<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>

class Solution:
    def minOperations(self, s1: str, s2: str, x: int) -> int:
        m = s1.count("1")
        n = s2.count("1")
        if m % 2 != n % 2:
            return -1

        s1 = [int(i) for i in s1]
        s2 = [int(i) for i in s2]

        @cache
        def dfs(i, s, p, t, f):
            if i == len(s2):
                return 0 if t == n and not p and not s else inf
            
            if f and s1[i]:
                t -= 1
            elif f:
                t += 1

            one = (s1[i] and not f) or (not s1[i] and f)

            a = b = c = inf
            if one and not s2[i]:
                if not s:
                    a = dfs(i + 1, s + 1, p, t - 1, False) + x
                else:
                    a = dfs(i + 1, s - 1, p, t - 1, False)
                if i < len(s2) - 1:
                    b = dfs(i + 1, s, p, t - 1, True) + 1
                if p:
                    c = dfs(i + 1, s, p - 1, t - 1, False)

            elif s2[i] and not one:
                if not p:
                    a = dfs(i + 1, s, p + 1, t + 1, False) + x
                else:
                    a = dfs(i + 1, s, p - 1, t + 1, False)
                if i < len(s2) - 1:
                    b = dfs(i + 1, s, p, t + 1, True) + 1
                if s:
                    c = dfs(i + 1, s - 1, p, t + 1, False)
            else:
                return dfs(i + 1, s, p, t, False)

            return min(a, b, c)

        ans = dfs(0, 0, 0, m, 0)
        return ans if ans != inf else -1

18100