计算数组arr []和brr []中的对(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 [] ，任务是计算两个数组中对(i，j)的对数，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 

天真的方法：解决问题的最简单方法是从两个给定的数组中生成所有对，并检查所需条件。对于每一个的量，条件被发现是真对，增加这种对计数。最后，打印获得的计数。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：想法是将给定的表达式(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。步骤如下：

1. 转换表达式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为对任何对(i，j)的两个数组的每个对应索引处的值之和进行计数。
2. 初始化辅助数组c []以在每个索引i处存储对应的和c [i] = a [i] + b [i] 。
3. 现在，问题减少了，以找到具有相同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，则可以成对。
5. 使用公式计算以上步骤中的有效对数：

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

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

输出：

4

时间复杂度： O(N)
辅助空间： O(N)