You are given two?0-indexed?strings?word1?and?word2.

A?move?consists of choosing two indices?i?and?j?such that?0 <= i < word1.length?and?0 <= j < word2.length?and swapping?word1[i]?with?word2[j].

Return?true?if it is possible to get the number of distinct characters in?word1?and?word2?to be equal with?exactly one?move.?Return?false?otherwise.

?

Example 1:

Input: word1 = "ac", word2 = "b"

Output: false

Explanation: Any pair of swaps would yield two distinct characters in the first string, and one in the second string.

Example 2:

Input: word1 = "abcc", word2 = "aab"

Output: true

Explanation: We swap index 2 of the first string with index 0 of the second string. The resulting strings are word1 = "abac" and word2 = "cab", which both have 3 distinct characters.

Example 3:

Input: word1 = "abcde", word2 = "fghij"

Output: true

Explanation: Both resulting strings will have 5 distinct characters, regardless of which indices we swap.

?

Constraints:

• 1 <= word1.length, word2.length <= 105

• word1?and?word2?consist of only lowercase English letters.

一開始想到的是用hashmap，然而。。。還是直接用數(shù)組快啊，畢竟就是常數(shù)的復(fù)雜度。

Runtime:?11 ms, faster than?67.88%?of?Java?online submissions for?Make Number of Distinct Characters Equal.

Memory Usage:?43.5 MB, less than?64.25%?of?Java?online submissions for?Make Number of Distinct Characters Equal.