Addition of two numbers without carry
You are given two positive numbers n and m. You have to find a simple addition of both numbers but with a given condition that there is not any carry system in this addition. That is no carry is added at higher MSBs.
Examples :
Input : m = 456, n = 854 Output : 200 Input : m = 456, n = 4 Output : 450
Algorithm :
Input n, m while(n||m) { // Add each bits bit_sum = (n%10) + (m%10); // Neglect carry bit_sum %= 10; // Update result // multiplier to maintain place value res = (bit_sum * multiplier) + res; n /= 10; m /= 10; // Update multiplier multiplier *=10; } print res
Approach :
To solve this problem we will need the bit by bit addition of numbers where we start adding two numbers from rightmost bit (LSB) and add integers from both numbers with the same position. Also, we will neglect carry at each position so that carry will not affect further higher bit position.
Start adding both numbers bit by bit and for each bit take the sum of integers then neglect their carry by taking the modulo of bit_sum by 10 further add bit_sum to res by multiplying bit_sum with a multiplier specifying place value. (Multiplier got incremented 10 times on each iteration.)
Below is the implementation of the above approach :
C++
// CPP program for special // addition of two number #include <bits/stdc++.h> using namespace std; int xSum( int n, int m) { // variable to store result int res = 0; // variable to maintain // place value int multiplier = 1; // variable to maintain // each digit sum int bit_sum; // Add numbers till each // number become zero while (n || m) { // Add each bits bit_sum = (n % 10) + (m % 10); // Neglect carry bit_sum %= 10; // Update result res = (bit_sum * multiplier) + res; n /= 10; m /= 10; // Update multiplier multiplier *= 10; } return res; } // Driver program int main() { int n = 8458; int m = 8732; cout << xSum(n, m); return 0; } |
Java
// Java program for special // addition of two number import java.util.*; import java.lang.*; public class GfG { public static int xSum( int n, int m) { int res = 0 ; int multiplier = 1 ; int bit_sum; // Add numbers till each // number become zero while ( true ) { if (n== 0 && m== 0 ) break ; // Add each bits bit_sum = (n % 10 ) + (m % 10 ); // Neglect carry bit_sum %= 10 ; // Update result res = (bit_sum * multiplier) + res; n /= 10 ; m /= 10 ; // Update multiplier multiplier *= 10 ; } return res; } // Driver function public static void main(String args[]) { int n = 8458 ; int m = 8732 ; System.out.println(xSum(n, m)); } } /* This code is contributed by Sagar Shukla */ |
Python3
# Python3 program for special # addition of two number import math def xSum(n, m) : # variable to # store result res = 0 # variable to maintain # place value multiplier = 1 # variable to maintain # each digit sum bit_sum = 0 # Add numbers till each # number become zero while (n or m) : # Add each bits bit_sum = ((n % 10 ) + (m % 10 )) # Neglect carry bit_sum = bit_sum % 10 # Update result res = (bit_sum * multiplier) + res n = math.floor(n / 10 ) m = math.floor(m / 10 ) # Update multiplier multiplier = multiplier * 10 return res # Driver code n = 8458 m = 8732 print (xSum(n, m)) # This code is contributed by # Manish Shaw(manishshaw1) |
C#
// C# program for special // addition of two number using System; public class GfG { public static int xSum( int n, int m) { int res = 0; int multiplier = 1; int bit_sum; // Add numbers till each // number become zero while ( true ) { // Add each bits bit_sum = (n % 10) + (m % 10); // Neglect carry bit_sum %= 10; // Update result res = (bit_sum * multiplier) + res; n /= 10; m /= 10; // Update multiplier multiplier *= 10; if (n == 0) break ; if (m == 0) break ; } return res; } // Driver function public static void Main() { int n = 8458; int m = 8732; Console.WriteLine(xSum(n, m)); } } /* This code is contributed by Vt_m */ |
PHP
<?php // php program for special // addition of two number function xSum( $n , $m ) { // variable to store result $res = 0; // variable to maintain // place value $multiplier = 1; // variable to maintain // each digit sum $bit_sum ; // Add numbers till each // number become zero while ( $n || $m ) { // Add each bits $bit_sum = ( $n % 10) + ( $m % 10); // Neglect carry $bit_sum %= 10; // Update result $res = ( $bit_sum * $multiplier ) + $res ; $n = floor ( $n / 10); $m = floor ( $m / 10); // Update multiplier $multiplier *= 10; } return $res ; } // Driver code $n = 8458; $m = 8732; echo xSum( $n , $m ); //This code is contributed by mits ?> |
Javascript
<script> // Javascript program for special // addition of two number function xSum(n, m) { // variable to store result var res = 0; // variable to maintain // place value var multiplier = 1; // variable to maintain // each digit sum var bit_sum; // Add numbers till each // number become zero while (n || m) { // Add each bits bit_sum = (n % 10) + (m % 10); // Neglect carry bit_sum %= 10; // Update result res = (bit_sum * multiplier) + res; n = parseInt(n / 10); m = parseInt(m / 10); // Update multiplier multiplier *= 10; } return res; } // Driver program var n = 8458; var m = 8732; document.write(xSum(n, m)); // This code is contributed by noob2000. </script> |
Output :
6180
Time Complexity: O(MAX(len(n),len(m))), where len(X) is the number of digits in a number X.
Space Complexity: O(1)
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