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 {
    /**
     * @param Integer $n
     * @return Integer[]
     */
    public function countBits($num) {
        $result = array_fill(0, $num + 1, 0);
        $borderPos = 1;
        $incrPos = 1;
        for ($i = 1; $i < count($result); $i++) {
            // when we reach pow of 2 ,  reset borderPos and incrPos
            if ($incrPos == $borderPos) {
                $result[$i] = 1;
                $incrPos = 1;
                $borderPos = $i;
            } else {
                $result[$i] = 1 + $result[$incrPos++];
            }
        }
        return $result;
    }
}

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