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2571 - Minimum Operations to Reduce an Integer to 0 (Easy)

https://leetcode.com/problems/minimum-operations-to-reduce-an-integer-to-0/

Problem Statement​

You are given a positive integer n, you can do the following operation any number of times:

  • Add or subtract a power of 2 from n.

Return the minimum number of operations to makenequal to0.

A number x is power of 2 if x == 2i where i >= 0.

Example 1:

Input: n = 39
Output: 3
Explanation: We can do the following operations:
- Add 20 = 1 to n, so now n = 40.
- Subtract 23 = 8 from n, so now n = 32.
- Subtract 25 = 32 from n, so now n = 0.
It can be shown that 3 is the minimum number of operations we need to make n equal to 0.

Example 2:

Input: n = 54
Output: 3
Explanation: We can do the following operations:
- Add 21 = 2 to n, so now n = 56.
- Add 23 = 8 to n, so now n = 64.
- Subtract 26 = 64 from n, so now n = 0.
So the minimum number of operations is 3.

Constraints:

  • 1 <= n <= 10^5

Approach 1: DP​

Written by @wingkwong
class Solution:
    def minOperations(self, n: int) -> int:
        # precompute power of 2
        p = {1 << i for i in range(20)}
        def dp(x):
            # reach 0 -> 0 operation
            if x == 0: return 0
            # if x is a power of 2, 
            # we need 1 operation (i.e. subtract itself)
            if x in p: return 1
            # otherwise we either add / subtract the lsb to x
            # e.g. 0111 -> 1000 -> 0000
            # e.g. 1001 -> 1000 -> 0000
            return min(dp(x + (x & -x)), dp(x - (x & -x))) + 1
        return dp(n)