# Count of pairs in Array with difference equal to the difference with digits reversed

Given an array arr[] of N integers, the task is to find the number of pairs of array elements (arr[i], arr[j]) such that the difference between the pairs is equal to the difference when the digits of both the numbers are reversed.

Examples:

Input: arr[] = {42, 11, 1, 97}
Output: 2
Explanation:
The valid pairs of array elements are (42, 97), (11, 1) as:
1. 42 – 97 = 24 – 79 = (-55)
2. 11 – 1   = 11 – 1 = (10)

Input: arr[] = {1, 2, 3, 4}
Output: 6

Approach: The given problem can be solved by using Hashing which is based on the following observations:

A valid pair (i, j) will follow the equation as

=> arr[i] – arr[j] = rev(arr[i]) – rev(arr[j])
=> arr[i] – rev(arr[i]) = arr[j] – rev(arr[j])

Follow the below steps to solve the problem:

• Now, create a function reverseDigits, which will take an integer as an argument and reverse the digits of that integer.
• Store the frequency of values arr[i] – rev(arr[i]) in an unordered map, say mp.
• For each key(= difference) of frequency X the number of pairs that can be formed is given by .
• The total count of pairs is given by the sum of the value of the above expression for each frequency stored in the map mp.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std; // Function to reverse the digits// of an integerint reverseDigits(int n){    // Convert the given number    // to a string    string s = to_string(n);     // Reverse the string    reverse(s.begin(), s.end());     // Return the value of the string    return stoi(s);}int countValidPairs(vector<int> arr){    // Stores resultant count of pairs    long long res = 0;     // Stores the frequencies of    // differences    unordered_map<int, int> mp;    for (int i = 0; i < arr.size(); i++) {        mp[arr[i] - reverseDigits(arr[i])]++;    }     // Traverse the map and count pairs    // formed for all frequency values    for (auto i : mp) {        long long int t = i.second;        res += t * (t - 1) / 2;    }     // Return the resultant count    return res;} // Driver Codeint main(){    vector<int> arr = { 1, 2, 3, 4 };    cout << countValidPairs(arr);     return 0;}

## Java

 // Java program for the above approachimport java.util.HashMap; class GFG {     // Function to reverse the digits    // of an integer    public static int reverseDigits(int n)     {               // Convert the given number        // to a string        String s = String.valueOf(n);         // Reverse the string        s = new StringBuffer(s).reverse().toString();         // Return the value of the string        return Integer.parseInt(s);    }     public static int countValidPairs(int[] arr)    {               // Stores resultant count of pairs        int res = 0;         // Stores the frequencies of        // differences        HashMap mp = new HashMap();        for (int i = 0; i < arr.length; i++) {            if (mp.containsKey(arr[i] - reverseDigits(arr[i]))) {                mp.put(arr[i] - reverseDigits(arr[i]), mp.get(arr[i] - reverseDigits(arr[i])) + 1);            } else {                mp.put(arr[i] - reverseDigits(arr[i]), 1);            }        }         // Traverse the map and count pairs        // formed for all frequency values        for (int i : mp.keySet()) {            int t = mp.get(i);            res += t * (t - 1) / 2;        }         // Return the resultant count        return res;    }     // Driver Code    public static void main(String args[])     {        int[] arr = { 1, 2, 3, 4 };        System.out.println(countValidPairs(arr));    } } // This code is contributed by saurabh_jaiswal.

## Python3

 # python program for the above approach # Function to reverse the digits# of an integerdef reverseDigits(n):     # Convert the given number    # to a string    s = str(n)     # Reverse the string    s = "".join(reversed(s))     # Return the value of the string    return int(s) def countValidPairs(arr):     # Stores resultant count of pairs    res = 0     # Stores the frequencies of    # differences    mp = {}     for i in range(0, len(arr)):        if not arr[i] - reverseDigits(arr[i]) in mp:            mp[arr[i] - reverseDigits(arr[i])] = 1        else:            mp[arr[i] - reverseDigits(arr[i])] += 1         # Traverse the map and count pairs        # formed for all frequency values    for i in mp:        t = mp[i]        res += (t * (t - 1)) // 2         # Return the resultant count    return res # Driver Codeif __name__ == "__main__":     arr = [1, 2, 3, 4]    print(countValidPairs(arr))     # This code is contributed by rakeshsahni

## C#

 // C# program for the above approachusing System;using System.Collections.Generic; class GFG {     // Function to reverse the digits    // of an integer    public static int reverseDigits(int n)    {         // Convert the given number        // to a string        string s = n.ToString();         // Reverse the string        char[] arr = s.ToCharArray();        Array.Reverse(arr);        string st = new string(arr);         // Return the value of the string        return Int32.Parse(st);    }     public static int countValidPairs(int[] arr)    {         // Stores resultant count of pairs        int res = 0;         // Stores the frequencies of        // differences        Dictionary<int, int> mp            = new Dictionary<int, int>();        for (int i = 0; i < arr.Length; i++) {            if (mp.ContainsKey(arr[i]                               - reverseDigits(arr[i]))) {                mp[arr[i] - reverseDigits(arr[i])]                    = mp[arr[i] - reverseDigits(arr[i])]                      + 1;            }            else {                mp[arr[i] - reverseDigits(arr[i])] = 1;            }        }         // Traverse the map and count pairs        // formed for all frequency values        foreach(int i in mp.Keys)        {            int t = mp[i];            res += t * (t - 1) / 2;        }         // Return the resultant count        return res;    }     // Driver Code    public static void Main()    {        int[] arr = { 1, 2, 3, 4 };        Console.WriteLine(countValidPairs(arr));    }} // This code is contributed by ukasp.

## Javascript

 

Output:
6

Time Complexity: O(N)
Auxiliary Space: O(N)

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