Merge pull request #2491 from Cyjin-jani/main

[Cyjin-jani] WEEK 05 Solutions

Author: Cyjin-jani

best-time-to-buy-and-sell-stock/Cyjin-jani.js

@@ -0,0 +1,43 @@
+// naive하게 풀면, time limit 초과함
+// O(n^2) 풀이가 되기 때문..
+const maxProfit_naive = function (prices) {
+  let profit = 0;
+
+  for (let i = 0; i < prices.length - 1; i++) {
+    const buy = prices[i];
+    for (let j = i + 1; j < prices.length; j++) {
+      const newProfit = prices[j] - buy;
+      if (newProfit > profit) {
+        profit = newProfit;
+      }
+    }
+  }
+
+  return profit;
+};
+
+// 투포인터 사용하여 풀기
+// tc: O(n)
+// sc: O(1)
+const maxProfit = function (prices) {
+  let buyIdx = 0;
+  let sellIdx = 1;
+  let profit = 0;
+
+  while (sellIdx < prices.length) {
+    let buyPrice = prices[buyIdx];
+    let sellPrice = prices[sellIdx];
+
+    // 더 낮은 가격에 매수 가능한 날을 찾으면 바로 거기서부터 재탐색
+    if (buyPrice > sellPrice) {
+      buyIdx = sellIdx;
+    } else {
+      let newProfit = sellPrice - buyPrice;
+      profit = Math.max(profit, newProfit);
+    }
+
+    sellIdx++;
+  }
+
+  return profit;
+};

group-anagrams/Cyjin-jani.js

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+const isAnagram = function (s, t) {
+  if (s.length !== t.length) return false;
+
+  const data = new Map();
+
+  for (let char of s) {
+    data.set(char, (data.get(char) || 0) + 1);
+  }
+
+  for (let char of t) {
+    if (!data.get(char)) return false;
+    data.set(char, data.get(char) - 1);
+  }
+
+  return true;
+};
+
+// Time Limit Exceeded로 fail..
+// tc: O(n^2);
+// sc: O(n)
+const groupAnagrams_naive = function (strs) {
+  if (strs.length < 2) return [strs];
+
+  const answer = [];
+
+  while (strs.length > 0) {
+    let temp = [strs[0]];
+
+    if (strs.length < 2) return answer;
+
+    strs.splice(0, 1);
+
+    for (let j = 0; j < strs.length; j++) {
+      if (isAnagram(temp[0], strs[j])) {
+        // 같다면, temp 배열에 넣음
+        temp.push(strs[j]);
+      }
+    }
+    answer.push(temp);
+
+    temp.forEach((t) => {
+      const idx = strs.indexOf(t);
+      if (idx !== -1) {
+        strs.splice(idx, 1);
+      }
+    });
+
+    if (strs.length === 1) {
+      answer.push(strs);
+    }
+  }
+
+  return answer;
+};
+
+// splice 같은 로직이 없어서 겨우 TLE을 통과했지만 여전히 O(n²)인 점은 변함이 없음.
+const groupAnagrams_set = function (strs) {
+  const visited = new Set();
+  const answer = [];
+
+  for (let i = 0; i < strs.length; i++) {
+    if (visited.has(i)) continue;
+
+    const group = [strs[i]];
+
+    for (let j = i + 1; j < strs.length; j++) {
+      if (!visited.has(j) && isAnagram(strs[i], strs[j])) {
+        group.push(strs[j]);
+        visited.add(j);
+      }
+    }
+
+    answer.push(group);
+  }
+
+  return answer;
+};
+
+//! AI로부터 힌트를 얻어 풀어봤습니다..
+// tc: O(nlogn)
+// sc: O(n)
+const groupAnagrams = function (strs) {
+  const map = new Map();
+
+  for (const str of strs) {
+    const key = str.split('').sort().join('');
+
+    if (!map.has(key)) {
+      map.set(key, []);
+    }
+    map.get(key).push(str);
+  }
+
+  return [...map.values()];
+};