两个未排序数组之间的最小差值对

给定两个整数数组，请计算差异最小(非负)的一对值(每个数组中的一个值)。退还差额。

例子 ：

Input : A[] = {l, 3, 15, 11, 2}
        B[] = {23, 127, 235, 19, 8} 
Output : 3  
That is, the pair (11, 8) 

Input : A[] = {l0, 5, 40}
        B[] = {50, 90, 80} 
Output : 10
That is, the pair (40, 50)

一个简单的解决方案是使用两个具有时间复杂度O(n ² )的循环进行蛮力攻击。

更好的解决方案是对数组进行排序。对数组进行排序后，我们可以使用下文中讨论的方法遍历数组，从而找到最小的差异。

从两个排序的数组中找到最接近的对

考虑以下两个数组：
答：{l，2，11，15}
B：{4，12，19，23，127，235}

1.假设指针a指向A的开头，指针b指向B的开头。a和bis之间的当前差值。3.将其存储为最小值。

2.我们如何(可能)使这种差异变小?好吧，bis处的值大于a处的值，因此移动b只会使差值更大。因此，我们要移动一个。

3.现在a指向2，b(仍然)指向4。该差为2，因此我们应更新min。移动a，因为它较小。

4.现在a指向11，b指向4.移动b。

5.现在a指向11，b指向12。将min更新到1。移动b。等等。

以下是该想法的实现。

C++

// C++ Code to find Smallest 
// Difference between two Arrays
#include 
using namespace std;
  
// function to calculate Small 
// result between two arrays
int findSmallestDifference(int A[], int B[],
                           int m, int n)
{
    // Sort both arrays using
    // sort function
    sort(A, A + m);
    sort(B, B + n);
  
    int a = 0, b = 0;
  
    // Initialize result as max value
    int result = INT_MAX;
  
    // Scan Both Arrays upto 
    // sizeof of the Arrays
    while (a < m && b < n)
    {
        if (abs(A[a] - B[b]) < result)
            result = abs(A[a] - B[b]);
  
        // Move Smaller Value
        if (A[a] < B[b])
            a++;
  
        else
            b++;
    }
  
    // return final sma result
    return result; 
}
  
// Driver Code
int main()
{
    // Input given array A
    int A[] = {1, 2, 11, 5};
  
    // Input given array B
    int B[] = {4, 12, 19, 23, 127, 235};
  
  
    // Calculate size of Both arrays
    int m = sizeof(A) / sizeof(A[0]);
    int n = sizeof(B) / sizeof(B[0]);
  
    // Call function to print 
    // smallest result
    cout << findSmallestDifference(A, B, m, n);
  
    return 0;
}

Java

// Java Code to find Smallest 
// Difference between two Arrays
import java.util.*;
  
class GFG 
{
      
    // function to calculate Small 
    // result between two arrays
    static int findSmallestDifference(int A[], int B[],
                                      int m, int n)
    {
        // Sort both arrays 
        // using sort function
        Arrays.sort(A);
        Arrays.sort(B);
      
        int a = 0, b = 0;
      
        // Initialize result as max value
        int result = Integer.MAX_VALUE;
      
        // Scan Both Arrays upto 
        // sizeof of the Arrays
        while (a < m && b < n)
        {
            if (Math.abs(A[a] - B[b]) < result)
                result = Math.abs(A[a] - B[b]);
      
            // Move Smaller Value
            if (A[a] < B[b])
                a++;
      
            else
                b++;
        }
          
        // return final sma result
        return result; 
    }
      
    // Driver Code
    public static void main(String[] args) 
    {
        // Input given array A
        int A[] = {1, 2, 11, 5};
      
        // Input given array B
        int B[] = {4, 12, 19, 23, 127, 235};
      
      
        // Calculate size of Both arrays
        int m = A.length;
        int n = B.length;
      
        // Call function to 
        // print smallest result
        System.out.println(findSmallestDifference
                                   (A, B, m, n));
          
    }
}
// This code is contributed
// by Arnav Kr. Mandal.

Python3

# Python 3 Code to find
# Smallest Difference between
# two Arrays
import sys
  
# function to calculate
# Small result between
# two arrays
def findSmallestDifference(A, B, m, n):
  
    # Sort both arrays 
    # using sort function
    A.sort()
    B.sort()
  
    a = 0
    b = 0
  
    # Initialize result as max value
    result = sys.maxsize
  
    # Scan Both Arrays upto
    # sizeof of the Arrays
    while (a < m and b < n):
      
        if (abs(A[a] - B[b]) < result):
            result = abs(A[a] - B[b])
  
        # Move Smaller Value
        if (A[a] < B[b]):
            a += 1
  
        else:
            b += 1
    # return final sma result
    return result 
  
# Driver Code
  
# Input given array A
A = [1, 2, 11, 5]
  
# Input given array B
B = [4, 12, 19, 23, 127, 235]
  
# Calculate size of Both arrays
m = len(A)
n = len(B)
  
# Call function to 
# print smallest result
print(findSmallestDifference(A, B, m, n))
  
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# Code to find Smallest 
// Difference between two Arrays
using System;
  
class GFG 
{
      
    // function to calculate Small 
    // result between two arrays
    static int findSmallestDifference(int []A, int []B,
                                      int m, int n)
    {
          
        // Sort both arrays using
        // sort function
        Array.Sort(A);
        Array.Sort(B);
      
        int a = 0, b = 0;
      
        // Initialize result as max value
        int result = int.MaxValue;
      
        // Scan Both Arrays upto 
        // sizeof of the Arrays
        while (a < m && b < n)
        {
            if (Math.Abs(A[a] - B[b]) < result)
                result = Math.Abs(A[a] - B[b]);
      
            // Move Smaller Value
            if (A[a] < B[b])
                a++;
      
            else
                b++;
        }
          
        // return final sma result
        return result; 
    }
      
    // Driver Code
    public static void Main() 
    {
          
        // Input given array A
        int []A = {1, 2, 11, 5};
      
        // Input given array B
        int []B = {4, 12, 19, 23, 127, 235};
      
      
        // Calculate size of Both arrays
        int m = A.Length;
        int n = B.Length;
      
        // Call function to 
        // print smallest result
        Console.Write(findSmallestDifference
                              (A, B, m, n));
          
    }
}
  
// This code is contributed
// by nitin mittal.

输出 ：

1

该算法需要O(m log m + n log n)时间进行排序，并花费O(m + n)时间来找到最小差异。因此，总体运行时间为O(m log m + n log n)。