Given two non-negative numbers n and m. The problem is to find the largest number having n number of set bits and m number of unset bits in its binary representation.
Note : 0 bits before leading 1 (or leftmost 1) in binary representation are counted
Constraints: 1 <= n, 0 <= m, (m+n) <= 31
Examples :
Input : n = 2, m = 2 Output : 12 (12)10 = (1100)2 We can see that in the binary representation of 12 there are 2 set and 2 unsets bits and it is the largest number. Input : n = 4, m = 1 Output : 30
Following are the steps:
- Calculate num = (1 << (n + m)) – 1. This will produce a number num having (n + m) number of bits and all are set.
- Now, toggle the last m bits of num and then return the toggled number. Refer this post.
C++
// C++ implementation to find the largest number// with n set and m unset bits#include <bits/stdc++.h>using namespace std;// function to toggle the last m bitsunsigned int toggleLastMBits(unsigned int n, unsigned int m){ // if no bits are required to be toggled if (m == 0) return n; // calculating a number 'num' having 'm' bits // and all are set unsigned int num = (1 << m) - 1; // toggle the last m bits and return the number return (n ^ num);}// function to find the largest number// with n set and m unset bitsunsigned int largeNumWithNSetAndMUnsetBits(unsigned int n, unsigned int m){ // calculating a number 'num' having '(n+m)' bits // and all are set unsigned int num = (1 << (n + m)) - 1; // required largest number return toggleLastMBits(num, m);}// Driver program to test aboveint main(){ unsigned int n = 2, m = 2; cout << largeNumWithNSetAndMUnsetBits(n, m); return 0;} |
Java
// Java implementation to find the largest number// with n set and m unset bitsimport java.io.*;class GFG { // Function to toggle the last m bits static int toggleLastMBits(int n, int m) { // if no bits are required to be toggled if (m == 0) return n; // calculating a number 'num' having 'm' bits // and all are set int num = (1 << m) - 1; // toggle the last m bits and return the number return (n ^ num); } // Function to find the largest number // with n set and m unset bits static int largeNumWithNSetAndMUnsetBits(int n, int m) { // calculating a number 'num' having '(n+m)' bits // and all are set int num = (1 << (n + m)) - 1; // required largest number return toggleLastMBits(num, m); } // driver program public static void main (String[] args) { int n = 2, m = 2; System.out.println(largeNumWithNSetAndMUnsetBits(n, m)); }}// Contributed by Pramod Kumar |
Python3
# Python implementation to# find the largest number# with n set and m unset bits# function to toggle# the last m bitsdef toggleLastMBits(n,m): # if no bits are required # to be toggled if (m == 0): return n # calculating a number # 'num' having 'm' bits # and all are set num = (1 << m) - 1 # toggle the last m bits # and return the number return (n ^ num)# function to find# the largest number# with n set and m unset bitsdef largeNumWithNSetAndMUnsetBits(n,m): # calculating a number # 'num' having '(n+m)' bits # and all are set num = (1 << (n + m)) - 1 # required largest number return toggleLastMBits(num, m)# Driver coden = 2m = 2print(largeNumWithNSetAndMUnsetBits(n, m))# This code is contributed# by Anant Agarwal. |
C#
// C# implementation to find the largest number// with n set and m unset bitsusing System;class GFG{ // Function to toggle the last m bits static int toggleLastMBits(int n, int m) { // if no bits are required to be toggled if (m == 0) return n; // calculating a number 'num' having 'm' bits // and all are set int num = (1 << m) - 1; // toggle the last m bits and return the number return (n ^ num); } // Function to find the largest number // with n set and m unset bits static int largeNumWithNSetAndMUnsetBits(int n, int m) { // calculating a number 'num' having '(n+m)' bits // and all are set int num = (1 << (n + m)) - 1; // required largest number return toggleLastMBits(num, m); } // Driver program public static void Main () { int n = 2, m = 2; Console.Write(largeNumWithNSetAndMUnsetBits(n, m)); }}// This code is contributed by Sam007 |
PHP
<?php// PHP implementation to find // the largest number with n // set and m unset bits// function to toggle // the last m bitsfunction toggleLastMBits($n, $m){ // if no bits are required // to be toggled if ($m == 0) return $n; // calculating a number 'num' // having 'm' bits and all are set $num = (1 << $m) - 1; // toggle the last m bits // and return the number return ($n ^ $num);}// function to find the largest number// with n set and m unset bitsfunction largeNumWithNSetAndMUnsetBits($n, $m){ // calculating a number 'num' // having '(n+m)' bits and all are set $num = (1 << ($n + $m)) - 1; // required largest number return toggleLastMBits($num, $m);}// Driver Code$n = 2; $m = 2;echo largeNumWithNSetAndMUnsetBits($n, $m);// This code is contributed by vt_m.?> |
Javascript
<script>// Javascript implementation to find the largest number// with n set and m unset bits// function to toggle the last m bitsfunction toggleLastMBits(n, m){ // if no bits are required to be toggled if (m == 0) return n; // calculating a number 'num' having 'm' bits // and all are set var num = (1 << m) - 1; // toggle the last m bits and return the number return (n ^ num);}// function to find the largest number// with n set and m unset bitsfunction largeNumWithNSetAndMUnsetBits(n, m){ // calculating a number 'num' having '(n+m)' bits // and all are set num = (1 << (n + m)) - 1; // required largest number return toggleLastMBits(num, m);}// Driver program to test abovevar n = 2, m = 2;document.write( largeNumWithNSetAndMUnsetBits(n, m));</script> |
Output :
12
Time Complexity : O(1)
Auxiliary Space: O(1)
For greater values of n and m, you can use long int and long long int datatypes to generate the required number.
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