Python solution with detailed explanation


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    G

    Solutions

    Number of 1 Bits https://leetcode.com/problems/number-of-1-bits/

    Algorithm 1

    1. n & (n - 1) drops the** lowest set bit.** It's a neat little bit trick. https://discuss.leetcode.com/topic/20120/c-solution-n-n-1/2
      Let's use n = 00101100 as an example. This binary representation has three 1s.
    • If n = 00101100, then n - 1 = 00101011, so n & (n - 1) = 00101100 & 00101011 = 00101000. Count = 1.
    • If n = 00101000, then n - 1 = 00100111, so n & (n - 1) = 00101000 & 00100111 = 00100000. Count = 2.
    • If n = 00100000, then n - 1 = 00011111, so n & (n - 1) = 00100000 & 00011111 = 00000000. Count = 3.
    • n is now zero, so the while loop ends, and the final count (the numbers of set bits) is returned.
    class Solution(object):
        def hammingWeight(self, n):
            """
            :type n: int
            :rtype: int
            """
            cnt = 0
            while n:
                n, cnt = n & (n-1), cnt + 1
            return cnt
    

    Algorithm 2

    1. Can we solve this recursively? What is the recurrence here?
    2. Start with the last bit. If it is odd then it is 1 otherwise it is 0.
    3. Say we have k bits and N(k) denote the number of bits set to 1.
    4. Then N(k) = N(k-1) + status_of_last_bit. How do we write this recurrence in terms of number n which can be represented in say k bits?
    5. F(n) = F(n >> 1) + n % 2
    6. n%2 can be represented as int(n&1).
    class Solution(object):
        def hammingWeight(self, n):
            """
            :type n: int
            :rtype: int
            """
            if n == 0:
                return 0
            else:
                return self.hammingWeight(n>>1) + int(n&1)
    

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