human
I found an interesting problem on https://leetcode.com/problems/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.)
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 <= S.length <= 20000
S only consists of '0' and '1' characters.

I tried it in Python, but could not do it. Can you solve it?

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Give me a the Python program you requested?
assistant
#Here is the Python program you requested.

class Solution:
    def minFlipsMonoIncr(self, S: str) -> int:
        str_len = len(S)
        count_arr = [[0,0] for x in range(str_len)]
        one_start_idx = -1
        for i in range(len(S)):
            if S[i]=='0':
                if i==0:
                    count_arr[i][0] +=1
                    count_arr[i][1] = 0
                else:
                    count_arr[i][0] = count_arr[i-1][0]+1
                    count_arr[i][1] = count_arr[i-1][1]
            else:
                if i==0:
                    count_arr[i][1] +=1
                else:
                    count_arr[i][1] = count_arr[i-1][1]+1
                    count_arr[i][0] = count_arr[i-1][0]
                
                if one_start_idx == -1:
                    one_start_idx = i
       
        total_flips = []
        total_flips.append(min(count_arr[str_len-1][0], count_arr[str_len-1][1]))
        
        for i in range(one_start_idx, str_len):
            if i==0:
                total_flips.append(count_arr[str_len-1][0]-count_arr[i][0])
            elif i==str_len-1:                    
                total_flips.append(count_arr[str_len-1][0])
                        
            else:
                total_flips.append(count_arr[i-1][1]+count_arr[str_len-1][0]-count_arr[i][0])
                
                
                
                           
                           
   
                
            
            
        return min(total_flips)