Given a positive integer n as a dividend and another number m (a form of 2^k), find the quotient and remainder without performing actual division
Examples:
Input : n = 43, m = 8
Output : Quotient = 5, Remainder = 3Input : n = 58, m = 16
Output : Quotient = 3, Remainder = 10
An approach using bitwise operation
In this, we are using a bitwise representation of a number for understanding the role of division of any number by divisor of form 2^k. All numbers which are power of two include only 1 set bits in their representation and we will use this property.
For finding the remainder we will take logical AND of the dividend (n) and divisor minus 1 (m-1), this will give only the set bits of dividend right to the set bit of divisor which is our actual remainder in that case.
Further, the left part of the dividend (from the position of set bit in divisor) would be considered for quotient. So, from dividend (n) removing all bits right from the position of set bit of divisor will result in the quotient, and right shifting the dividend log2(m) times will do this job for finding the quotient.
- Remainder = n & (m-1)
- Quotient = (n >> log2(m) )
Note: Log2(m) will give the number of bits present in the binary representation of m.
Finding Quotient and Remainder
Below is the implementation:
C++
// CPP to find remainder and quotient#include<bits/stdc++.h>using namespace std;// function to print remainder and quotientvoid divide(int n,int m){ // print Remainder by // n AND (m-1) cout <<"Remainder = " << ((n) &(m-1)); // print quotient by // right shifting n by (log2(m)) times cout <<"\nQuotient = " <<(n >> (int)(log2(m))); }// driver programint main(){ int n = 43, m = 8; divide(n, m); return 0;} |
Java
// Java to find remainder and quotientimport java.io.*;public class GFG { // function to print remainder and // quotient static void divide(int n, int m) { // print Remainder by // n AND (m-1) System.out.println("Remainder = " + ((n) &(m-1))); // print quotient by right shifting // n by (log2(m)) times System.out.println("Quotient = " + (n >> (int)(Math.log(m) / Math.log(2)))); } // driver program static public void main (String[] args) { int n = 43, m = 8; divide(n, m); }}// This code is contributed by vt_m. |
Python 3
# Python 3 to find remainder and # quotientimport math# function to print remainder and # quotientdef divide(n, m): # print Remainder by # n AND (m-1) print("Remainder = ", ((n) &(m-1))) # print quotient by # right shifting n by # (log2(m)) times print("Quotient = " ,(n >> (int)(math.log2(m))))# driver programn = 43m = 8divide(n, m)# This code is contributed by# Smitha |
C#
// C# to find remainder and quotientusing System;public class GFG{ // function to print remainder and quotient static void divide(int n,int m) { // print Remainder by // n AND (m-1) Console.WriteLine("Remainder = " +((n) & (m - 1))); // print quotient by // right shifting n by (log2(m)) times Console.WriteLine("Quotient = " + (n >> (int)(Math.Log(m)))); } // Driver program static public void Main () { int n = 43, m = 8; divide(n, m); }}// This code is contributed by vt_m. |
PHP
<?php// PHP Code to find remainder // and quotient// function to print remainder// and quotientfunction divide($n,$m){ // print Remainder by // n AND (m-1) echo "Remainder = ". (($n) &($m - 1)); // print quotient by // right shifting n by // (log(m,2)) times 2 // is base echo "\nQuotient = ".($n >> (int)(log($m, 2)));} // Driver Code $n = 43; $m = 8; divide($n, $m);//This code is contributed by mits ?> |
Javascript
<script>// javascript program to find last index// of character x in given string. // function to print remainder and // quotient function divide(n, m) { // print Remainder by // n AND (m-1) document.write("Remainder = " + ((n) &(m-1)) + "<br/>"); // print quotient by right shifting // n by (log2(m)) times document.write("Quotient = " + (n >> (Math.log(m) / Math.log(2)))); } // Driver code let n = 43, m = 8; divide(n, m); // This code is contributed by sanjoy_62.</script> |
Remainder = 3 Quotient = 5
Time Complexity: O(1)
Auxiliary Space: O(1)
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