Apply Operations to Make Two Strings Equal
<p>You are given two <strong>0-indexed</strong> binary strings <code>s1</code> and <code>s2</code>, both of length <code>n</code>, and a positive integer <code>x</code>.</p>

<p>You can perform any of the following operations on the string <code>s1</code> <strong>any</strong> number of times:</p>

<ul>
	<li>Choose two indices <code>i</code> and <code>j</code>, and flip both <code>s1[i]</code> and <code>s1[j]</code>. The cost of this operation is <code>x</code>.</li>
	<li>Choose an index <code>i</code> such that <code>i &lt; n - 1</code> and flip both <code>s1[i]</code> and <code>s1[i + 1]</code>. The cost of this operation is <code>1</code>.</li>
</ul>

<p>Return <em>the <strong>minimum</strong> cost needed to make the strings </em><code>s1</code><em> and </em><code>s2</code><em> equal, or return </em><code>-1</code><em> if it is impossible.</em></p>

<p><strong>Note</strong> that flipping a character means changing it from <code>0</code> to <code>1</code> or vice-versa.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;1100011000&quot;, s2 = &quot;0101001010&quot;, x = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> We can do the following operations:
- Choose i = 3 and apply the second operation. The resulting string is s1 = &quot;110<u><strong>11</strong></u>11000&quot;.
- Choose i = 4 and apply the second operation. The resulting string is s1 = &quot;1101<strong><u>00</u></strong>1000&quot;.
- Choose i = 0 and j = 8 and apply the first operation. The resulting string is s1 = &quot;<u><strong>0</strong></u>1010010<u><strong>1</strong></u>0&quot; = s2.
The total cost is 1 + 1 + 2 = 4. It can be shown that it is the minimum cost possible.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;10110&quot;, s2 = &quot;00011&quot;, x = 4
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is not possible to make the two strings equal.
</pre>

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

<ul>
	<li><code>n == s1.length == s2.length</code></li>
	<li><code>1 &lt;= n, x &lt;= 500</code></li>
	<li><code>s1</code> and <code>s2</code> consist only of the characters <code>&#39;0&#39;</code> and <code>&#39;1&#39;</code>.</li>
</ul>
String; Dynamic Programming
func minOperations(s1 string, s2 string, x int) int {
	idx := []int{}
	for i := range s1 {
		if s1[i] != s2[i] {
			idx = append(idx, i)
		}
	}
	m := len(idx)
	if m&1 == 1 {
		return -1
	}
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, m)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i > j {
			return 0
		}
		if f[i][j] != -1 {
			return f[i][j]
		}
		f[i][j] = dfs(i+1, j-1) + x
		f[i][j] = min(f[i][j], dfs(i+2, j)+idx[i+1]-idx[i])
		f[i][j] = min(f[i][j], dfs(i, j-2)+idx[j]-idx[j-1])
		return f[i][j]
	}
	return dfs(0, m-1)
}