<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 solveTab(string s) {
        int n = s.length();
        vector<vector<int>> dp(n + 1, vector<int>(2, 0));

        // Fill the dp table from the end to the beginning
        for (int curr_index = n - 1; curr_index >= 0; curr_index--) {
            for (int prev = 1; prev >= 0; prev--) {
                int flip = INT_MAX, no_flip = INT_MAX;

                if (s[curr_index] == '0') {
                    if (prev == 0) {
                        flip = 1 + dp[curr_index + 1][1];
                        no_flip = dp[curr_index + 1][0];
                    } else { // prev == 1
                        flip = 1 + dp[curr_index + 1][1];
                    }
                } else { // s[curr_index] == '1'
                    if (prev == 0) {
                        flip = 1 + dp[curr_index + 1][0];
                        no_flip = dp[curr_index + 1][1];
                    } else { // prev == 1
                        no_flip = dp[curr_index + 1][1];
                    }
                }
                dp[curr_index][prev] = min(flip, no_flip);
            }
        }

        // The result is the minimum flips needed starting from the beginning
        return dp[0][0];
    }

    int minFlipsMonoIncr(string s) {
        if (s.length() == 1) return 0;

        return solveTab(s);
    }
};
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