Count total unset bits in all the numbers from 1 to N Last Updated : 12 Nov, 2021 Comments Improve Suggest changes Like Article Like Report Given a positive integer N, the task is to count the total number of unset bits in the binary representation of all the numbers from 1 to N. Note that leading zeroes will not be counted as unset bits.Examples: Input: N = 5 Output: 4 IntegerBinary RepresentationCount of unset bits110210131104100251011 0 + 1 + 0 + 2 + 1 = 4Input: N = 14 Output: 17 Approach: Iterate the loop from 1 to N.While number is greater than 0 divide it by 2 and check the remainder.If remainder is equal to 0 then increase the value of count by 1. Below is the implementation of the above approach: C++ // C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count of unset // bits in the binary representation of // all the numbers from 1 to n int countUnsetBits(int n) { // To store the count of unset bits int cnt = 0; // For every integer from the range [1, n] for (int i = 1; i <= n; i++) { // A copy of the current integer int temp = i; // Count of unset bits in // the current integer while (temp) { // If current bit is unset if (temp % 2 == 0) cnt++; temp = temp / 2; } } return cnt; } // Driver code int main() { int n = 5; cout << countUnsetBits(n); return 0; } Java // Java implementation of the approach class GFG { // Function to return the count of unset // bits in the binary representation of // all the numbers from 1 to n static int countUnsetBits(int n) { // To store the count of unset bits int cnt = 0; // For every integer from the range [1, n] for (int i = 1; i <= n; i++) { // A copy of the current integer int temp = i; // Count of unset bits in // the current integer while (temp > 0) { // If current bit is unset if (temp % 2 == 0) { cnt = cnt + 1; } temp = temp / 2; } } return cnt; } // Driver code public static void main(String[] args) { int n = 5; System.out.println(countUnsetBits(n)); } } // This code is contributed by 29AjayKumar Python3 # Python3 implementation of the approach # Function to return the count of unset # bits in the binary representation of # all the numbers from 1 to n def countUnsetBits(n) : # To store the count of unset bits cnt = 0; # For every integer from the range [1, n] for i in range(1, n + 1) : # A copy of the current integer temp = i; # Count of unset bits in # the current integer while (temp) : # If current bit is unset if (temp % 2 == 0) : cnt += 1; temp = temp // 2; return cnt; # Driver code if __name__ == "__main__" : n = 5; print(countUnsetBits(n)); # This code is contributed by AnkitRai01 C# // C# implementation of the approach using System; class GFG { // Function to return the count of unset // bits in the binary representation of // all the numbers from 1 to n static int countUnsetBits(int n) { // To store the count of unset bits int cnt = 0; // For every integer from the range [1, n] for (int i = 1; i <= n; i++) { // A copy of the current integer int temp = i; // Count of unset bits in // the current integer while (temp > 0) { // If current bit is unset if (temp % 2 == 0) cnt = cnt + 1; temp = temp / 2; } } return cnt; } // Driver code public static void Main() { int n = 5; Console.Write(countUnsetBits(n)); } } // This code is contributed by Sanjit_Prasad JavaScript <script> // Javascript implementation of the approach // Function to return the count of unset // bits in the binary representation of // all the numbers from 1 to n function countUnsetBits(n) { // To store the count of unset bits var cnt = 0; // For every integer from the range [1, n] for (var i = 1; i <= n; i++) { // A copy of the current integer var temp = i; // Count of unset bits in // the current integer while (temp) { // If current bit is unset if (temp % 2 == 0) cnt++; temp = parseInt(temp / 2); } } return cnt; } // Driver code var n = 5; document.write( countUnsetBits(n)); </script> Output: 4 Time Complexity: O(n * log n) Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Types of Asymptotic Notations in Complexity Analysis of Algorithms N Naman_Garg Follow Improve Article Tags : Bit Magic Mathematical DSA Practice Tags : Bit MagicMathematical Similar Reads Basics & PrerequisitesTime Complexity and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. 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