Objective: Write Given two integers ‘number’ and ‘divisor’, Write an algorithm to find the remainder if ‘number’ is divided by ‘divisor’.
Condition: You are not allowed to use modulo or % operator.
Example:
num = 10, divisor = 4 remainder = 2 num = 11, divisor = 2 remainder = 1
This is fun puzzle which is asked in the interview.
Approach:
1. This problem will become very trivial if use of modulo or % operator is allowed.
2. Idea is Keep subtracting the divisor from number till number>=divisor.
3. Once the step above is done, remaining value of number will be the remainder.
Example:
number = 10, divisor = 4 number = number – divisor => 10 – 4 = 6 number = number – divisor => 6 – 4 = 2 remainder = 2
Code:
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| public class FindRemainder { | |
| public static void remainder(int n, int divisor){ | |
| if(divisor==0){ | |
| System.out.println("Cannot divide by 0"); | |
| return; | |
| } | |
| //if either number or divisor is negative | |
| if(n<0) | |
| n *=–1; | |
| if(divisor<0) | |
| divisor *= –1; | |
| int number = n; | |
| //subtract divisor from n till n>=divisor | |
| while(n>=divisor){ | |
| n -= divisor; | |
| } | |
| System.out.println("Number: " + number + " , divisor: " + divisor + ". remainder: " + n); | |
| } | |
| public static void main(String[] args) { | |
| int n = 10; | |
| int divisor = 4; | |
| remainder(n, divisor); | |
| } | |
| } |
Output:
Number: 10, divisor: 4. remainder: 2
Continuously subtracting is not such an effective way to go. Instead, we can do it like this:
mul = int(number/divisor);
remainder = number – mul * divisor;