be of same size.
        long a[] = {4, 1, 8, 7};
        long b[] = {2, 3, 6, 5};
        int n = a.length;
        System.out.println(findMinSum(a, b, n));
    }   
}
  
// This code is contributed by Raghav Sharma

Python3

# Python3 program to find minimum sum
# of absolute differences of two arrays.
def findMinSum(a, b, n):
 
    # Sort both arrays
    a.sort()
    b.sort()
 
    # Find sum of absolute differences
    sum = 0
     
    for i in range(n):
        sum = sum + abs(a[i] - b[i])
 
    return sum
 
# Driver program
     
# Both a[] and b[] must be of same size.
a = [4, 1, 8, 7]
b = [2, 3, 6, 5]
n = len(a)
 
print(findMinSum(a, b, n))
 
# This code is contributed by Anant Agarwal.

C#

// C# program to find minimum sum of
// absolute differences of two arrays.
using System;
 
class MinSum {
     
    // Returns minimum possible pairwise
    // absolute difference of two arrays.
    static long findMinSum(long []a, long []b,
                           long n)
    {
         
        // Sort both arrays
        Array.Sort(a);
        Array.Sort(b);
     
        // Find sum of absolute differences
        long sum = 0 ;
        for (int i = 0; i < n; i++)
            sum = sum + Math.Abs(a[i] - b[i]);
     
        return sum;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        // Both a[] and b[] must be of same size.
        long []a = {4, 1, 8, 7};
        long []b = {2, 3, 6, 5};
        int n = a.Length;
        Console.Write(findMinSum(a, b, n));
    }
}
 
// This code is contributed by parashar...

Javascript

 
// JavaScript program for the above approach
 
// Function to return the minimum number
 
// Returns minimum possible pairwise
// absolute difference of two arrays.
    function findMinSum(a, b, n)
    {
        // Sort both arrays
        a.sort();
        b.sort();
      
        // Find sum of absolute differences
        let sum = 0 ;
        for (let i = 0; i < n; i++)
            sum = sum + Math.abs(a[i] - b[i]);
      
        return sum;
    }
     
// Driver Code
 
    // Both a[] and b[] must be of same size.
    let a = [4, 1, 8, 7];
    let b = [2, 3, 6, 5];
    let n = a.length;
    document.write(findMinSum(a, b, n));
 
// This code is contributed by code_hunt.

输出 ：

6