LeetCode-in-All

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

class Solution {
    fun countBits(num: Int): IntArray {
        val result = IntArray(num + 1)
        var borderPos = 1
        var incrPos = 1
        for (i in 1 until result.size) {
            // when we reach pow of 2, reset borderPos and incrPos
            if (incrPos == borderPos) {
                result[i] = 1
                incrPos = 1
                borderPos = i
            } else {
                result[i] = 1 + result[incrPos++]
            }
        }
        return result
    }
}

This site is open source. Improve this page.