【leetcode】1247. Minimum Swaps to Make Strings Equal

题目如下：

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'.

解题思路：这个题目还是有点考巧劲的。如果s1[i] != s2[i]，只有这两种取值 s1[i] = x,s2[i] = y(记为xy)和s1[i] = y,s2[i] = x(记为yx)。理想情况下，进行s1[i]和s2[j]的交换，使得s1[i] = s2[i] 与 s1[j] = s2[j]同时成立，那么一定是交换次数最少的场景。而题目中例子1的交换则符合这个场景，只要分别求出(xy)和(yx)的个数，然后分别进行(xy)和(yx)的组内交换，剩余的配对再进行例子2的组间交换，即可求得最小的交换次数。

代码如下：

class Solution(object):
    def minimumSwap(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        dic = {}
        dic[('x','y')] = 0
        dic[('y','x')] = 0

        for (i,j) in zip(s1,s2):
            if i != j:
                dic[(i,j)]  += 1
        if (dic[('x','y')] + dic[('y','x')] ) % 2 != 0:
            return -1
        elif dic[('x','y')] % 2 == 0 and dic[('y','x')] % 2 == 0:
            return dic[('x','y')] / 2  + dic[('y','x')] / 2
        return dic[('x','y')] / 2  + dic[('y','x')] / 2 + 2