Given a positive integer ‘l‘ and ‘r‘. Find the smallest number ‘n‘ such that l <= n <= r and count of the number of set bits(number of ‘1’s in binary representation) is as maximum as possible.
Examples :
Input: 1 4
Output: 3
Explanation:
Binary representation from ‘1’ to ‘4’:
110 = 0012
210 = 0102
310 = 0112
110 = 1002
Thus number ‘3’ has maximum set bits = 2Input: 1 10
Output: 7
Simple approach is to traverse from ‘l’ to ‘r’ and count the set bits for each ‘x'(l <= n <= r) and print the number whose count is maximum among them. Time complexity of this approach is O(n*log(r)).
C++
// C++ program to find number whose set// bits are maximum among 'l' and 'r'#include <bits/stdc++.h>using namespace std;// Returns smallest number whose set bits// are maximum in given range.int countMaxSetBits(int left, int right){ // Initialize the maximum count and // final answer as 'num' int max_count = -1, num; for (int i = left; i <= right; ++i) { int temp = i, cnt = 0; // Traverse for every bit of 'i' // number while (temp) { if (temp & 1) ++cnt; temp >>= 1; } // If count is greater than previous // calculated max_count, update it if (cnt > max_count) { max_count = cnt; num = i; } } return num;}// Driver codeint main(){ int l = 1, r = 5; cout << countMaxSetBits(l, r) << "\n"; l = 1, r = 10; cout << countMaxSetBits(l, r); return 0;} |
Java
// Java program to find number whose set// bits are maximum among 'l' and 'r'class gfg{ // Returns smallest number whose set bits// are maximum in given range.static int countMaxSetBits(int left, int right){ // Initialize the maximum count and // final answer as 'num' int max_count = -1, num = 0; for (int i = left; i <= right; ++i) { int temp = i, cnt = 0; // Traverse for every bit of 'i' // number while (temp > 0) { if (temp % 2 == 1) ++cnt; temp >>= 1; } // If count is greater than previous // calculated max_count, update it if (cnt > max_count) { max_count = cnt; num = i; } } return num;}// Driver codepublic static void main(String[] args) { int l = 1, r = 5; System.out.println(countMaxSetBits(l, r)); l = 1; r = 10; System.out.print(countMaxSetBits(l, r));}}// This code has been contributed by 29AjayKumar |
Python3
# Python code to find number whose set# bits are maximum among 'l' and 'r'def countMaxSetBits( left, right): max_count = -1 for i in range(left, right+1): temp = i cnt = 0 # Traverse for every bit of 'i' # number while temp: if temp & 1: cnt +=1 temp = temp >> 1 # If count is greater than previous # calculated max_count, update it if cnt > max_count: max_count = cnt num=i return num# driver code l = 1r = 5print(countMaxSetBits(l, r))l = 1r = 10print(countMaxSetBits(l, r)) # This code is contributed by "Abhishek Sharma 44" |
C#
// C# program to find number whose set// bits are maximum among 'l' and 'r'using System; class gfg{ // Returns smallest number whose set bits// are maximum in given range.static int countMaxSetBits(int left, int right){ // Initialize the maximum count and // final answer as 'num' int max_count = -1, num = 0; for (int i = left; i <= right; ++i) { int temp = i, cnt = 0; // Traverse for every bit of 'i' // number while (temp > 0) { if (temp % 2 == 1) ++cnt; temp >>= 1; } // If count is greater than previous // calculated max_count, update it if (cnt > max_count) { max_count = cnt; num = i; } } return num;}// Driver codepublic static void Main(String[] args) { int l = 1, r = 5; Console.WriteLine(countMaxSetBits(l, r)); l = 1; r = 10; Console.Write(countMaxSetBits(l, r));}}/* This code contributed by PrinciRaj1992 */ |
PHP
<?php// PHP program to find number // whose set bits are maximum// among 'l' and 'r'// Returns smallest number // whose set bits are maximum// in given range.function countMaxSetBits($left, $right){ // Initialize the maximum // count and final answer // as 'num' $max_count = -1; $num; for ($i = $left; $i <= $right; ++$i) { $temp = $i; $cnt = 0; // Traverse for every // bit of 'i' number while ($temp) { if ($temp & 1) ++$cnt; $temp >>= 1; } // If count is greater than // previous calculated // max_count, update it if ($cnt > $max_count) { $max_count = $cnt; $num = $i; } } return $num;}// Driver code$l = 1; $r = 5;echo countMaxSetBits($l, $r), "\n";$l = 1; $r = 10;echo countMaxSetBits($l, $r); // This code is contributed by m_kit?> |
