2085. Count Common Words With One Occurrence

Description

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.

 

Solutions

Solution: Hash Table

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {string[]} words1
 * @param {string[]} words2
 * @return {number}
 */
const countWords = function (words1, words2) {
  const wordMap = words1.reduce((map, word) => {
    const count = map.get(word) ?? 0;

    return map.set(word, count + 1);
  }, new Map());
  let result = 0;

  for (const word of words2) {
    const count = wordMap.get(word) ?? 0;

    if (count > 1 || count < 0) continue;
    wordMap.set(word, count - 1);
  }
  for (const count of wordMap.values()) {
    if (count === 0) result += 1;
  }
  return result;
};