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[moonjonghoo] Week 05 Solutions #1400

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May 3, 2025
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24 changes: 24 additions & 0 deletions best-time-to-buy-and-sell-stock/moonjonghoo.js
Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function (prices) {
let minPrices = Infinity;
let profit = 0;
for (price of prices) {
minPrices = Math.min(price, minPrices);
profit = Math.max(profit, price - minPrices);
}
return profit;
};

var maxProfit = function (prices) {
let maxProfit = 0;
let currentProfit = 0;
for (let i = 1; i < prices.length; i++) {
let diff = prices[i] - prices[i - 1];
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어제와 오늘의 가격 차이를 계산하고 그 합이 최대가 되는 구간을 찾는 로직이군여!

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맞습니다!

currentProfit = Math.max(0, currentProfit + diff);
maxProfit = Math.max(maxProfit, currentProfit);
}
return maxProfit;
};
22 changes: 22 additions & 0 deletions group-anagrams/moonjonghoo.js
Original file line number Diff line number Diff line change
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/**
* @param {string[]} strs
* @return {string[][]}
*/
var groupAnagrams = function (strs) {
let hashMap = new Map();
for (let i = 0; i < strs.length; i++) {
let key = strs[i].split("").sort().join("");
if (!hashMap.has(key)) {
hashMap.set(key, [strs[i]]);
} else {
hashMap.set(key, [...hashMap.get(key), strs[i]]);
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hashMap.get(key).push(strs[i]);

을 활용하면 새 배열을 생성하지 않고 바로 push 할 수 있습니다.

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아 맞습니다. 반영을 못했네요 ㅠㅠ

}
}
let answer = [];
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Array.from(hashMap.values()) 를 활용할 수도 있습니다.

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오 그러네요 감사합니다.

for ([key, value] of hashMap) {
answer.push(value);
}
return answer.sort((a, b) => a.length - b.length).map((arr) => arr.sort());
};

console.log(groupAnagrams(["eat", "tea", "tan", "ate", "nat", "bat"]));
31 changes: 31 additions & 0 deletions word-break/moonjonghoo.js
Original file line number Diff line number Diff line change
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/**
* @param {string} s
* @param {string[]} wordDict
* @return {boolean}
*/
var wordBreak = function (s, wordDict) {
const memo = new Map(); // 실패/성공 여부 기억
const wordSet = new Set(wordDict); // lookup 최적화

function canBreak(sub) {
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재귀 + 메모이제이션 풀이 잘 보고 갑니다!

if (sub === "") return true;

if (memo.has(sub)) return memo.get(sub);

for (let i = 1; i <= sub.length; i++) {
const prefix = sub.slice(0, i);
if (wordSet.has(prefix)) {
const suffix = sub.slice(i);
if (canBreak(suffix)) {
memo.set(sub, true);
return true;
}
}
}

memo.set(sub, false); // 실패한 경우도 기억
return false;
}

return canBreak(s);
};