Count total set bits in all numbers from 1 to N | Set 3
Given a positive integer N, the task is to count the total number of set bits in binary representation of all the numbers from 1 to N.
Examples:
Input: N = 3
Output: 4
setBits(1) + setBits(2) + setBits(3) = 1 + 1 + 2 = 4Input: N = 6
Output: 9
Approach: Solution to this problem has been published in the Set 1 and the Set 2 of this article. Here, a dynamic programming based approach is discussed.
- Base case: Number of set bits in 0 and 1 are 0 and 1 respectively.
- Now for every element i from the range [2, N], if i is even then it will have the same number of set bits as i / 2 because to get the number i we just shift the number i / 2 by one. While shifting, the number of set bits does not change.
- Similarly, if i is odd then it will have 1 additional set bit at 0th position than i – 1 which was even.
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 // set bits in all the integers // from the range [1, n] int countSetBits(int n) { // To store the required count // of the set bits int cnt = 0; // To store the count of set // bits in every integer vector<int> setBits(n + 1); // 0 has no set bit setBits[0] = 0; // 1 has a single set bit setBits[1] = 1; // For the rest of the elements for (int i = 2; i <= n; i++) { // If current element i is even then // it has set bits equal to the count // of the set bits in i / 2 if (i % 2 == 0) { setBits[i] = setBits[i / 2]; } // Else it has set bits equal to one // more than the previous element else { setBits[i] = setBits[i - 1] + 1; } } // Sum all the set bits for (int i = 0; i <= n; i++) { cnt = cnt + setBits[i]; } return cnt; } // Driver code int main() { int n = 6; cout << countSetBits(n); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the count of // set bits in all the integers // from the range [1, n] static int countSetBits(int n) { // To store the required count // of the set bits int cnt = 0; // To store the count of set // bits in every integer int []setBits = new int[n + 1]; // 0 has no set bit setBits[0] = 0; // 1 has a single set bit setBits[1] = 1; // For the rest of the elements for (int i = 2; i <= n; i++) { // If current element i is even then // it has set bits equal to the count // of the set bits in i / 2 if (i % 2 == 0) { setBits[i] = setBits[i / 2]; } // Else it has set bits equal to one // more than the previous element else { setBits[i] = setBits[i - 1] + 1; } } // Sum all the set bits for (int i = 0; i <= n; i++) { cnt = cnt + setBits[i]; } return cnt; } // Driver code public static void main(String[] args) { int n = 6; System.out.println(countSetBits(n)); } } // This code is contributed by Princi Singh |
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Python3
# Python3 implementation of the approach # Function to return the count of # set bits in all the integers # from the range [1, n] def countSetBits(n): # To store the required count # of the set bits cnt = 0 # To store the count of set # bits in every integer setBits = [0 for x in range(n + 1)] # 0 has no set bit setBits[0] = 0 # 1 has a single set bit setBits[1] = 1 # For the rest of the elements for i in range(2, n + 1): # If current element i is even then # it has set bits equal to the count # of the set bits in i / 2 if (i % 2 == 0): setBits[i] = setBits[i // 2] # Else it has set bits equal to one # more than the previous element else: setBits[i] = setBits[i - 1] + 1 # Sum all the set bits for i in range(0, n + 1): cnt = cnt + setBits[i] return cnt # Driver code n = 6print(countSetBits(n)) # This code is contributed by Sanjit Prasad |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the count of // set bits in all the integers // from the range [1, n] static int countSetBits(int n) { // To store the required count // of the set bits int cnt = 0; // To store the count of set // bits in every integer int []setBits = new int[n + 1]; // 0 has no set bit setBits[0] = 0; // 1 has a single set bit setBits[1] = 1; // For the rest of the elements for (int i = 2; i <= n; i++) { // If current element i is even then // it has set bits equal to the count // of the set bits in i / 2 if (i % 2 == 0) { setBits[i] = setBits[i / 2]; } // Else it has set bits equal to one // more than the previous element else { setBits[i] = setBits[i - 1] + 1; } } // Sum all the set bits for (int i = 0; i <= n; i++) { cnt = cnt + setBits[i]; } return cnt; } // Driver code static public void Main () { int n = 6; Console.WriteLine(countSetBits(n)); } } // This code is contributed by AnkitRai01 |
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Output:
9
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