Description (leetcode)

Given two string arrays words1 and words2, return the number of strings that appear exactly once in each of the two arrays.

Example 1:

Input: words1 = [“leetcode”,“is”,“amazing”,“as”,“is”], words2 = [“amazing”,“leetcode”,“is”]

Output: 2

Explanation:

  • “leetcode” appears exactly once in each of the two arrays. We count this string.

  • “amazing” appears exactly once in each of the two arrays. We count this string.

  • “is” appears in each of the two arrays, but there are 2 occurrences of it in words1. We do not count this string.

  • “as” appears once in words1, but does not appear in words2. We do not count this string.

Thus, there are 2 strings that appear exactly once in each of the two arrays.

Example 2:

Input: words1 = [“b”,“bb”,“bbb”], words2 = [“a”,“aa”,“aaa”]

Output: 0

Explanation: There are no strings that appear in each of the two arrays.

Example 3:

Input: words1 = [“a”,“ab”], words2 = [“a”,“a”,“a”,“ab”]

Output: 1

Explanation: The only string that appears exactly once in each of the two arrays is “ab”.

Constraints:

  • 1 <= words1.length, words2.length <= 1000
  • 1 <= words1[i].length, words2[j].length <= 30
  • words1[i] and words2[j] consists only of lowercase English letters.

submission

click to reveal
use std::collections::HashMap;

impl Solution {
    // one-liner, pls don't write code like this
    pub fn count_words(words1: Vec<String>, words2: Vec<String>) -> i32 {
        words2
            .into_iter()
            .fold(
                words1.into_iter().fold(HashMap::new(), |mut cnt, s| {
                    cnt.entry(s)
                        .and_modify(|(c, _)| {
                            *c += 1;
                        })
                        .or_insert((1, 0));
                    cnt
                }),
                |mut cnt, s| {
                    cnt.entry(s)
                        .and_modify(|(_, c)| {
                            *c += 1;
                        })
                        .or_insert((0, 1));
                    cnt
                },
            )
            .into_iter()
            .filter(|&(_, (c1, c2))| c1 == 1 && c2 == 1)
            .count() as _
    }
}