Given an array arr[] of size N. You need to find the total number of good pairs such that the if you chose two numbers arr[i] and arr[j] then the following condition holds arr[i] + SpecialNumber(arr[j]) = = arr[j] + SpecialNumber(arr[i]). Here Special Number indicates the number after removing the digits at the even position of the given number like if there is a number 1234 then the SpecialNumber(1234) will be 13 which comes after removing 2 and 4 from even positions.

Examples:

Input:- N = 3, arr[] = [4, 5, 6]
Output:- 3
Explanation: There is no number whose length is two so numbers remain the same after converting them to special numbers and 4 + SpecialNumber(6) == 6 + SpecialNumber(4), similarly with (4, 5) and with (5, 6).

Input: N = 4, arr[] = [12, 13, 12, 20]
Output: 1
Explanation: only one pair occurs which is (12, 12).

Approach: This can be solved with the following idea:

• As we have to find such pairs which hold the condition arr[i] + SpecialNumber(arr[j]) == arr[j] + SpecialNumber(arr[i]).
• Let’s make this simple by changing the equation to arr[i] – SpecialNumber(arr[i]) == arr[j] – SpecialNumber(arr[j]).
• Now we just have to find out the numbers whose difference with their special number is equal.
• After finding the count of such a number we can simply apply the formula to find the total pairs which is, let’s say that there are n numbers with the same difference of their value and their special number. Then the total pairs we can form by use of them are (n-1)*(n)/2.
• We will find the total pairs in linear time by using this formula.

Below are the steps involved in the implementation of the code:

• For finding the difference between the number and its special number first of all we need to find the special number for every number.
• After this, we will find the difference between each number with its special number and store it in a unordered_map.
• After this, we will just traverse the map and count the pairs which we can form for a particular value. 

Below is the implementation of the code:

// C++ code for the above approac
#include <bits/stdc++.h>
using namespace std;

// Function to find special number
// of a number
int SpecialNumber(int n)
{

    // Coverting number to string
    // for easy use
    string num = to_string(n);

    int ans = 0;

    for (int i = 0; i < num.length(); i++) {

        // Taking only odd position
        // digits in 1-based indexing
        if (i % 2 == 0) {
            ans = ans * 10 + (num[i] - '0');
        }
    }

    // Returning special number
    return ans;
}

// Function to count pairs
int CountPairs(int n, vector<int> arr)
{

    // To store total pairs
    int ans = 0;

    // Map to store difference of
    // number and its special number
    unordered_map<int, int> mm;

    for (int i = 0; i < n; i++) {

        // Taking specialnumber
        int temp = SpecialNumber(arr[i]);

        mm[arr[i] - temp]++;
    }

    // Iterating over map
    for (auto x : mm) {

        // Taking frequency of a
        // particular difference
        int temp = x.second;

        // Counting total pairs of a
        // particular difference
        ans += (temp * (temp - 1)) / 2;
    }

    return ans;
}

// Driver code
int main()
{
    int N = 3;

    vector<int> arr = { 4, 5, 6 };

    // Function call
    cout << CountPairs(N, arr);

    return 0;
}

3

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