从数组 arr[] 和 brr[] 中计算对 (i, j) 使得 arr[i] – brr[j] = arr[j] – brr[i]

给定两个由N 个整数组成的数组arr[]和brr[] ，任务是计算数组中的(i, j)对的数量，使得(arr[i] – brr[j])和(arr[ j] – brr[i])相等。

例子：

Input: A[] = {1, 2, 3, 2, 1}, B[] = {1, 2, 3, 2, 1} 
Output: 2 
Explanation: The pairs satisfying the condition are: 

1. (1, 5): arr[1] – brr[5] = 1 – 1 = 0, arr[5[ – brr[1] = 1 – 1 = 0
2. (2, 4): arr[2] – brr[4] = 2 – 2 = 0, arr[4] – brr[2] = 2 – 2 = 0

Input: A[] = {1, 4, 20, 3, 10, 5}, B[] = {9, 6, 1, 7, 11, 6} 
Output: 4 

朴素方法：解决问题的最简单方法是从两个给定数组生成所有对并检查所需条件。对于发现条件为真的每一对，增加这些对的计数。最后，打印获得的计数。

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：思想是将给定的表达式(a[i] – b[j] = a[j] – b[i])转换为形式(a[i] + b[i] = a[j] + b[j])然后计算满足条件的对。以下是步骤：

1. 变换表达式a[i] – b[j] = a[j] – b[i] ==> a[i] + b[i] = a[j] +b[j] 。表达式的一般形式变成计算任何对(i, j)的两个数组的每个对应索引处的值的总和。
2. 初始化一个辅助数组c[]以在每个索引i处存储相应的总和c[i] = a[i] + b[i] 。
3. 现在问题减少到找到具有相同c[i]值的可能对的数量。
4. 计算数组c[]中每个元素的频率，如果任何c[i]频率值大于 1，则可以配对。
5. 使用公式计算上述步骤中有效对的数量：

$\frac{c[i]*(c[i] - 1)}{2}$

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the pairs such that
// given condition is satisfied
int CountPairs(int* a, int* b, int n)
{
    // Stores the sum of element at
    // each corresponding index
    int C[n];
 
    // Find the sum of each index
    // of both array
    for (int i = 0; i < n; i++) {
        C[i] = a[i] + b[i];
    }
 
    // Stores frequency of each element
    // present in sumArr
    map freqCount;
 
    for (int i = 0; i < n; i++) {
        freqCount[C[i]]++;
    }
 
    // Initialize number of pairs
    int NoOfPairs = 0;
 
    for (auto x : freqCount) {
        int y = x.second;
 
        // Add possible vaid pairs
        NoOfPairs = NoOfPairs
                    + y * (y - 1) / 2;
    }
 
    // Return Number of Pairs
    cout << NoOfPairs;
}
 
// Driver Code
int main()
{
    // Given array arr[] and brr[]
    int arr[] = { 1, 4, 20, 3, 10, 5 };
 
    int brr[] = { 9, 6, 1, 7, 11, 6 };
 
    // Size of given array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function calling
    CountPairs(arr, brr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.io.*;
 
class GFG{
     
// Function to find the minimum number
// needed to be added so that the sum
// of the digits does not exceed K
static void CountPairs(int a[], int b[], int n)
{
     
    // Stores the sum of element at
    // each corresponding index
    int C[] = new int[n];
   
    // Find the sum of each index
    // of both array
    for(int i = 0; i < n; i++)
    {
        C[i] = a[i] + b[i];
    }
     
    // Stores frequency of each element
    // present in sumArr
    // map freqCount;
    HashMap freqCount = new HashMap<>();
   
    for(int i = 0; i < n; i++)
    {
        if (!freqCount.containsKey(C[i]))
            freqCount.put(C[i], 1);
        else
            freqCount.put(C[i],
            freqCount.get(C[i]) + 1);
    }
   
    // Initialize number of pairs
    int NoOfPairs = 0;
   
    for(Map.Entry x : freqCount.entrySet())
    {
        int y = x.getValue();
   
        // Add possible vaid pairs
        NoOfPairs = NoOfPairs +
                  y * (y - 1) / 2;
    }
   
    // Return Number of Pairs
   System.out.println(NoOfPairs);
}
 
// Driver Code
public static void main(String args[])
{
   
    // Given array arr[] and brr[]
    int arr[] = { 1, 4, 20, 3, 10, 5 };
    int brr[] = { 9, 6, 1, 7, 11, 6 };
     
