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

/**
 * @param {number} n
 * @return {number[]}
 */
var countBits = function(num) {
    const result = new Array(num + 1).fill(0)
    let borderPos = 1
    let incrPos = 1

    for (let i = 1; i <= num; i++) {
        if (incrPos === borderPos) {
            result[i] = 1
            incrPos = 1
            borderPos = i
        } else {
            result[i] = 1 + result[incrPos++]
        }
    }

    return result
}

export { countBits }

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