Divide two integers without using multiplication, division and mod operator

Difficulty Level : Medium
Last Updated : 08 Feb, 2021

Given a two integers say a and b. Find the quotient after dividing a by b without using multiplication, division and mod operator.

Example:

Input : a = 10, b = 3
Output : 3

Input : a = 43, b = -8
Output :  -5

Approach : Keep subtracting the divisor from the dividend until dividend becomes less than divisor. The dividend becomes the remainder, and the number of times subtraction is done becomes the quotient. Below is the implementation of above approach :

C++

// C++ implementation to Divide two
// integers without using multiplication,
// division and mod operator
#include <bits/stdc++.h>
using namespace std;
 
// Function to divide a by b and
// return floor value it
int divide(int dividend, int divisor) {
 
  // Calculate sign of divisor i.e.,
  // sign will be negative only iff
  // either one of them is negative
  // otherwise it will be positive
  int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
 
  // Update both divisor and
  // dividend positive
  dividend = abs(dividend);
  divisor = abs(divisor);
 
  // Initialize the quotient
  int quotient = 0;
  while (dividend >= divisor) {
    dividend -= divisor;
    ++quotient;
  }
 
  return sign * quotient;
}
 
// Driver code
int main() {
  int a = 10, b = 3;
  cout << divide(a, b) << "\n";
 
  a = 43, b = -8;
  cout << divide(a, b);
 
  return 0;
}

Java

/*Java implementation to Divide two
integers without using multiplication,
division and mod operator*/
 
import java.io.*;
 
class GFG 
{
     
    // Function to divide a by b and
    // return floor value it
    static int divide(int dividend, int divisor) 
    {
         
        // Calculate sign of divisor i.e.,
        // sign will be negative only iff
        // either one of them is negative
        // otherwise it will be positive
        int sign = ((dividend < 0) ^ 
                   (divisor < 0)) ? -1 : 1;
     
        // Update both divisor and
        // dividend positive
        dividend = Math.abs(dividend);
        divisor = Math.abs(divisor);
     
        // Initialize the quotient
        int quotient = 0;
         
        while (dividend >= divisor)
        {
            dividend -= divisor;
            ++quotient;
        }
     
        return sign * quotient;
    }    
     
    public static void main (String[] args) 
    {
        int a = 10;
        int b = 3;
         
        System.out.println(divide(a, b));
         
        a = 43;
        b = -8;
         
        System.out.println(divide(a, b));
    }
}
 
// This code is contributed by upendra singh bartwal.

Python3

# Python 3 implementation to Divide two
# integers without using multiplication,
# division and mod operator
 
# Function to divide a by b and
# return floor value it
def divide(dividend, divisor): 
 
    # Calculate sign of divisor i.e.,
    # sign will be negative only iff
    # either one of them is negative
    # otherwise it will be positive
    sign = -1 if ((dividend < 0) ^  (divisor < 0)) else 1
     
    # Update both divisor and
    # dividend positive
    dividend = abs(dividend)
    divisor = abs(divisor)
     
    # Initialize the quotient
    quotient = 0
    while (dividend >= divisor): 
        dividend -= divisor
        quotient += 1
     
     
    return sign * quotient
 
 
# Driver code
a = 10
b = 3
print(divide(a, b))
a = 43
b = -8
print(divide(a, b))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# implementation to Divide two without 
// using multiplication, division and mod
// operator
using System;
 
class GFG {
     
    // Function to divide a by b and
    // return floor value it
    static int divide(int dividend, int divisor) 
    {
         
        // Calculate sign of divisor i.e.,
        // sign will be negative only iff
        // either one of them is negative
        // otherwise it will be positive
        int sign = ((dividend < 0) ^ 
                (divisor < 0)) ? -1 : 1;
     
        // Update both divisor and
        // dividend positive
        dividend = Math.Abs(dividend);
        divisor = Math.Abs(divisor);
     