    // Size of given array
    int N = arr.length;
     
    // Function calling
    CountPairs(arr, brr, N);
}
}
 
// This code is contributed by bikram2001jha

Python3

# Python3 program for the above approach
 
# Function to count the pairs such that
# given condition is satisfied
def CountPairs(a, b, n):
     
    # Stores the sum of element at
    # each corresponding index
    C = [0] * n
  
    # Find the sum of each index
    # of both array
    for i in range(n):
        C[i] = a[i] + b[i]
     
    # Stores frequency of each element
    # present in sumArr
    freqCount = dict()
  
    for i in range(n):
        if C[i] in freqCount.keys():
            freqCount[C[i]] += 1
        else:
            freqCount[C[i]] = 1
  
    # Initialize number of pairs
    NoOfPairs = 0
  
    for x in freqCount:
        y = freqCount[x]
  
        # Add possible vaid pairs
        NoOfPairs = (NoOfPairs + y *
                       (y - 1) // 2)
     
    # Return Number of Pairs
    print(NoOfPairs)
 
# Driver Code
 
# Given array arr[] and brr[]
arr = [ 1, 4, 20, 3, 10, 5 ]
brr = [ 9, 6, 1, 7, 11, 6 ]
  
# Size of given array
N = len(arr)
  
# Function calling
CountPairs(arr, brr, N)
 
# This code is contributed by code_hunt

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the minimum number
// needed to be added so that the sum
// of the digits does not exceed K
static void CountPairs(int []a, int []b,
                       int n)
{
     
    // Stores the sum of element at
    // each corresponding index
    int []C = new int[n];
   
    // Find the sum of each index
    // of both array
    for(int i = 0; i < n; i++)
    {
        C[i] = a[i] + b[i];
    }
     
    // Stores frequency of each element
    // present in sumArr
    // map freqCount;
    Dictionary freqCount = new Dictionary();
   
    for(int i = 0; i < n; i++)
    {
        if (!freqCount.ContainsKey(C[i]))
            freqCount.Add(C[i], 1);
        else
            freqCount[C[i]] = freqCount[C[i]] + 1;
    }
   
    // Initialize number of pairs
    int NoOfPairs = 0;
   
    foreach(KeyValuePair x in freqCount)
    {
        int y = x.Value;
   
        // Add possible vaid pairs
        NoOfPairs = NoOfPairs +
                  y * (y - 1) / 2;
    }
     
    // Return Number of Pairs
    Console.WriteLine(NoOfPairs);
}
 
// Driver Code
public static void Main(String []args)
{
     
    // Given array []arr and brr[]
    int []arr = { 1, 4, 20, 3, 10, 5 };
    int []brr = { 9, 6, 1, 7, 11, 6 };
     
    // Size of given array
    int N = arr.Length;
     
    // Function calling
    CountPairs(arr, brr, N);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// JavaScript program for the above approach
 
// Function to count the pairs such that
// given condition is satisfied
function CountPairs(a,b, n)
{
    // Stores the sum of element at
    // each corresponding index
    var C = Array(n);
 
    // Find the sum of each index
    // of both array
    for (var i = 0; i < n; i++) {
        C[i] = a[i] + b[i];
    }
 
    // Stores frequency of each element
    // present in sumArr
    var freqCount = new Map();
 
    for (var i = 0; i < n; i++) {
        if(freqCount.has(C[i]))
            freqCount.set(C[i], freqCount.get(C[i])+1)
        else
            freqCount.set(C[i], 1)
    }
 
    // Initialize number of pairs
    var NoOfPairs = 0;
 
    freqCount.forEach((value, key) => {
         
        var y = value;
 
        // Add possible vaid pairs
        NoOfPairs = NoOfPairs
                    + y * (y - 1) / 2;
    });
 
    // Return Number of Pairs
    document.write( NoOfPairs);
}
 
// Driver Code
 
// Given array arr[] and brr[]
var arr = [1, 4, 20, 3, 10, 5];
var brr = [ 9, 6, 1, 7, 11, 6 ];
 
// Size of given array
var N = arr.length;
 
// Function calling
CountPairs(arr, brr, N);
 

输出：

4

时间复杂度： O(N)
辅助空间： O(N)

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