# 1247. Minimum Swaps to Make Strings Equal

You are given two strings `s1` and `s2` of equal length consisting of letters `"x"` and `"y"` **only**. Your task is to make these two strings equal to each other. You can swap any two characters that belong to **different** strings, which means: swap `s1[i]` and `s2[j]`.

Return the minimum number of swaps required to make `s1` and `s2` equal, or return `-1` if it is impossible to do so.

**Example 1:**

```
Input: s1 = "xx", s2 = "yy"
Output: 1
Explanation: 
Swap s1[0] and s2[1], s1 = "yx", s2 = "yx".
```

**Example 2:** 

```
Input: s1 = "xy", s2 = "yx"
Output: 2
Explanation: 
Swap s1[0] and s2[0], s1 = "yy", s2 = "xx".
Swap s1[0] and s2[1], s1 = "xy", s2 = "xy".
Note that you can't swap s1[0] and s1[1] to make s1 equal to "yx", cause we can only swap chars in different strings.
```

**Example 3:**

```
Input: s1 = "xx", s2 = "xy"
Output: -1
```

**Example 4:**

```
Input: s1 = "xxyyxyxyxx", s2 = "xyyxyxxxyx"
Output: 4
```

**Constraints:**

* `1 <= s1.length, s2.length <= 1000`
* `s1, s2` only contain `'x'` or `'y'`.

给定由x和y构成的等长字符串s1和s2，问最少需要多少在s1和s2间交换（且只能s1和s2间交换）多少个字符能使得s1 == s2。这道题虽然标为easy，但难道了很多英雄好汉。由于只牵扯两个字符，实际并不需要上复杂的算法，只需要总结给定的例子的pattern。首先当s1\[i] == s2\[i]时，i可忽略，因为已经一致，再怎么换都是瞎折腾。对剩下的，两两一对的看有两种情况，xx | yy 或xy | yx，前者只需要交换一次，后者需要先换成前者，再交换一次。当剩下的是奇数时，无论怎么换都不可能一致。因此统计有多少x和y pair，最后可能再剩下个xy | yx。

```cpp
class Solution {
public:
    int minimumSwap(string s1, string s2) {
        int x1 = 0, x2 = 0, y1 = 0, y2 = 0;
        for (int i = 0; i < s1.length(); ++i) {
            if (s1[i] == s2[i]) continue;
            if (s1[i] == 'x') ++x1;
            else ++y1;
            if (s2[i] == 'x') ++x2;
            else ++y2;
        }
        if ((x1 + y1) & 1 == 1) return -1;
        return x1 / 2 + y1 / 2 + (x1 % 2) * 2;
    }
};
```