Counting Bits LeetCode Solution

Problem – Counting Bits LeetCode Solution

Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1‘s in the binary representation of i.

Example 1:

Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10

Example 2:

Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

Constraints:

• 0 <= n <= 105

Follow up:

• It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
• Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

Counting Bits LeetCode Solution in C++

class Solution {
public:
    vector<int> countBits(int n) {
        vector<int> ans;
		
		// iterating fromt 0 to n
        for(int i = 0; i<=n; i++)
        {
			// sum is initialised as 0
            int sum = 0;
            int num = i;
			// while num not equals zero
            while(num != 0)
            {
				// we have to count 1's in binary representation of i, therefore % 2
                sum += num%2;
                num = num/2;
            }
			// add sum to ans vector
            ans.push_back(sum);
        }
		// return 
        return ans;
    }
};

Counting Bits LeetCode Solution in Java

public int[] countBits(int num) {
    int[] f = new int[num + 1];
    for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
    return f;
}

Counting Bits LeetCode Solution in Python

def countBits(self, num: int) -> List[int]:
    counter = [0]
    for i in range(1, num+1):
        counter.append(counter[i >> 1] + i % 2)
    return counter

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