# Find Quotient and Remainder of two integer without using division operators

• Difficulty Level : Expert
• Last Updated : 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++

 `// C++ implementation to Find Quotient``// and Remainder of two integer without``// using / and % operator using Binary search` `#include ``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;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## Javascript

 ``

Output:

`3, 1`

Time Complexity: O(logN)
Space Complexity: O(n)
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