<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 {
public:
    int dp[2][100001];
    
    int solve(string &s, char prev, int idx) {
        if(idx >= s.size())
            return 0;
        
        if(dp[prev - '0'][idx] != -1)
            return dp[prev - '0'][idx];
        
        int minFlips = INT_MAX;
        int flip = INT_MAX, noFlip = INT_MAX;
        if(prev == '0') {
            if(s[idx] == '0') {
                flip   = 1 + solve(s, '1', idx+1);
                noFlip = solve(s, '0', idx+1);
            } else {
                flip   = 1 + solve(s, '0', idx+1);
                noFlip = solve(s, '1', idx+1);
            }
            minFlips = min(flip, noFlip);
        } else {
            if(s[idx] == '0') {
                flip = 1 + solve(s, '1', idx+1);
            } else {
                noFlip = solve(s, '1', idx+1);
            }
            minFlips = min(flip, noFlip);
        }
        return dp[prev - '0'][idx] = minFlips;
    }
    
    int minFlipsMonoIncr(string s) {
        memset(dp, -1, sizeof(dp));
        return solve(s, '0', 0);
    }
};
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