        // Initialize the quotient
        int quotient = 0;
         
        while (dividend >= divisor)
        {
            dividend -= divisor;
            ++quotient;
        }
     
        return sign * quotient;
    } 
     
    public static void Main () 
    {
         
        int a = 10;
        int b = 3;
        Console.WriteLine(divide(a, b));
         
        a = 43;
        b = -8;
        Console.WriteLine(divide(a, b));
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// PHP implementation to Divide two
// integers without using multiplication,
// division and mod operator
 
// Function to divide a by b and
// return floor value it
function divide($dividend, $divisor) {
 
    // Calculate sign of divisor i.e.,
    // sign will be negative only iff
    // either one of them is negative
    // otherwise it will be positive
    $sign = (($dividend < 0) ^ 
             ($divisor < 0)) ? -1 : 1;
     
    // Update both divisor and
    // dividend positive
    $dividend = abs($dividend);
    $divisor = abs($divisor);
     
    // Initialize the quotient
    $quotient = 0;
    while ($dividend >= $divisor)
    {
        $dividend -= $divisor;
        ++$quotient;
    }
     
    return $sign * $quotient;
}
 
// Driver code
$a = 10;
$b = 3;
echo divide($a, $b)."\n";
 
$a = 43;
$b = -8;
echo divide($a, $b);
 
// This code is contributed by Sam007
?>

Output :

3
-5

Time complexity : O(a)
Auxiliary space : O(1)
Efficient Approach : Use bit manipulation in order to find the quotient. The divisor and dividend can be written as

dividend = quotient * divisor + remainder

As every number can be represented in base 2(0 or 1), represent the quotient in binary form by using shift operator as given below :

Determine the most significant bit in the quotient. This can easily be calculated by iterating on the bit position i from 31 to 1.
Find the first bit for which $divisor << i$ is less than dividend and keep updating the i^th bit position for which it is true.
Add the result in temp variable for checking the next position such that (temp + (divisor << i) ) is less than dividend.
Return the final answer of quotient after updating with corresponding sign.

Below is the implementation of above approach :

C++

// C++ implementation to Divide two
// integers without using multiplication,
// division and mod operator
#include <bits/stdc++.h>
using namespace std;
 
// Function to divide a by b and
// return floor value it
int divide(long long dividend, long long divisor) {
 
  // Calculate sign of divisor i.e.,
  // sign will be negative only iff
  // either one of them is negative
  // otherwise it will be positive
  int sign = ((dividend < 0) ^ 
              (divisor < 0)) ? -1 : 1;
 
  // remove sign of operands
  dividend = abs(dividend);
  divisor = abs(divisor);
 
  // Initialize the quotient
  long long quotient = 0, temp = 0;
 
  // test down from the highest bit and
  // accumulate the tentative value for
  // valid bit
  for (int i = 31; i >= 0; --i) {
 
    if (temp + (divisor << i) <= dividend) {
      temp += divisor << i;
      quotient |= 1LL << i;
    }
  }
 
  return sign * quotient;
}
 
// Driver code
int main() {
  int a = 10, b = 3;
  cout << divide(a, b) << "\n";
 
  a = 43, b = -8;
  cout << divide(a, b);
 
  return 0;
}

Java

// Java implementation to Divide 
// two integers without using 
// multiplication, division 
// and mod operator 
import java.io.*; 
import java.util.*; 
 
// Function to divide a by b 
// and return floor value it 
class GFG 
{ 
public static long divide(long dividend, 
                        long divisor) 
{ 
 
// Calculate sign of divisor 
// i.e., sign will be negative 
// only iff either one of them 
// is negative otherwise it 
// will be positive 
long sign = ((dividend < 0) ^ 
            (divisor < 0)) ? -1 : 1; 
 
// remove sign of operands 
dividend = Math.abs(dividend); 
divisor = Math.abs(divisor); 
 
// Initialize the quotient 
long quotient = 0, temp = 0; 
 
