[leetcode, JS] 2085. Count Common Words With One Occurrence

문제

문제 링크 : Count Common Words With One Occurrence

풀이

/**
 * @param {string[]} words1
 * @param {string[]} words2
 * @return {number}
 */
var countWords = function(words1, words2) {
    let cnt = 0
    for(let i=0; i<words1.length; i++) {
        if( words1.indexOf(words1[i]) === words1.lastIndexOf(words1[i])
            && words2.indexOf(words1[i]) > -1 && words2.indexOf(words1[i]) === words2.lastIndexOf(words1[i])
        ) cnt ++
    }

    return cnt
};

1. 배열 words1에 존재하면서 하나만 존재(indexOf === lastIndexOf), 배열 words2에 존재(indexOf > -1)하면서 하나만 존재(indexOf ==== lastIndexOf)를 찾아 cnt++

• Runtime 80 ms, Memory 44.22 MB

다른 풀이

/**
 * @param {string[]} words1
 * @param {string[]} words2
 * @return {number}
 */
var countWords = function(words1, words2) {
    let map1=new Map()
    let map2=new Map()
    
    for(let i=0;i<words1.length;i++)
    {
        map1.set(words1[i],map1.get(words1[i])+1||1)
    }
    for(let i=0;i<words2.length;i++)
    {
        map2.set(words2[i],map2.get(words2[i])+1||1)
    }

    let count=0;
    for(const key of map1)
    {
        if(key[1]===1)
        {
            if(map2.get(key[0])===1) count++
        }
    }

    return count
};

1. Map을 사용하여 카운팅

• Runtime 66 ms, Memory 45.92 MB