Divide two integers without using multiplication, division and mod operator

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : Keep subtracting the divisor from 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)

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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