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338. Counting Bits

Easy

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

object Solution {
    def countBits(num: Int): Array[Int] = {
        val result = new Array[Int](num + 1)
        var borderPos = 1
        var incrPos = 1

        for (i <- 1 until result.length) {
            if (incrPos == borderPos) {
                result(i) = 1
                incrPos = 1
                borderPos = i
            } else {
                result(i) = 1 + result(incrPos)
                incrPos += 1
            }
        }

        result
    }
}

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