Find Quotient and Remainder of two integer without using division operators - GeeksforGeeks

07 Apr, 2021
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++
Java
Python3
C#
Javascript
// C++ 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 remainderpair<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 codeint 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;}
// JAVA implementation to Find Quotient// and Remainder of two integer without// using / and % operator using Binary search class GFG{  // Function to the quotient and remainderstatic 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 codepublic 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
# Python3 implementation to Find Quotient# and Remainder of two integer without# using / and % operator using Binary search # Function to the quotient and remainderdef 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 codeif __name__ == "__main__" :     dividend = 10; divisor = 3;     ans = divide(dividend, divisor);     print(ans[0],", ",ans[1]) # This code is contributed by Yash_R
// 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 remainderstatic 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 codepublic 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
<script> // Javascript implementation to Find Quotient// and Remainder of two integer without// using / and % operator using Binary search // Function to the quotient and remainderfunction find(dividend, divisor, start, end){         // Check if start is greater than the end    if (start > end)        return [0, dividend];     // Calculate mid    var mid = start + parseInt((end - start) / 2);     var 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);} function divide(dividend, divisor){    return find(dividend, divisor, 1, dividend);} // Driver codevar dividend = 10, divisor = 3;var ans = divide(dividend, divisor); document.write(ans[0] + ", ");document.write(ans[1] + "\n"); // This code is contributed by gauravrajput1 </script>
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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