Calculate the number of set bits for every number from 0 to N

Given a non-negative integer N, the task is to find the count of set bits for every number from 0 to N.

Examples:

Input: N = 3
Output: 0 1 1 2
0, 1, 2 and 3 can be written in binary as 0, 1, 10 and 11.
The number of 1’s in their binary representation are 0, 1, 1 and 2.



Input: N = 5
Output: 0 1 1 2 1 2

Naive approach: Run a loop from 0 to N and using inbuilt bit count function __builtin_popcount(), find the number of set bits in all the required integers.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the count
// of set bits in all the
// integers from 0 to n
void findSetBits(int n)
{
    for (int i = 0; i <= n; i++)
        cout << __builtin_popcount(i) << " ";
}
  
// Driver code
int main()
{
    int n = 5;
  
    findSetBits(n);
  
    return 0;
}
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// Java implementation of the approach
class GFG 
{
  
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
    for (int i = 0; i <= n; i++)
        System.out.print(Integer.bitCount(i) + " ");
}
  
// Driver code
public static void main(String[] args) 
{
    int n = 5;
  
    findSetBits(n);
}
}
  
// This code is contributed by Rajput-Ji
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# Python 3 implementation of the approach
def count(n):
    count = 0
    while (n): 
        count += n & 1
        n >>= 1
    return count
  
# Function to find the count
# of set bits in all the
# integers from 0 to n
def findSetBits(n):
    for i in range(n + 1):
        print(count(i), end = " ")
      
# Driver code
if __name__ == '__main__':
    n = 5
  
    findSetBits(n)
  
# This code is contributed by Surendra_Gangwar
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Output:
0 1 1 2 1 2

Efficient approach: Let us write the binary representation of numbers in the range (0, 6).

0 in binary – 000
1 in binary – 001
2 in binary – 010
3 in binary – 011
4 in binary – 100
5 in binary – 101
6 in binary – 110

Since, any even number can be written as (2 * i) and any odd number can be written as (2 * i + 1) where i is a natural number.
2, 4 and 3, 6 have equal number of 1’s in their binary representation as multiplying any number is equivalent to shifting it left by 1 (read here).
Similarly, any even number 2 * i and i will have equal number of 1’s in its binary representation.
Number of 1’s in 5(101) is equal to number of 1’s in 2’s binary representation + 1. So in case of any odd number (2 * i + 1), it will be (number of 1’s in the binary representation of i) + 1.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the count
// of set bits in all the
// integers from 0 to n
void findSetBits(int n)
{
  
    // dp[i] will store the count
    // of set bits in i
    int dp[n + 1];
  
    // Initialise the dp array
    memset(dp, 0, sizeof(dp));
  
    // Count of set bits in 0 is 0
    cout << dp[0] << " ";
  
    // For every number starting from 1
    for (int i = 1; i <= n; i++) {
  
        // If current number is even
        if (i % 2 == 0) {
  
            // Count of set bits in i is equal to
            // the count of set bits in (i / 2)
            dp[i] = dp[i / 2];
        }
  
        // If current element is odd
        else {
  
            // Count of set bits in i is equal to
            // the count of set bits in (i / 2) + 1
            dp[i] = dp[i / 2] + 1;
        }
  
        // Print the count of set bits in i
        cout << dp[i] << " ";
    }
}
  
// Driver code
int main()
{
    int n = 5;
  
    findSetBits(n);
  
    return 0;
}
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// Java implementation of the approach
class GFG
{
  
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
  
    // dp[i] will store the count
    // of set bits in i
    int []dp = new int[n + 1];
  
    // Count of set bits in 0 is 0
    System.out.print(dp[0] + " ");
  
    // For every number starting from 1
    for (int i = 1; i <= n; i++) 
    {
  
        // If current number is even
        if (i % 2 == 0
        {
  
            // Count of set bits in i is equal to
            // the count of set bits in (i / 2)
            dp[i] = dp[i / 2];
        }
  
        // If current element is odd
        else
        {
  
            // Count of set bits in i is equal to
            // the count of set bits in (i / 2) + 1
            dp[i] = dp[i / 2] + 1;
        }
  
        // Print the count of set bits in i
        System.out.print(dp[i] + " ");
    }
}
  
// Driver code
public static void main(String []args)
{
    int n = 5;
  
    findSetBits(n);
}
}
  
// This code is contributed by Rajput-Ji
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# Python3 implementation of the approach 
  
# Function to find the count of set bits 
# in all the integers from 0 to n 
def findSetBits(n) :
  
    # dp[i] will store the count 
    # of set bits in i 
    # Initialise the dp array
    dp = [0] * (n + 1); 
      
    # Count of set bits in 0 is 0 
    print(dp[0], end = " "); 
  
    # For every number starting from 1 
    for i in range(1, n + 1) :
  
        # If current number is even 
        if (i % 2 == 0) :
  
            # Count of set bits in i is equal to 
            # the count of set bits in (i / 2) 
            dp[i] = dp[i // 2]; 
  
        # If current element is odd 
        else :
  
            # Count of set bits in i is equal to 
            # the count of set bits in (i / 2) + 1 
            dp[i] = dp[i // 2] + 1
  
        # Print the count of set bits in i 
        print(dp[i], end = " "); 
  
# Driver code 
if __name__ == "__main__"
  
    n = 5
  
    findSetBits(n); 
  
# This code is contributed by AnkitRai01
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// C# implementation of the approach
using System;
      
class GFG
{
  
// Function to find the count
// of set bits in all the
// integers from 0 to n
static void findSetBits(int n)
{
  
    // dp[i] will store the count
    // of set bits in i
    int []dp = new int[n + 1];
  
    // Count of set bits in 0 is 0
    Console.Write(dp[0] + " ");
  
    // For every number starting from 1
    for (int i = 1; i <= n; i++) 
    {
  
        // If current number is even
        if (i % 2 == 0) 
        {
  
            // Count of set bits in i is equal to
            // the count of set bits in (i / 2)
            dp[i] = dp[i / 2];
        }
  
        // If current element is odd
        else
        {
  
            // Count of set bits in i is equal to
            // the count of set bits in (i / 2) + 1
            dp[i] = dp[i / 2] + 1;
        }
  
        // Print the count of set bits in i
        Console.Write(dp[i] + " ");
    }
}
  
// Driver code
public static void Main(String []args)
{
    int n = 5;
  
    findSetBits(n);
}
}
  
// This code is contributed by 29AjayKumar
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Output:
0 1 1 2 1 2



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