<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>

# The idea is that, we transform the two operation:

# flip only s[i], cost is x / 2.
# flip s[i] and s[j], cost is j - i, where i < j.
# The first operation is equal to original one,
# as long as it will be done in even times,
# and we will do a joint order(拼多多).

# We use odd to count the number of s1[i] != s2[i],
# After one operation, the odd/even of difference won't change,
# so if the number of diff is odd, we return -1.

# res is the current result,
# pre is the last time result plus the distance.
# To keep using integer, we double the cost during calculation.

# Then we iterate each pair of s1[i] and s2[i],
# we increment pre += 2, the double cost of 2nd operation.
# We update res = min(res + x, pre),
# since we can take the either operation,
# and update pre to the previous res.

# Finally, we check if we have odd diff,
# if so we return -1,
# otherwise we return res / 2, since the cost was doubled.

class Solution:
    def minOperations(self, s1: str, s2: str, x: int) -> int:
        res, pre, c = 0, inf, 0
        for a,b in zip(s1, s2):
            pre += 2
            if a != b:
                res, pre, c = min(res + x, pre), res, c ^ 1
        return res // 2 if c == 0 else -1


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