Find the largest number with n set and m unset bits Last Updated : 31 May, 2022 Comments Improve Suggest changes Like Article Like Report 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 countedConstraints: 1 <= n, 0 <= m, (m+n) <= 31Examples : 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 bits unsigned 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 bits unsigned 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 above int 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 bits import 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 bits def 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 bits def 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 print(largeNumWithNSetAndMUnsetBits(n, m)) # This code is contributed # by Anant Agarwal. C# // C# implementation to find the largest number // with n set and m unset bits using 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 bits function 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 ^ $num); } // function to find the largest number // with n set and m unset bits function largeNumWithNSetAndMUnsetBits($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 bits function 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 bits function 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 above var 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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