Given two numbers, return a sum of them without using operators + and/or -, and using ++ and/or –.
Examples:
Input: x = 10, y = 5 Output: 15 Input: x = 10, y = -5 Output: 10
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The idea is to do y times x++, if y is positive, and do y times x– if y is negative.
C++
// C++ program to add two numbers using ++#include <bits/stdc++.h>using namespace std;// Returns value of x+y without using +int add(int x, int y){ // If y is positive, y times add 1 to x while (y > 0 && y--) x++; // If y is negative, y times subtract 1 from x while (y < 0 && y++) x--; return x;}int main(){ cout << add(43, 23) << endl; cout << add(43, -23) << endl; return 0;} |
Java
// java program to add two numbers// using ++public class GFG { // Returns value of x+y without // using + static int add(int x, int y) { // If y is positive, y times // add 1 to x while (y > 0 && y != 0) { x++; y--; } // If y is negative, y times // subtract 1 from x while (y < 0 && y != 0) { x--; y++; } return x; } // Driver code public static void main(String args[]) { System.out.println(add(43, 23)); System.out.println(add(43, -23)); }}// This code is contributed by Sam007. |
Python3
# python program to add two # numbers using ++# Returns value of x + y # without using + def add(x, y): # If y is positive, y # times add 1 to x while (y > 0 and y): x = x + 1 y = y - 1 # If y is negative, y # times subtract 1 # from x while (y < 0 and y) : x = x - 1 y = y + 1 return x# Driver codeprint(add(43, 23))print(add(43, -23))# This code is contributed # by Sam007. |
C#
// C# program to add two numbers// using ++using System;public class GFG { // Returns value of x+y without // using + static int add(int x, int y) { // If y is positive, y times // add 1 to x while (y > 0 && y != 0) { x++; y--; } // If y is negative, y times // subtract 1 from x while (y < 0 && y != 0) { x--; y++; } return x; } // Driver code public static void Main() { Console.WriteLine(add(43, 23)); Console.WriteLine(add(43, -23)); }}// This code is contributed by Sam007. |
PHP
<?php// PHP program to add two// numbers using ++// Returns value of// x+y without using +function add($x, $y) { // If y is positive, // y times add 1 to x while ($y > 0 && $y--) $x++; // If y is negative, // y times subtract // 1 from x while ($y < 0 && $y++) $x--; return $x;} // Driver Code echo add(43, 23), "\n"; echo add(43, -23), "\n";// This code is contributed by ajit.?> |
Javascript
<script> // Javascript program to // add two numbers using ++ // Returns value of x+y without // using + function add(x, y) { // If y is positive, y times // add 1 to x while (y > 0 && y != 0) { x++; y--; } // If y is negative, y times // subtract 1 from x while (y < 0 && y != 0) { x--; y++; } return x; } document.write(add(43, 23) + "</br>"); document.write(add(43, -23) + "</br>"); </script> |
Output:
66 20
Time Complexity: O(y)
Auxiliary Space: O(1)
Thanks to Gaurav Ahirwar for suggesting the above solution.
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