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.
Efficient Approach:
- Define a function find that takes four parameters – dividend, divisor, start, and end, and returns a pair of integers – quotient and remainder.
- Check if the start is greater than the end. If yes, return {0, dividend}, where 0 is the quotient and dividend is the remainder.
- Calculate the mid value as (start + end) / 2.
- Subtract the product of divisor and mid from the dividend and store it in a variable n.
- Check if n is greater than divisor. If yes, increment the mid by 1 and update the start to mid+1.
- Check if n is less than 0. If yes, decrement the mid by 1 and update the end to mid-1.
- If n is equal to the divisor, increment the mid by 1 and set n to 0.
- Return the pair {mid, n} as the quotient and remainder.
- Recursively call the find function with updated values of start and end until start becomes greater than end.
- Define a function divide that takes two parameters – dividend and divisor, and returns the quotient and remainder using the find function with start = 1 and end = dividend.
- In the main function, call the divide function with the given values of dividend and divisor, and store the returned pair in a variable ans.
- Print the quotient and remainder by accessing the first and second elements of ans respectively.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
pair<int, int> find(int dividend, int divisor,
int start, int end)
{
if (start > end)
return { 0, dividend };
int mid = start + (end - start) / 2;
int n = dividend - divisor * mid;
if (n > divisor)
start = mid + 1;
else if (n < 0)
end = mid - 1;
else {
if (n == divisor) {
++mid;
n = 0;
}
return { mid, n };
}
return find(dividend, divisor, start, end);
}
pair<int, int> divide(int dividend, int divisor)
{
return find(dividend, divisor, 1, dividend);
}
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
class GFG{
static int[] find(int dividend, int divisor,
int start, int end)
{
if (start > end)
return new int[] { 0, dividend };
int mid = start + (end - start) / 2;
int n = dividend - divisor * mid;
if (n > divisor)
start = mid + 1;
else if (n < 0)
end = mid - 1;
else {
if (n == divisor) {
++mid;
n = 0;
}
return new int[] { mid, n };
}
return find(dividend, divisor, start, end);
}
static int[] divide(int dividend, int divisor)
{
return find(dividend, divisor, 1, dividend);
}
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");
}
}
|
Python3
def find(dividend, divisor, start, end) :
if (start > end) :
return ( 0, dividend );
mid = start + (end - start) // 2;
n = dividend - divisor * mid;
if (n > divisor) :
start = mid + 1;
elif (n < 0) :
end = mid - 1;
else :
if (n == divisor) :
mid += 1;
n = 0;
return ( mid, n );
return find(dividend, divisor, start, end);
def divide(dividend, divisor) :
return find(dividend, divisor, 1, dividend);
if __name__ == "__main__" :
dividend = 10; divisor = 3;
ans = divide(dividend, divisor);
print(ans[0],", ",ans[1])
|
C#
using System;
public class GFG{
static int[] find(int dividend, int divisor,
int start, int end)
{
if (start > end)
return new int[] { 0, dividend };
int mid = start + (end - start) / 2;
int n = dividend - divisor * mid;
if (n > divisor)
start = mid + 1;
else if (n < 0)
end = mid - 1;
else {
if (n == divisor) {
++mid;
n = 0;
}
return new int[] { mid, n };
}
return find(dividend, divisor, start, end);
}
static int[] divide(int dividend, int divisor)
{
return find(dividend, divisor, 1, dividend);
}
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");
}
}
|
Javascript
<script>
function find(dividend, divisor, start, end)
{
if (start > end)
return [0, dividend];
var mid = start + parseInt((end - start) / 2);
var n = dividend - divisor * mid;
if (n > divisor)
start = mid + 1;
else if (n < 0)
end = mid - 1;
else
{
if (n == divisor)
{
++mid;
n = 0;
}
return [ mid, n];
}
return find(dividend, divisor, start, end);
}
function divide(dividend, divisor)
{
return find(dividend, divisor, 1, dividend);
}
var dividend = 10, divisor = 3;
var ans = divide(dividend, divisor);
document.write(ans[0] + ", ");
document.write(ans[1] + "\n");
</script>
|
Time Complexity: O(logN)
Space Complexity: O(n)
Approach: Brute force approach based on repeated subtraction
One approach to finding the quotient and remainder of two integers without using division or mod operators is by repeated subtraction. The basic idea is to subtract the divisor from the dividend until the dividend becomes less than the divisor. The number of times the subtraction is performed is the quotient, and the final value of the dividend is the remainder.
Here are the steps for this approach:
- Initialize the quotient to 0 and the remainder to the value of the dividend.
- While the remainder is greater than or equal to the divisor, subtract the divisor from the remainder and increment the quotient by 1.
- Return the quotient and remainder.
C++
#include <iostream>
using namespace std;
void findQuotientAndRemainder(int dividend, int divisor, int& quotient, int& remainder) {
quotient = 0;
remainder = dividend;
while (remainder >= divisor) {
remainder -= divisor;
quotient += 1;
}
}
int main() {
int dividend = 10;
int divisor = 3;
int quotient, remainder;
findQuotientAndRemainder(dividend, divisor, quotient, remainder);
cout << "Quotient: " << quotient << endl;
cout << "Remainder: " << remainder << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
static void findQuotientAndRemainder(int dividend,
int divisor,
int[] qr)
{
qr[0] = 0;
qr[1] = dividend;
while (qr[1] >= divisor) {
qr[1] -= divisor;
qr[0] += 1;
}
}
public static void main(String[] args)
{
int dividend = 10;
int divisor = 3;
int[] qr = new int[2];
findQuotientAndRemainder(dividend, divisor, qr);
System.out.println("Quotient: " + qr[0]);
System.out.println("Remainder: " + qr[1]);
}
}
|
Python3
def find_quotient_and_remainder(dividend, divisor):
quotient = 0
remainder = dividend
while remainder >= divisor:
remainder -= divisor
quotient += 1
return quotient, remainder
dividend = 10
divisor = 3
quotient, remainder = find_quotient_and_remainder(dividend, divisor)
print("Quotient:", quotient)
print("Remainder:", remainder)
|
C#
using System;
class Program {
static void FindQuotientAndRemainder(int dividend, int divisor, out int quotient, out int remainder) {
quotient = 0;
remainder = dividend;
while (remainder >= divisor) {
remainder -= divisor;
quotient += 1;
}
}
static void Main(string[] args) {
int dividend = 10;
int divisor = 3;
int quotient, remainder;
FindQuotientAndRemainder(dividend, divisor, out quotient, out remainder);
Console.WriteLine("Quotient: {0}", quotient);
Console.WriteLine("Remainder: {0}", remainder);
Console.ReadKey();
}
}
|
Javascript
<script>
function findQuotientAndRemainder(dividend, divisor) {
let quotient = 0;
let remainder = dividend;
while (remainder >= divisor) {
remainder -= divisor;
quotient += 1;
}
return { quotient, remainder };
}
const dividend = 10;
const divisor = 3;
const { quotient, remainder } = findQuotientAndRemainder(dividend, divisor);
document.write("Quotient: " + quotient + "<br>");
document.write("Remainder: " + remainder);
</script>
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OutputQuotient: 3
Remainder: 1
Time Complexity: O(dividend/divisor)
Auxiliary Space: O(1)