Javascript
<script>// JavaScript program to find number whose set// bits are maximum among 'l' and 'r'// Returns smallest number whose set bits// are maximum in given range.function countMaxSetBits(left, right){ // Initialize the maximum count and // final answer as 'num' let max_count = -1, num = 0; for(let i = left; i <= right; ++i) { let temp = i, cnt = 0; // Traverse for every bit of 'i' // number while (temp > 0) { if (temp % 2 == 1) ++cnt; temp >>= 1; } // If count is greater than previous // calculated max_count, update it if (cnt > max_count) { max_count = cnt; num = i; } } return num;}// Driver Codelet l = 1, r = 5;document.write(countMaxSetBits(l, r) + "<br/>");l = 1; r = 10;document.write(countMaxSetBits(l, r));// This code is contributed by code_hunt</script> |
Output :
3 7
Time Complexity: O(n*log(r))
Auxiliary Space: O(1)
Efficient approach is to use bit-manipulation. Instead of iterating for every number from ‘l’ to ‘r’, iterate only after updating the desired number(‘num’) i.e., take the bitwise ‘OR’ of number with the consecutive number. For instance,
Let l = 2, and r = 10
1. num = 2
2. x = num OR (num + 1)
= 2 | 3 = 010 | 011 = 011
num = 3(011)
3. x = 3 | 4 = 011 | 100 = 111
num = 7(111)
4. x = 7 | 8 = 0111 | 1000 = 1111
Since 15(11112) is greater than
10, thus stop traversing for next number.
5. Final answer = 7
C++
// C++ program to find number whose set// bits are maximum among 'l' and 'r'#include <bits/stdc++.h>using namespace std;// Returns smallest number whose set bits// are maximum in given range.int countMaxSetBits(int left, int right){ while ((left | (left + 1)) <= right) left |= left + 1; return left;}// Driver codeint main(){ int l = 1, r = 5; cout << countMaxSetBits(l, r) << "\n"; l = 1, r = 10; cout << countMaxSetBits(l, r) ; return 0;} |
Java
// Java program to find number // whose set bits are maximum // among 'l' and 'r'import java.io.*;class GFG { // Returns smallest number // whose set bits are // maximum in given range. static int countMaxSetBits(int left, int right) { while ((left | (left + 1)) <= right) left |= left + 1; return left; }// Driver codepublic static void main (String[] args) { int l = 1; int r = 5; System.out.println(countMaxSetBits(l, r)); l = 1; r = 10; System.out.println(countMaxSetBits(l, r));}}// This code is contributed by @ajit |
Python3
# Python code to find number whose set# bits are maximum among 'l' and 'r'def countMaxSetBits( left, right): while(left | (left+1)) <= right: left |= left+1 return left # driver codel = 1r = 5print(countMaxSetBits(l, r))l = 1r = 10print(countMaxSetBits(l, r)) # This code is contributed by "Abhishek Sharma 44" |
C#
// C# program to find number // whose set bits are maximum // among 'l' and 'r'using System;class GFG{ // Returns smallest number // whose set bits are // maximum in given range. static int countMaxSetBits(int left, int right) { while ((left | (left + 1)) <= right) left |= left + 1; return left; }// Driver codestatic public void Main (){ int l = 1; int r = 5; Console.WriteLine(countMaxSetBits(l, r)); l = 1; r = 10; Console.WriteLine(countMaxSetBits(l, r));}}// This code is contributed by @ajit |
PHP
<?php// PHP program to find number // whose set bits are maximum// among 'l' and 'r'// Returns smallest number// whose set bits are// maximum in given range.function countMaxSetBits($left, $right){ while (($left | ($left + 1)) <= $right) $left |= $left + 1; return $left;}// Driver code$l = 1 ; $r = 5;echo countMaxSetBits($l, $r) , "\n";$l = 1; $r = 10;echo countMaxSetBits($l, $r) ;// This code is contributed by aj_36?> |
Javascript
<script>// Javascript program to find number whose set// bits are maximum among 'l' and 'r'// Returns smallest number whose set bits// are maximum in given range.function countMaxSetBits( left, right){ while ((left | (left + 1)) <= right) left |= left + 1; return left;}// driver code let l = 1, r = 5; document.write(countMaxSetBits(l, r) + "</br>"); l = 1, r = 10; document.write(countMaxSetBits(l, r)) ;</script> |
Output :
3 7
Time complexity: O(log(n))
Auxiliary space: O(1)
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