<p>A binary string is monotone increasing if it consists of some number of <code>0</code>&#39;s (possibly none), followed by some number of <code>1</code>&#39;s (also possibly none).</p>

<p>You are given a binary string <code>s</code>. You can flip <code>s[i]</code> changing it from <code>0</code> to <code>1</code> or from <code>1</code> to <code>0</code>.</p>

<p>Return <em>the minimum number of flips to make </em><code>s</code><em> monotone increasing</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;00110&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> We flip the last digit to get 00111.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;010110&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> We flip to get 011111, or alternatively 000111.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;00011000&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> We flip to get 00000000.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution:
    def minFlipsMonoIncr(self, s: str) -> int:
        n = len(s)
        
        mem = collections.defaultdict(dict)
        
        def minFlip(c_prev, i):
            if i >= n:
                return 0
            if c_prev in mem and i in mem[c_prev]:
                return mem[c_prev][i]
            if c_prev == '0':
                if s[i] == '0':
                    flip =  minFlip('0', i+1)
                else:
                    flip =  min(minFlip('1', i+1), minFlip('0', i+1) + 1)
            else:
                if s[i] == '0':
                    flip = 1 + minFlip('1', i+1)
                else:
                    flip = minFlip('1', i+1)
            mem[c_prev][i] = flip
            return flip
                
                
        return minFlip('0',0)
            
#         mem = {}
        
#         def count_zero_chars(s):
#             count = 0
#             for char in s:
#                 if char == '0':
#                     count += 1
#             return count
        
#         def dfs(s, i):
#             # print(s)
#             if i == len(s):
#                 return 0
#             if i in mem:
#                 return mem[i]
#             if s[i] == '0':
#                 mem[i] = dfs(s, i+1)
#                 return dfs(s, i+1)
#             if s[i] == '1':
#                 mem[i] = min(1 + dfs(s, i+1), count_zero_chars(s[i+1:]))
#                 return min(1 + dfs(s, i+1), count_zero_chars(s[i+1:]))
            
            # since we are calling count_zero_chars, it takes O(n), and the recursion call with memory at most O(n), so it is actually O(N^2). We can add one variable (prev_variable) to the dfs to reduce to O(n)
        
        
        
        # return dfs(s, 0)
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