Given a number n, count number of integers smaller than or equal to n that have odd number of set bits.
Examples:
Input : 5 Output : 3 Explanation : Integers with odd number of set bits in range 1 to 5 : 0 contains 0 set bits 1 contains 1 set bits 2 contains 1 set bits 3 contains 2 set bits 4 contains 1 set bits 5 contains 2 set bits Input : 10 Output : 5 Explanation : Integers with odd set bits are 1, 2, 4, 7 and 8.
Prerequisites: Count number of set bits
The idea is based on below fact.
If n is odd then there are total n+1 integers smaller than or equal to n (0, 1, 2 … n) and half of these integers contain odd number of set bits.
How to handle case when n is even? We know result for n-1. We count set bits in n and add 1 to n/2 if the count is odd. Else we return n/2.
C++
// CPP code to find numbers with// odd number of set bits#include <bits/stdc++.h>using namespace std;// function that returns the number// of integers with odd number of// set bitsint countWithOddSetBits(int n){ // If n is odd, then half of the // integers in (0, 1, .. n) contain // odd number of set bits. if (n % 2 != 0) return (n + 1) / 2; // If n is even, we know result for // n-1. We explicitly compute set bit // count in n. int count = __builtin_popcount(n); int ans = n / 2; if (count % 2 != 0) ans++; return ans;}// Driver codeint main(){ int n = 10; cout << countWithOddSetBits(n); return 0;} |
C
// C code to find numbers with// odd number of set bits#include <stdio.h>// function that returns the number// of integers with odd number of// set bitsint countWithOddSetBits(int n){ // If n is odd, then half of the // integers in (0, 1, .. n) contain // odd number of set bits. if (n % 2 != 0) return (n + 1) / 2; // If n is even, we know result for // n-1. We explicitly compute set bit // count in n. int count = __builtin_popcount(n); int ans = n / 2; if (count % 2 != 0) ans++; return ans;}// Driver codeint main(){ int n = 10; printf("%d",countWithOddSetBits(n)); return 0;}// This code is contributed by kothavvsaaash. |
Java
// Java code to find numbers // with odd number of set bitsimport java.io.*;class GFG { // function that returns the// number of integers with // odd number of set bitsstatic int countWithOddSetBits(int n){ // If n is odd, then half // of the integers in // (0, 1, .. n) contain // odd number of set bits. if (n % 2 != 0) return (n + 1) / 2; // If n is even, we know // result for n-1. We // explicitly compute set // bit count in n. int count = (n); int ans = n / 2; if (count % 2 != 0) ans++; return ans;}// Driver Codepublic static void main (String[] args) { int n = 10; System.out.println( countWithOddSetBits(n));}}// This code is contributed by aj_36 |
Python3
# Python 3 code to find numbers with# odd number of set bits# function that returns the number# of integers with odd number of# set bitsdef countWithOddSetBits(n): # If n is odd, then half of the # integers in (0, 1, .. n) contain # odd number of set bits. if (n % 2 != 0): return (n + 1) / 2 # If n is even, we know result for # n-1. We explicitly compute set # bit count in n. count = bin(n).count('1') ans = n / 2 if (count % 2 != 0): ans += 1 return ans# Driver codeif __name__ == '__main__': n = 10 print(int(countWithOddSetBits(n)))# This code is contributed by# Surendra_Gangwar |
C#
// C# code to find numbers // with odd number of set bitsusing System;class GFG{ // function that returns the// number of integers with // odd number of set bitsstatic int countWithOddSetBits(int n){ // If n is odd, then half // of the integers in // (0, 1, .. n) contain // odd number of set bits. if (n % 2 != 0) return (n + 1) / 2; // If n is even, we know // result for n-1. We // explicitly compute set // bit count in n. int count = (n); int ans = n / 2; if (count % 2 != 0) ans++; return ans;}// Driver Codestatic public void Main (){ int n = 10; Console.WriteLine(countWithOddSetBits(n));}}// This code is contributed by ajit |
PHP
<?php// PHP code to find numbers with// odd number of set bits// function that returns the number// of integers with odd number of// set bitsfunction countWithOddSetBits($n){ // If n is odd, then half of // the integers in (0, 1, .. n) // contain odd number of set bits. if ($n % 2 != 0) return ($n + 1) / 2; // If n is even, we know result // for n-1. We explicitly compute // set bit count in n. $count = ($n); $ans = $n / 2; if ($count % 2 != 0) $ans++; return $ans;}// Driver code$n = 10;echo countWithOddSetBits($n);// This code is contributed by aj_36?> |
Javascript
<script> // Javascript code to find numbers // with odd number of set bits // function that returns the // number of integers with // odd number of set bits function countWithOddSetBits(n) { // If n is odd, then half // of the integers in // (0, 1, .. n) contain // odd number of set bits. if (n % 2 != 0) return parseInt((n + 1) / 2, 10); // If n is even, we know // result for n-1. We // explicitly compute set // bit count in n. let count = (n); let ans = parseInt(n / 2, 10); if (count % 2 != 0) ans++; return ans; } let n = 10; document.write(countWithOddSetBits(n)); </script> |
Output :
5
Time complexity: O(k), where k is the max number of bits in a number.
Auxiliary space: O(1)
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