Given two integer m and n, the task is to find the sum of distinct digits of both the numbers and print YES if the both the sums are equal else print NO.
Examples:
Input: m = 2452, n = 9222
Output: YES
The sum of distinct digits of 2452 is 11 (2 + 4 + 5)
And of 9222 is 11 (9 + 2)
Input: m = 121, n = 3035
Output: NO
Approach: Find the sum of unique digits of m and n and store them in sumM and sumN respectively. If sumM = sumN then print YES else print NO.
Below is the implementation of the above approach:
C++
// C++ program to check if the sum of distinct // digits of two integers are equal#include <iostream>using namespace std; // Function to return the sum of // distinct digits of a number int distinctDigitSum(int n) { bool used[10]; int sum = 0; while (n > 0) { // Take last digit int digit = n % 10; // If digit has not been used before if (!used[digit]) { // Set digit as used used[digit] = true; sum += digit; } // Remove last digit n = (int)n / 10; } return sum; } // Function to check whether the sum of // distinct digits of two numbers are equal string checkSum(int m, int n) { int sumM = distinctDigitSum(m); int sumN = distinctDigitSum(n); if (sumM != sumN) return "YES"; return "NO"; } // Driver code int main() { int m = 2452, n = 9222; cout << (checkSum(m, n)); return 0;} |
Java
// Java program to check if the sum of distinct // digits of two integers are equalpublic class HelloWorld { // Function to return the sum of // distinct digits of a number static int distinctDigitSum(int n) { boolean used[] = new boolean[10]; int sum = 0; while (n > 0) { // Take last digit int digit = n % 10; // If digit has not been used before if (!used[digit]) { // Set digit as used used[digit] = true; sum += digit; } // Remove last digit n = n / 10; } return sum; } // Function to check whether the sum of // distinct digits of two numbers are equal static String checkSum(int m, int n) { int sumM = distinctDigitSum(m); int sumN = distinctDigitSum(n); if (sumM == sumN) return "YES"; return "NO"; } // Driver code public static void main(String[] args) { int m = 2452, n = 9222; System.out.println(checkSum(m, n)); }} |
Python3
# Python3 program to check if the sum of # distinct digits of two integers are equal # Function to return the sum of # distinct digits of a number def distinctDigitSum(n) : used = [False] * 10 sum = 0 while (n > 0) : # Take last digit digit = n % 10 # If digit has not been used before if (not used[digit]) : # Set digit as used used[digit] = True sum += digit # Remove last digit n = n // 10 return sum # Function to check whether the sum of # distinct digits of two numbers are equal def checkSum(m, n) : sumM = distinctDigitSum(m) sumN = distinctDigitSum(n) if (sumM == sumN) : return "YES" return "NO" # Driver code if __name__ == "__main__" : m = 2452 n = 9222 print(checkSum(m, n)) # This code is contributed by Ryuga |
C#
// C# program to check if the sum of distinct // digits of two integers are equal// Function to return the sum of// distinct digits of a numberusing System;public class GFG{ static int distinctDigitSum(int n) { bool []used = new bool[10]; int sum = 0; while (n > 0) { // Take last digit int digit = n % 10; // If digit has not been used before if (!used[digit]) { // Set digit as used used[digit] = true; sum += digit; } // Remove last digit n = n / 10; } return sum; } // Function to check whether the sum of // distinct digits of two numbers are equal static String checkSum(int m, int n) { int sumM = distinctDigitSum(m); int sumN = distinctDigitSum(n); if (sumM == sumN) return "YES"; return "NO"; } // Driver code static public void Main (){ int m = 2452, n = 9222; Console.WriteLine(checkSum(m, n)); }//This code is contributed by akt_mit } |
PHP
<?php// PHP program to check if the sum of distinct // digits of two integers are equal// Function to return the sum of// distinct digits of a numberfunction distinctDigitSum($n){ $used[10] = array(); $sum = 0; while ($n > 0) { // Take last digit $digit = $n % 10; // If digit has not been used before if ($used > 0) { // Set digit as used $used[$digit] = true; $sum += $digit; } // Remove last digit $n = (int)$n / 10; } return $sum;}// Function to check whether the sum of// distinct digits of two numbers are equalfunction checkSum($m, $n){ $sumM = distinctDigitSum($m); $sumN = distinctDigitSum($n); if ($sumM != $sumN) return "YES"; return "NO";}// Driver code$m = 2452;$n = 9222;echo (checkSum($m, $n));// This code is contributed by ajit..?> |
Javascript
<script>// javascript program to check if the sum of distinct // digits of two integers are equal // Function to return the sum of // distinct digits of a number function distinctDigitSum(n) { var used = Array(10).fill(false); var sum = 0; while (n > 0) { // Take last digit var digit = n % 10; // If digit has not been used before if (!used[digit]) { // Set digit as used used[digit] = true; sum += digit; } // Remove last digit n = parseInt(n / 10); } return sum; } // Function to check whether the sum of // distinct digits of two numbers are equal function checkSum(m , n) { var sumM = distinctDigitSum(m); var sumN = distinctDigitSum(n); if (sumM == sumN) return "YES"; return "NO"; } // Driver code var m = 2452, n = 9222; document.write(checkSum(m, n));// This code is contributed by todaysgaurav </script> |
Output:
YES
Time Complexity: O(log10m + log10n) since at the end of the loop the number gets divided by 10 so the algorithm takes logarithmic time to perform all operations
Auxiliary Space: O(10) since an array of constant length 10 is used so overall space occupied by the algorithm is constant
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