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

  • Last Updated : 10 Jun, 2021

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 
 

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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:
 

C++




// 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;
}

Java




// 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

Python 3




# 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

C#




// C# implementation of the approach
using System;
     
class GFG
{
 
static int count(int n)
    {
        int count = 0;
        while (n > 0)
        {
            count += n & 1;
            n >>= 1;
        }
        return count;
    }
     
// 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++)
        Console.Write(count(i)+" ");
}
 
// Driver code
public static void Main(String []args)
{
    int n = 5;
 
    findSetBits(n);
}
}
 
// This code is contributed by SHUBHAMSINGH10

Javascript




<script>
    // Javascript implementation of the approach
     
    function count(n)
    {
        let count = 0;
        while (n > 0)
        {
            count += n & 1;
            n >>= 1;
        }
        return count;
    }
       
    // Function to find the count
    // of set bits in all the
    // integers from 0 to n
    function findSetBits(n)
    {
        for (let i = 0; i <= n; i++)
            document.write(count(i)+" ");
    }
     
    let n = 5;
   
    findSetBits(n);
 
</script>
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: 
 

C++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to find the count
    // of set bits in all the
    // integers from 0 to n
    function findSetBits(n)
    {
 
        // dp[i] will store the count
        // of set bits in i
        let dp = new Array(n + 1);
        dp.fill(0);
 
        // Count of set bits in 0 is 0
        document.write(dp[0] + " ");
 
        // For every number starting from 1
        for (let 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[parseInt(i / 2, 10)];
            }
 
            // 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[parseInt(i / 2, 10)] + 1;
            }
 
            // Print the count of set bits in i
            document.write(dp[i] + " ");
        }
    }
     
    let n = 5;
  
    findSetBits(n);
     
// This code is contributed by divyeshrabadiya07   
</script>
Output: 
0 1 1 2 1 2

 




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