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++

// 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;
}

Java

// Java program to count total number of
// pairs having elements with same
// sum of digits
import java.util.*;

class GFG
{

static class pair
{
    int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function for returning
// sum of digits of a number
static 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
static int totalPairs(int arr1[], int arr2[],
                      int n, int m)
{

    // set is used to avoid duplicate pairs
    Set<pair> s = new HashSet<>();

    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.add(new pair(arr1[i], arr2[j]));
                else
                    s.add(new pair(arr2[j], arr1[i]));
            }
        }
    }

    // return size of the set
    return s.size();
}

// Driver code
public static void main(String[] args)
{
    int arr1[] = { 100, 3, 7, 50 };
    int arr2[] = { 5, 1, 10, 4 };
    int n = arr1.length;
    int m = arr2.length;

    System.out.println(totalPairs(arr1, arr2, n, m));
}
}

// This code is contributed by Rajput-Ji

Python3

# 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

C#

// C# program to count total number of
// pairs having elements with same
// sum of digits
using System;
using System.Collections.Generic;

class GFG
{

public class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function for returning
// sum of digits of a number
static 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
static int totalPairs(int []arr1, int []arr2,
                      int n, int m)
{

    // set is used to avoid duplicate pairs
    HashSet<pair> s = new HashSet<pair>();

    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.Add(new pair(arr1[i], arr2[j]));
                else
                    s.Add(new pair(arr2[j], arr1[i]));
            }
        }
    }

    // return size of the set
    return s.Count;
}

// Driver code
public static void Main(String[] args)
{
    int []arr1 = { 100, 3, 7, 50 };
    int []arr2 = { 5, 1, 10, 4 };
    int n = arr1.Length;
    int m = arr2.Length;

    Console.WriteLine(totalPairs(arr1, arr2, n, m));
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program to count total number of
// pairs having elements with same
// sum of digits

// Function for returning
// sum of digits of a number
function digitSum(n)
{
    var sum = 0;
    while (n > 0) {
        sum += n % 10;
        n = parseInt(n / 10);
    }
    return sum;
}

// Function to return the total pairs
// of elements with equal sum of digits
function totalPairs(arr1, arr2, n, m)
{

    // set is used to avoid duplicate pairs
    var s = new Set();

    for (var i = 0; i < n; i++) {
        for (var 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.add([arr1[i], arr2[j]]);
                else
                    s.add([arr2[j], arr1[i]]);
            }
        }
    }

    // return size of the set
    return s.size;
}

// Driver code
var arr1 = [100, 3, 7, 50 ];
var arr2 = [5, 1, 10, 4 ];
var n = arr1.length;
var m = arr2.length;
document.write( totalPairs(arr1, arr2, n, m));

</script>

Output

3

Time complexity: O(n*m), where n is the number of elements in the first input array and m is the number of elements in the second input array. This is because the code uses nested loops to iterate through both arrays and check each pair of elements. Within each iteration, the digitSum() function is called which takes constant time.
Auxiliary Space:  O(n), where n is the number of unique pairs that have the same sum of digits. This is because the code uses a set to store unique pairs of elements. The set will contain at most n elements.