Counting Bits

Problem

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

Solution

/**
 * @param {number} n
 * @return {number[]}
 */
var countBits = function(n) {
    const dp = [0];
    for (let i = 1; i <= n; i++) {
        dp.push(1 + dp[i & (i - 1)]);
    }
    return dp;
};

We will implement a DP solution. Let dp[i] be the number of 1's in the binary representation of i.

• For the base case, if i = 0, dp[0] = 0 since there are no 0s.
• For the recursive case, notice that i & (i - 1) removes the least significant 1 bit from i (and produces a smaller value if i is not 0). This means dp[i] = 1 + dp[i & (i - 1)].

For example, if n = 14, its binary representation is 1110, so n - 1 is 1101, n & (n - 1) = 1100.