Find Quotient and Remainder of two integer without using division operators

Given two positive integers dividend and divisor, our task is to find quotient and remainder. The use of division or mod operator is not allowed.

Examples:

Input : dividend = 10, divisor = 3
Output : 3, 1
Explanation:
The quotient when 10 is divided by 3 is 3 and the remainder is 1.

Input : dividend = 11, divisor = 5
Output : 2, 1
Explanation:
The quotient when 11 is divided by 5 is 2 and the remainder is 1.

Approach:



To solve the problem mentioned above we will use the Binary Search technique. We can implement the search method in range 1 to N where N is the dividend. Here we will use multiplication to decide the range. As soon as we break out of the while loop of binary search we get our quotient and the remainder can be found using the multiplication and subtraction operator. Handle the special case, when the dividend is less than or equal to the divisor, without the use of binary search.

Below is the implementation of the above approach:

C++

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// CPP implementation to Find Quotient
// and Remainder of two integer without
// using / and % operator using Binary search
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to the quotient and remainder
pair<int, int> find(int dividend, int divisor,
                    int start, int end)
{
  
    // Check if start is greater than the end
    if (start > end)
        return { 0, dividend };
  
    // Calculate mid
    int mid = start + (end - start) / 2;
  
    int n = dividend - divisor * mid;
  
    // Check if n is greater than divisor
    // then increment the mid by 1
    if (n > divisor)
        start = mid + 1;
  
    // Check if n is less than 0
    // then decrement the mid by 1
    else if (n < 0)
        end = mid - 1;
  
    else {
        // Check if n equals to divisor
        if (n == divisor) {
            ++mid;
            n = 0;
        }
  
        // Return the final answer
        return { mid, n };
    }
  
    // Recursive calls
    return find(dividend, divisor, start, end);
}
  
pair<int, int> divide(int dividend, int divisor)
{
    return find(dividend, divisor, 1, dividend);
}
  
// Driver code
int main(int argc, char* argv[])
{
    int dividend = 10, divisor = 3;
  
    pair<int, int> ans;
  
    ans = divide(dividend, divisor);
  
    cout << ans.first << ", ";
    cout << ans.second << endl;
  
    return 0;
}

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Java

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// JAVA implementation to Find Quotient
// and Remainder of two integer without
// using / and % operator using Binary search
  
class GFG{
   
// Function to the quotient and remainder
static int[] find(int dividend, int divisor,
                    int start, int end)
{
   
    // Check if start is greater than the end
    if (start > end)
        return new int[] { 0, dividend };
   
    // Calculate mid
    int mid = start + (end - start) / 2;
   
    int n = dividend - divisor * mid;
   
    // Check if n is greater than divisor
    // then increment the mid by 1
    if (n > divisor)
        start = mid + 1;
   
    // Check if n is less than 0
    // then decrement the mid by 1
    else if (n < 0)
        end = mid - 1;
   
    else {
        // Check if n equals to divisor
        if (n == divisor) {
            ++mid;
            n = 0;
        }
   
        // Return the final answer
        return new int[] { mid, n };
    }
   
    // Recursive calls
    return find(dividend, divisor, start, end);
}
   
static int[]  divide(int dividend, int divisor)
{
    return find(dividend, divisor, 1, dividend);
}
   
// Driver code
public static void main(String[] args)
{
    int dividend = 10, divisor = 3;
    
    int []ans = divide(dividend, divisor);
   
    System.out.print(ans[0]+ ", ");
    System.out.print(ans[1] +"\n");
   
}
}
  
// This code contributed by sapnasingh4991

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Python3

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# Python3 implementation to Find Quotient 
# and Remainder of two integer without 
# using / and % operator using Binary search 
  
# Function to the quotient and remainder 
def find(dividend, divisor,  start,  end) :
  
    # Check if start is greater than the end 
    if (start > end) :
        return ( 0, dividend ); 
  
    # Calculate mid 
    mid = start + (end - start) // 2
  
    n = dividend - divisor * mid; 
  
    # Check if n is greater than divisor 
    # then increment the mid by 1 
    if (n > divisor) :
        start = mid + 1
  
    # Check if n is less than 0 
    # then decrement the mid by 1 
    elif (n < 0) :
        end = mid - 1
  
    else :
        # Check if n equals to divisor 
        if (n == divisor) : 
            mid += 1
            n = 0
  
        # Return the final answer 
        return ( mid, n ); 
      
    # Recursive calls 
    return find(dividend, divisor, start, end); 
  
def divide(dividend, divisor) : 
  
    return find(dividend, divisor, 1, dividend); 
  
# Driver code 
if __name__ == "__main__" :
  
    dividend = 10; divisor = 3
  
    ans = divide(dividend, divisor); 
  
    print(ans[0],", ",ans[1])
  
# This code is contributed by Yash_R

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C#

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// C# implementation to Find Quotient
// and Remainder of two integer without
// using / and % operator using Binary search
   
using System;
  
public class GFG{
    
// Function to the quotient and remainder
static int[] find(int dividend, int divisor,
                    int start, int end)
{
    
    // Check if start is greater than the end
    if (start > end)
        return new int[] { 0, dividend };
    
    // Calculate mid
    int mid = start + (end - start) / 2;
    
    int n = dividend - divisor * mid;
    
    // Check if n is greater than divisor
    // then increment the mid by 1
    if (n > divisor)
        start = mid + 1;
    
    // Check if n is less than 0
    // then decrement the mid by 1
    else if (n < 0)
        end = mid - 1;
    
    else {
        // Check if n equals to divisor
        if (n == divisor) {
            ++mid;
            n = 0;
        }
    
        // Return the readonly answer
        return new int[] { mid, n };
    }
    
    // Recursive calls
    return find(dividend, divisor, start, end);
}
    
static int[]  divide(int dividend, int divisor)
{
    return find(dividend, divisor, 1, dividend);
}
    
// Driver code
public static void Main(String[] args)
{
    int dividend = 10, divisor = 3;
     
    int []ans = divide(dividend, divisor);
    
    Console.Write(ans[0]+ ", ");
    Console.Write(ans[1] +"\n");
    
}
}
// This code contributed by Princi Singh

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Output:

3, 1

Time Complexity: O(logN)

Space Complexity: O(n)

Similar article: Divide two integers without using multiplication, division and mod operator

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