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
is less than dividend and keep updating the ith 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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