LeetCode 926. Flip String to Monotone Increasing (Medium)

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.)

We are given a string S of '0's and '1's, and we may flip any '0' to a '1' or a '1' to a '0'.

Return the minimum number of flips to make S monotone increasing.

 

Example 1:

Input: "00110"
Output: 1
Explanation: We flip the last digit to get 00111.

Example 2:

Input: "010110"
Output: 2
Explanation: We flip to get 011111, or alternatively 000111.

Example 3:

Input: "00011000"
Output: 2
Explanation: We flip to get 00000000.

 

Note:

1. 1 <= S.length <= 20000
2. S only consists of '0' and '1' characters.

 方法：dp

 思路：

 1 用两个变量维护当遇到当前character的时候，flip成0或者flip成1时候的数量，cnt1，cnt0。

 2 当当前character是0的时候，我们可以把它flip成1，cnt1 = min（cnt1，cnt0）+1。因为这个0之前的可能是翻成0的，也可能是翻成1的。所以取两种情况下的最小值。如果character是1的话，我们可以吧它flip成0. 这时候记得要先update

cnt1 = min（cnt1， cnt0）。然后再update cnt0+=1 

 3 最后返回cnt0和cnt1的最小值即可。

time complexity: O(n) space complexity:O(1) 

class Solution:
    def minFlipsMonoIncr(self, S: str) -> int:
        if not S or len(S) == 0: return 0 
        
        count0 = 0 
        count1 = 0 
        
        for i in range(len(S)):
            if S[i] == '0': # flip to 1 => two choices, coming from count0 (count0+1) or coming from count1(count1+1)
                count1 = min(count0, count1) + 1 
            else: # flip to 0 => one choice, coming from count0, so count0 += 1 
                count1 = min(count0, count1)
                count0 += 1 
                
                
        return min(count0, count1)