916. Word Subsets
선택한 UI 언어에 맞게 문제 텍스트를 러시아어에서 번역합니다. 코드는 변경하지 않습니다.
Вам given два 배열а строк words1 и words2. 문자열 b является подмножеством строки a, если каждая буква в b встречается в ней, включая кратность. На예제, "wrr" является подмножеством "warrior", но не является подмножеством "world". 문자열 a из words1 является универсальной, если для каждой строки b в words2, b является подмножеством a. return 배열 всех универсальных строк в words1. Вы можете вернуть ответ в любом порядке.
예제:
Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"]
Output: ["facebook","google","leetcode"]
C# 해법
매칭됨/원본using System;
using System.Collections.Generic;
using System.Linq;
public class Solution {
public IList<string> WordSubsets(string[] words1, string[] words2) {
int[] maxCount = new int[26];
foreach (string word in words2) {
int[] count = GetCount(word);
for (int i = 0; i < 26; i++) {
maxCount[i] = Math.Max(maxCount[i], count[i]);
}
}
List<string> result = new List<string>();
foreach (string word in words1) {
int[] count = GetCount(word);
if (IsUniversal(count, maxCount)) {
result.Add(word);
}
}
return result;
}
private int[] GetCount(string word) {
int[] count = new int[26];
foreach (char c in word) {
count[c - 'a']++;
}
return count;
}
private bool IsUniversal(int[] count, int[] maxCount) {
for (int i = 0; i < 26; i++) {
if (count[i] < maxCount[i]) {
return false;
}
}
return true;
}
}C++ 해법
자동 초안, 제출 전 검토#include <bits/stdc++.h>
using namespace std;
// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
public vector<string> WordSubsets(vector<string> words1, vector<string> words2) {
vector<int>& maxCount = new int[26];
foreach (string word in words2) {
vector<int>& count = GetCount(word);
for (int i = 0; i < 26; i++) {
maxCount[i] = max(maxCount[i], count[i]);
}
}
List<string> result = new List<string>();
foreach (string word in words1) {
vector<int>& count = GetCount(word);
if (IsUniversal(count, maxCount)) {
result.push_back(word);
}
}
return result;
}
private vector<int>& GetCount(string word) {
vector<int>& count = new int[26];
foreach (char c in word) {
count[c - 'a']++;
}
return count;
}
private bool IsUniversal(vector<int>& count, vector<int>& maxCount) {
for (int i = 0; i < 26; i++) {
if (count[i] < maxCount[i]) {
return false;
}
}
return true;
}
}Java 해법
매칭됨/원본import java.util.*;
class Solution {
public List<String> wordSubsets(String[] words1, String[] words2) {
int[] maxCount = new int[26];
for (String word : words2) {
int[] count = getCount(word);
for (int i = 0; i < 26; i++) {
maxCount[i] = Math.max(maxCount[i], count[i]);
}
}
List<String> result = new ArrayList<>();
for (String word : words1) {
int[] count = getCount(word);
if (isUniversal(count, maxCount)) {
result.add(word);
}
}
return result;
}
private int[] getCount(String word) {
int[] count = new int[26];
for (char c : word.toCharArray()) {
count[c - 'a']++;
}
return count;
}
private boolean isUniversal(int[] count, int[] maxCount) {
for (int i = 0; i < 26; i++) {
if (count[i] < maxCount[i]) {
return false;
}
}
return true;
}
}JavaScript 해법
매칭됨/원본var wordSubsets = function(words1, words2) {
const getCount = (word) => {
const count = {};
for (const char of word) {
count[char] = (count[char] || 0) + 1;
}
return count;
};
const maxCount = {};
for (const word of words2) {
const count = getCount(word);
for (const char in count) {
maxCount[char] = Math.max(maxCount[char] || 0, count[char]);
}
}
return words1.filter(word => {
const count = getCount(word);
return Object.keys(maxCount).every(char => count[char] >= maxCount[char]);
});
};Python 해법
매칭됨/원본from collections import Counter
def wordSubsets(words1, words2):
def count(word):
return Counter(word)
max_count = Counter()
for word in words2:
for char, cnt in count(word).items():
max_count[char] = max(max_count[char], cnt)
result = []
for word in words1:
word_count = count(word)
if all(word_count[char] >= max_count[char] for char in max_count):
result.append(word)
return resultGo 해법
매칭됨/원본package main
func wordSubsets(words1 []string, words2 []string) []string {
maxCount := make([]int, 26)
for _, word := range words2 {
count := getCount(word)
for i := 0; i < 26; i++ {
if count[i] > maxCount[i] {
maxCount[i] = count[i]
}
}
}
result := []string{}
for _, word := range words1 {
count := getCount(word)
if isUniversal(count, maxCount) {
result = append(result, word)
}
}
return result
}
func getCount(word string) []int {
count := make([]int, 26)
for _, char := range word {
count[char-'a']++
}
return count
}
func isUniversal(count []int, maxCount []int) bool {
for i := 0; i < 26; i++ {
if count[i] < maxCount[i] {
return false
}
}
return true
}Algorithm
1⃣Подсчитать максимальное количество каждой буквы в каждом слове из words2.
2⃣Проверить каждое слово из words1, если оно содержит не менее максимального количества каждой буквы, которая встречается в словах из words2.
3⃣Вернуть 배열 слов из words1, которые удовлетворяют этому условию.
😎
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