// test down from the highest 
// bit and accumulate the 
// tentative value for 
// valid bit 
// 1<<31 behaves incorrectly and gives Integer
// Min Value which should not be the case, instead 
  // 1L<<31 works correctly. 
for (int i = 31; i >= 0; --i) 
{ 
 
    if (temp + (divisor << i) <= dividend) 
    { 
        temp += divisor << i; 
        quotient |= 1L << i; 
    } 
} 
 
return (sign * quotient); 
} 
 
// Driver code 
public static void main(String args[]) 
{ 
int a = 10, b = 3; 
System.out.println(divide(a, b)); 
 
int a1 = 43, b1 = -8; 
System.out.println(divide(a1, b1)); 
 
 
} 
} 
 
// This code is contributed 
// by Akanksha Rai(Abby_akku) 

Python3

# Python3 implementation to 
# Divide two integers 
# without using multiplication,
# division and mod operator
 
# Function to divide a by 
# b and return floor value it
def divide(dividend, divisor):
     
    # Calculate sign of divisor 
    # i.e., sign will be negative
    # either one of them is negative
    # only iff otherwise it will be
    # positive
     
    sign = (-1 if((dividend < 0) ^ 
                  (divisor < 0)) else 1);
     
    # remove sign of operands
    dividend = abs(dividend);
    divisor = abs(divisor);
     
    # Initialize
    # the quotient
    quotient = 0;
    temp = 0;
     
    # test down from the highest 
    # bit and accumulate the 
    # tentative value for valid bit
    for i in range(31, -1, -1):
        if (temp + (divisor << i) <= dividend):
            temp += divisor << i;
            quotient |= 1 << i;
     
    return sign * quotient;
 
# Driver code
a = 10;
b = 3;
print(divide(a, b));
 
a = 43;
b = -8;
print(divide(a, b));
 
# This code is contributed by mits

C#

// C# implementation to Divide
// two integers without using 
// multiplication, division 
// and mod operator
using System;
 
// Function to divide a by b
// and return floor value it
class GFG
{
public static long divide(long dividend, 
                          long divisor) 
{
 
// Calculate sign of divisor 
// i.e., sign will be negative 
// only iff either one of them 
// is negative otherwise it 
// will be positive
long sign = ((dividend < 0) ^ 
             (divisor < 0)) ? -1 : 1;
 
// remove sign of operands
dividend = Math.Abs(dividend);
divisor = Math.Abs(divisor);
 
// Initialize the quotient
long quotient = 0, temp = 0;
 
// test down from the highest 
// bit and accumulate the 
// tentative value for
// valid bit
for (int i = 31; i >= 0; --i) 
{
 
    if (temp + (divisor << i) <= dividend) 
    {
        temp += divisor << i;
        quotient |= 1LL << i;
    }
}
 
return (sign * quotient);
}
 
// Driver code
public static void Main() 
{
int a = 10, b = 3;
Console.WriteLine(divide(a, b));
 
int a1 = 43, b1 = -8;
Console.WriteLine(divide(a1, b1));
 
}
}
 
// This code is contributed by mits

PHP

<?php
// PHP implementation to 
// Divide two integers 
// without using multiplication,
// division and mod operator
 
// Function to divide a by 
// b and return floor value it
function divide($dividend, 
                $divisor) 
{
 
// Calculate sign of divisor 
// i.e., sign will be negative
// either one of them is negative
// only iff otherwise it will be
// positive
$sign = (($dividend < 0) ^ 
         ($divisor < 0)) ? -1 : 1;
 
// remove sign of operands
$dividend = abs($dividend);
$divisor = abs($divisor);
 
// Initialize
// the quotient
$quotient = 0;
$temp = 0;
 
// test down from the highest 
// bit and accumulate the 
// tentative value for valid bit
for ($i = 31; $i >= 0; --$i) 
{
 
    if ($temp + ($divisor << $i) <= $dividend) 
    {
        $temp += $divisor << $i;
        $quotient |= (double)(1) << $i;
    }
}
 
return $sign * $quotient;
}
 
// Driver code
$a = 10;
$b = 3;
echo divide($a, $b). "\n";
 
$a = 43;
$b = -8;
echo divide($a, $b);
 
// This code is contributed by mits
?>

Output :

3
-5

Time complexity : O(log(a))
Auxiliary space : O(1)

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