<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[100005][2];

  int func(string& s, int ind, int curr) {
    if(ind >= s.size()) return 0;

    if(dp[ind][curr] != -1) return dp[ind][curr];

    if(s[ind]-'0' == curr) {
      int ans = func(s, ind+1, curr);
      if(s[ind]-'0' == 0) {
        ans = min(ans, 1+func(s, ind+1, 1));
      }
      return dp[ind][curr] = ans;
    }

    int ans = 1+func(s, ind+1, curr);
    if(curr == 0) { //can afford to not flip
      ans = min(ans, func(s, ind+1, 1));
    }

    return dp[ind][curr] = ans;
  }

  int minFlipsMonoIncr(string s) {
    memset(dp, -1, sizeof(dp));
    return min(func(s, 0, 0), func(s, 0, 1));
  }
};
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