Problem URL: Flip string to monotone increasing

A string of '0's and '1's is monotone increasing if it consists of some number of '0's (possibly 0), followed by some number of '1's (also possibly 0.)

• So we iterate throughs, but for all0before the initial 1, we count nothing. It only becomes interesting once we hit a1. We could rephrase the problem to: )Discard all the initial0, then determine which requires fewer flips: changing the remaining string to all0, or changing it to all1.
• For each 1 we hit, we incrementones.
• For each 0 we hit after the initial 1, we decrementones and increment res, but only if onesis positive. Here's the reason why–in order to achieve a monotonic string, we need to flip either this 0 or the most recent 1. We won't know which flip would actually happen until the end, but either way it would cause a flip.
• Once we iterate through the string, we return res.

class Solution:
    def minFlipsMonoIncr(self, s: str) -> int:
        ones, res = 0, 0
        for num in s:
            if num == '1':
                ones += 1
            elif ones:
                ones -= 1
                res += 1
        return res

Time complexity: O(n), n is the length of the string.
Space complexity: O(1)