<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:
        #two operations are
        # opr1 = do index ko lo i and j uska cost is x
        # opr2 = index i and i+1 ko lo and flip kar do cost is 1

        n=len(s1)
        d=[]
        for i in range(n):
            if s1[i]!=s2[i]:
                d.append(i)
        
        #agara odd hai toh -1 return kar do
        if len(d)%2==1: return -1
        dp={}
        def solve(d,i,x,n):
            #base case
            if i==len(d):
                if n==0:
                    return 0
                else: 
                    return int(1e9+7)
            
            if (i,n) in dp: return dp[(i,n)]
            res=float("inf")
            if i<len(d)-1:
                res=min(res,d[i+1]-d[i]+solve(d,i+2,x,n))
            
            res=min(res,x+solve(d,i+1,x,n+1))

            if n>0:
                res=min(res,solve(d,i+1,x,n-1))
            dp[(i,n)]=res
            return res

        return solve(d,0,x,0)



191