338. Counting Bits(Solution || Leetcode easy || Java)

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++)?

SOLUTION:

class Solution {
    public int[] countBits(int n) {
        //create an output array to return
        int[] output = new int[n+1];

        //start running the loop
        for(int i = 1; i < output.length; i++){

        //number of bits at i is equal to number of bits at
        // i/2 and this i%2 is for the msb(most significant bit
        //which tells us whether the number is odd or even)

            output[i] = output[i/2] + i%2;
        }

        return output;
    }
}

Runtime1 ms

Beats

99.98%

Memory46.4 MB

Beats

90.1%

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