Skip to content

338. Counting Bits

Description

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.

Do not solve it with built-in functions (i.e., like __builtin_popcount in C++).

 

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?

Solutions

Solution 1

Thinking

Count set bits of every integer in \(0\ldots n\). Per-number bit scans cost \(O(\log i)\). \(n\le 10^5\) still allows that, but we can do better later.

Call bit_count (or repeated lowbit) on each \(i\) to fill the array.

1
2
3
class Solution:
    def countBits(self, n: int) -> List[int]:
        return [i.bit_count() for i in range(n + 1)]

1
2
3
4
5
6
7
8
9
class Solution {
    public int[] countBits(int n) {
        int[] ans = new int[n + 1];
        for (int i = 0; i <= n; ++i) {
            ans[i] = Integer.bitCount(i);
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution {
public:
    vector<int> countBits(int n) {
        vector<int> ans(n + 1);
        for (int i = 0; i <= n; ++i) {
            ans[i] = __builtin_popcount(i);
        }
        return ans;
    }
};

1
2
3
4
5
6
7
func countBits(n int) []int {
    ans := make([]int, n+1)
    for i := 0; i <= n; i++ {
        ans[i] = bits.OnesCount(uint(i))
    }
    return ans
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
function countBits(n: number): number[] {
    const ans: number[] = Array(n + 1).fill(0);
    for (let i = 0; i <= n; ++i) {
        ans[i] = bitCount(i);
    }
    return ans;
}

function bitCount(n: number): number {
    let count = 0;
    while (n) {
        n &= n - 1;
        ++count;
    }
    return count;
}

Solution 2

Thinking

Method 1 counts each integer from scratch. Clearing the lowest set bit yields \(i\&(i-1)\), whose answer is already known, so \(ans[i]=ans[i\&(i-1)]+1\). Fill upward in \(O(n)\).

1
2
3
4
5
6
class Solution:
    def countBits(self, n: int) -> List[int]:
        ans = [0] * (n + 1)
        for i in range(1, n + 1):
            ans[i] = ans[i & (i - 1)] + 1
        return ans

1
2
3
4
5
6
7
8
9
class Solution {
    public int[] countBits(int n) {
        int[] ans = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            ans[i] = ans[i & (i - 1)] + 1;
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution {
public:
    vector<int> countBits(int n) {
        vector<int> ans(n + 1);
        for (int i = 1; i <= n; ++i) {
            ans[i] = ans[i & (i - 1)] + 1;
        }
        return ans;
    }
};

1
2
3
4
5
6
7
func countBits(n int) []int {
    ans := make([]int, n+1)
    for i := 1; i <= n; i++ {
        ans[i] = ans[i&(i-1)] + 1
    }
    return ans
}

1
2
3
4
5
6
7
function countBits(n: number): number[] {
    const ans: number[] = Array(n + 1).fill(0);
    for (let i = 1; i <= n; ++i) {
        ans[i] = ans[i & (i - 1)] + 1;
    }
    return ans;
}

Comments