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