数组中唯一对 (i, j) 的计数，使得 A[i] 和 A[j] 的反向之和等于 A[i] 和 A[j] 的反向之和

给定一个由N 个正整数组成的数组arr[] ，任务是找到唯一对(i, j)的计数，使得arr[i]和反向 (arr[j])的总和与总和相同reverse(arr[i])和arr[j] 。

例子：

Input: arr[] = {2, 15, 11, 7}
Output: 3
Explanation:
The pairs are (0, 2), (0, 3) and (2, 3).

(0, 2): arr[0] + reverse(arr[2]) (= 2 + 11 = 13) and reverse(arr[0]) + arr[2](= 2 + 11 = 13).
(0, 3): arr[0] + reverse(arr[3]) (= 2 + 7 = 9) and reverse(arr[0]) + arr[3](= 2 + 7 = 9).
(2, 3): arr[2] + reverse(arr[3]) (= 11 + 7 = 18) and reverse(arr[2]) + arr[3](= 11 + 7 = 18).

Input: A[] = {22, 115, 7, 313, 17, 23, 22}
Output: 6

编程需要懂一点英语

朴素方法：最简单的方法是生成给定数组的所有可能对，如果任何一对元素满足给定条件，则计算这些对。完成上述步骤m后，打印count的值作为结果。

时间复杂度： O(N ² *log M)，其中 M 是 A[] 中的最大元素
辅助空间： O(1)

高效的方法：上述方法可以通过使用哈希技术并将等式重写为：

A[i] + reverse(A[j]) = reverse(A[i]) + A[j]
=> A[i] – reverse(A[i]) = A[j] – reverse(A[j])

编程需要懂一点英语

现在，我们的想法是为每个元素arr[i]计算(A[i] – reverse(A[i]))的频率，然后计算满足给定条件的有效对的可能数量。请按照以下步骤解决问题：

维护一个 Hashmap，比如u_map来存储A[i] – reverse(A[i])对于任何索引i的频率计数。
初始化一个变量对来存储满足给定条件的对的数量。
使用变量i遍历给定数组A[]并执行以下操作：
- 将A[i] – reverse(A[i])的频率存储在val 中。
- 按val递增对。
- 更新u_map中val的频率。
完成上述步骤后，打印pairs的值作为结果。

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
// reverse of the number n
int reverse(int n)
{
    int temp = n, rev = 0, r;
 
    // Iterate until temp is 0
    while (temp) {
 
        r = temp % 10;
        rev = rev * 10 + r;
        temp /= 10;
    }
 
    // Return the reversed number
    return rev;
}
 
// Function to count number of unique
// pairs (i, j) from the array A[]
// which satisfies the given condition
void countPairs(int A[], int N)
{
    // Store the frequency of keys
    // as A[i] - reverse(A[i])
    unordered_map u_map;
 
    // Stores count of desired pairs
    int pairs = 0;
 
    // Iterate the array A[]
    for (int i = 0; i < N; i++) {
 
        int val = A[i] - reverse(A[i]);
 
        // Add frequency of val
        // to the required answer
        pairs += u_map[val];
 
        // Increment frequency of val
        u_map[val]++;
    }
 
    // Print the number of pairs
    cout << pairs;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 15, 11, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    countPairs(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find the
// reverse of the number n
static int reverse(int n)
{
    int temp = n, rev = 0, r;
 
    // Iterate until temp is 0
    while (temp > 0)
    {
        r = temp % 10;
        rev = rev * 10 + r;
        temp /= 10;
    }
 
    // Return the reversed number
    return rev;
}
 
// Function to count number of unique
// pairs (i, j) from the array A[]
// which satisfies the given condition
static void countPairs(int A[], int N)
{
     
    // Store the frequency of keys
    // as A[i] - reverse(A[i])
    HashMap map = new HashMap<>();
 
    // Stores count of desired pairs
    int pairs = 0;
 
    // Iterate the array A[]
    for(int i = 0; i < N; i++)
    {
        int val = A[i] - reverse(A[i]);
 
        // Add frequency of val
        // to the required answer
        pairs += map.getOrDefault(val, 0);
 
        // Increment frequency of val
        map.put(val, map.getOrDefault(val, 0) + 1);
    }
 
    // Print the number of pairs
    System.out.println(pairs);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 15, 11, 7 };
    int N = arr.length;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
from collections import defaultdict
 
# Function to find the
# reverse of the number n
def reverse(n):
    temp = n
    rev = 0
 
    # Iterate until temp is 0
    while (temp):
        r = temp % 10
        rev = rev * 10 + r
        temp //= 10
 
    # Return the reversed number
    return rev
 
# Function to count number of unique
# pairs (i, j) from the array A[]
# which satisfies the given condition
def countPairs(A, N):
 
    # Store the frequency of keys
    # as A[i] - reverse(A[i])
    u_map = defaultdict(int)
 
    # Stores count of desired pairs
    pairs = 0
 
    # Iterate the array A[]
    for i in range(N):
        val = A[i] - reverse(A[i])
 
        # Add frequency of val
        # to the required answer
        pairs += u_map[val]
 
        # Increment frequency of val
        u_map[val] += 1
 
    # Print the number of pairs
    print(pairs)
 
 
# Driver Code
if __name__ == "__main__":
 
    arr = [2, 15, 11, 7]
    N = len(arr)
 
    # Function Call
    countPairs(arr, N)
 
    # This code is contributed by chitranayal.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the
// reverse of the number n
static int reverse(int n)
{
    int temp = n, rev = 0, r;
 
    // Iterate until temp is 0
    while (temp > 0)
    {
        r = temp % 10;
        rev = rev * 10 + r;
        temp /= 10;
    }
 
    // Return the reversed number
    return rev;
}
 
// Function to count number of unique
// pairs (i, j) from the array A[]
// which satisfies the given condition
static void countPairs(int []A, int N)
{
     
    // Store the frequency of keys
    // as A[i] - reverse(A[i])
    Dictionary u_map = new Dictionary();
                                            
    // Stores count of desired pairs
    int pairs = 0;
 
    // Iterate the array A[]
    for(int i = 0; i < N; i++)
    {
        int val = A[i] - reverse(A[i]);
 
        // Add frequency of val
        // to the required answer
        if (u_map.ContainsKey(val))
            pairs += u_map[val];
 
        // Increment frequency of val
        if (u_map.ContainsKey(val))
            u_map[val]++;
        else
            u_map.Add(val, 1);
    }
 
    // Print the number of pairs
    Console.Write(pairs);
}
 
// Driver Code
public static void Main()
{
    int []arr = { 2, 15, 11, 7 };
    int N = arr.Length;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

输出：

时间复杂度： O(N*log M)，其中M是数组中最大的元素。
辅助空间： O(1)

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