Count distinct pairs from two arrays having same sum of digits

Given two arrays arr1[] and arr2[]. The task is to find the total number of distinct pairs(formed by picking 1 element from arr1 and one element from arr2), such that both the elements of the pair have the sum of digits.

Note: Pairs occurring more than once must be counted only once.

Examples:

Input : arr1[] = {33, 41, 59, 1, 3}
        arr2[] = {3, 32, 51, 3}
Output : 3
Possible pairs are:
(33, 51), (41, 32), (3, 3)

Input : arr1[] = {1, 6, 4, 22}
        arr2[] = {1, 3, 24}
Output : 2
Possible pairs are:
(1, 1), (6, 24)


Approach:

  • Run two nested loops to generate all possible pairs from the two arrays taking one element from arr1[] and one from arr2[].
  • If sum of digits is equal, then insert the pair(a, b) into a set, in order to avoid duplicates where a is the smaller element and b is the larger one.
  • Total pairs will be the size of the final set.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count total number of
// pairs having elements with same
// sum of digits
  
#include <bits/stdc++.h>
using namespace std;
  
// Function for returning
// sum of digits of a number
int digitSum(int n)
{
    int sum = 0;
    while (n > 0) {
        sum += n % 10;
        n = n / 10;
    }
    return sum;
}
  
// Function to return the total pairs
// of elements with equal sum of digits
int totalPairs(int arr1[], int arr2[], int n, int m)
{
  
    // set is used to avoid duplicate pairs
    set<pair<int, int> > s;
  
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
  
            // check sum of digits
            // of both the elements
            if (digitSum(arr1[i]) == digitSum(arr2[j])) {
  
                if (arr1[i] < arr2[j])
                    s.insert(make_pair(arr1[i], arr2[j]));
                else
                    s.insert(make_pair(arr2[j], arr1[i]));
            }
        }
    }
  
    // return size of the set
    return s.size();
}
  
// Driver code
int main()
{
    int arr1[] = { 100, 3, 7, 50 };
    int arr2[] = { 5, 1, 10, 4 };
    int n = sizeof(arr1) / sizeof(arr1[0]);
    int m = sizeof(arr2) / sizeof(arr2[0]);
  
    cout << totalPairs(arr1, arr2, n, m);
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to count total number of 
# pairs having elements with same sum of digits 
  
# Function for returning 
# sum of digits of a number 
def digitSum(n): 
   
    Sum = 0 
    while n > 0:  
        Sum += n % 10 
        n = n // 10 
       
    return Sum 
  
# Function to return the total pairs 
# of elements with equal sum of digits 
def totalPairs(arr1, arr2, n, m): 
  
    # set is used to avoid duplicate pairs 
    s = set() 
  
    for i in range(0, n):  
        for j in range(0, m):  
  
            # check sum of digits 
            # of both the elements 
            if digitSum(arr1[i]) == digitSum(arr2[j]):  
  
                if arr1[i] < arr2[j]: 
                    s.add((arr1[i], arr2[j])) 
                else:
                    s.add((arr2[j], arr1[i])) 
               
    # return size of the set 
    return len(s) 
  
# Driver code 
if __name__ == "__main__":
   
    arr1 = [100, 3, 7, 50]  
    arr2 = [5, 1, 10, 4
    n = len(arr1) 
    m = len(arr2) 
  
    print(totalPairs(arr1, arr2, n, m)) 
      
# This code is contributed by Rituraj Jain

chevron_right


Output:

3


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : rituraj_jain



Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.