Find the Quotient and Remainder

Easy5 min

Topics

BasicsVariablesMath

Companies

TCSInfosysWipro

Given a non-negative dividend and a positive divisor, return [quotient, remainder]: how many whole times the divisor fits into the dividend, and what is left over. Division with a remainder is the long division you learnt at school, and it underlies many later problems (digit extraction, even/odd checks, clock arithmetic). The quotient and remainder always satisfy dividend = quotient × divisor + remainder. The first solution literally removes groups one at a time; the second lets the / and % operators do it instantly.

Test Case 1:

Input:dividend = 17, divisor = 5

Output:[3, 2]

Explanation:Three groups of 5 make 15, and 2 is left over.

Test Case 2:

Input:dividend = 4, divisor = 9

Output:[0, 4]

Explanation:The divisor is bigger, so no whole group fits and everything is left over.

Constraints

◆0 ≤ dividend ≤ 1000000
◆1 ≤ divisor ≤ 1000000

Try the Dry Run

Mode

Approach

Approach & Solutions

Repeated SubtractionBrute

Division is repeated subtraction. Keep taking divisor away from what remains, counting each time, until what remains is smaller than divisor. The count is the quotient and what remains is the remainder. It is slow for large quotients, but it shows exactly what / and % mean.

TimeO(dividend / divisor)

SpaceO(1)

Java

Try it out

class Solution {
    public int[] quotientAndRemainder(int dividend, int divisor) {
        int quotient = 0;
        int remaining = dividend;
        while (remaining >= divisor) {
            remaining -= divisor;
            quotient++;
        }
        return new int[] {quotient, remaining};
    }
}

Division and Modulo OperatorsOptimal

The / operator on two ints gives the quotient directly and the % operator gives the remainder, each in one step no matter how big the numbers are. Python writes the whole-number division as //, and JavaScript uses Math.floor(dividend / divisor).

TimeO(1)