338.Counting Bits

Tags: [bit_manipulation], [math], [dp]

Link: https://leetcode.com/problems/counting-bits/\#/description

Given a non negative integer numbernum. For every numbersiin the range0 ≤ i ≤ numcalculate the number of 1's in their binary representation and return them as an array.

Example:
For num = 5you should return[0,1,1,2,1,2].

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n)/possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

Solution: DP

class Solution(object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        if num < 0:
            raise ValueError('The input is invalid')

        if num == 0:
            return [0]
        if num == 1:
            return [0, 1]
        if num == 2:
            return [0, 1, 1]

        counters = [0 for _ in xrange(num + 1)]
        counters[0] = 0
        counters[1] = 1
        counters[2] = 1

        for n in xrange(3, num + 1):
            if n % 2 != 0:
                counters[n] = counters[n - 1] + 1
            else:
                counters[n] = counters[n / 2]

        return counters

Revelation:

• We can first use straight forward method to solve this problem, (the straight forward solution list below).

• For this kind of question, if we think there exists some tricks, we can list some small number example:

• num = 0, result = [0], n = 0, num of '1's bits = 0

• num = 1, result = [0, 1], n = 1, num of '1's bits = 1
• num = 2, result = [0, 1, 1], n = 2, num of '1's bits = 1
• num = 3, result = [0, 1, 1, 2], n = 3, num of '1's bits = 2
• num = 4, result = [0, 1, 1, 2, 1], n = 4, num of '1's bits = 1
• num = 5, result = [0, 1, 1, 2, 1, 2], n = 5, num of '1's bits = 2
• num = 6, result = [0, 1, 1, 2, 1, 2, 2], n = 6, num of '1's bits = 2
• num = 7, result = [0, 1, 1, 2, 1, 2, 2, 3], n = 7, num of '1's bits = 3
• num = 8, result = [0, 1, 1, 2, 1, 2, 2, 3, 1], n = 8, num of '1's bits = 1
• num = 9, result = [0, 1, 1, 2, 1, 2, 2, 3, 1, 2], n = 9, num of '1's bits = 2
• num = 10, result = [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2], n = 10, num of '1's bits = 2
• num = 11, result = [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3], n = 11, num of '1's bits = 3
• num = 12, result = [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2], n = 12, num of '1's bits = 2
• num = 13, result = [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3], n = 13, num of '1's bits = 3

• num = 14, result = [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3], n = 14, num of '1's bits = 3

• From the above example, when num >= 3, there exists the regulation, which is that: if current n is odd number, the number of '1's bits = the number of '1's bits of (n - 1). If the current n is even number, the number of '1's bits = the number of '1's bits of (n/2).

Note:

• Time complexity = O(num).

Straight forward solution:

class Solution(object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        if num < 0:
            raise ValueError('the input should be less than 0')

        counters = [0 for _ in xrange(num + 1)]
        for i in xrange(32):
            for n in xrange(num + 1):
                if (n >> i) & 1 == 1:
                    counters[n] += 1

        return counters

Note:

• Time complexity = O(num).