Merge pull request #457 from JEONGHWANMIN/main

[환미니니] Week5 문제풀이

Author: JEONGHWANMIN

3sum/hwanmini.js

@@ -0,0 +1,39 @@
+// 시간복잡도 O(n2)
+// 공간복잡도 O(k) (k는 결과 개수)
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var threeSum = function(nums) {
+    nums.sort((a, b) => a - b);
+
+    let result = []
+
+    for (let i = 0 ; i <= nums.length - 2; i++) {
+        if (i > 0 && nums[i] === nums[i-1]) continue
+
+        const curNum = nums[i]
+        let leftIdx = i+1;
+        let rightIdx = nums.length - 1;
+
+        while (leftIdx < rightIdx) {
+            const leftNum = nums[leftIdx]
+            const rightNum = nums[rightIdx]
+            const threeSum = curNum + leftNum + rightNum
+            if (threeSum === 0) {
+                result.push([curNum, leftNum, rightNum])
+                while (leftIdx < rightIdx && nums[leftIdx] === nums[leftIdx+1]) leftIdx++
+                while (leftIdx < rightIdx && nums[rightIdx] === nums[rightIdx-1]) rightIdx--
+                leftIdx++
+                rightIdx--
+            } else if (threeSum < 0) {
+                leftIdx = leftIdx + 1
+            } else if (threeSum > 0) {
+                rightIdx = rightIdx - 1
+            }
+        }
+
+    }
+
+    return result
+};

best-time-to-buy-and-sell-stock/hwanmini.js

@@ -0,0 +1,23 @@
+// 시간복잡도: O(n)
+// 공간복잡도: O(1)
+/**
+ * @param {number[]} prices
+ * @return {number}
+ */
+var maxProfit = function(prices) {
+    if (!prices || !prices.length) return 0
+
+    let maxBenefit = 0;
+    let buyPrice = Infinity;
+
+    for (let i = 0 ; i < prices.length ; i++) {
+        buyPrice = Math.min(buyPrice, prices[i]);
+        maxBenefit = Math.max(maxBenefit, prices[i] - buyPrice)
+    }
+
+
+    return maxBenefit
+};
+
+console.log(maxProfit([7,1,5,3,6,4]))
+console.log(maxProfit([7,6,4,3,1]))

group-anagrams/hwanmini.js

@@ -0,0 +1,27 @@
+// n 배열길이, k 단어길이
+// 시간복잡도: O(n * k log k)
+// 공간복잡도: O(n * k)
+/**
+ * @param {string[]} strs
+ * @return {string[][]}
+ */
+var groupAnagrams = function(strs) {
+    const map = new Map()
+
+    for (const word of strs) {
+        const sortedWord = word.split("").sort().join("")
+
+        if (!map.has(sortedWord)) {
+            map.set(sortedWord, [])
+        }
+
+        map.set(sortedWord, [...map.get(sortedWord), word]);
+    }
+
+
+    return [...map.values()]
+};
+
+console.log(groupAnagrams(["eat","tea","tan","ate","nat","bat"]))
+console.log(groupAnagrams([""]))
+console.log(groupAnagrams(["a"]))

implement-trie-prefix-tree/hwanmini.js

@@ -0,0 +1,60 @@
+// 시간복잡도
+// insert: O(L), search: O(L), startsWith: O(L)
+// 공간복잡도: O(N * L)
+
+var TrieNode = function(){
+    this.children = {};
+    this.endOfWord = false;
+}
+
+var Trie = function() {
+    this.root = new TrieNode();
+};
+
+/**
+ * @param {string} word
+ * @return {void}
+ */
+Trie.prototype.insert = function(word) {
+    let curNode = this.root;
+    for (let i = 0; i < word.length; i++) {
+        const char = word[i];
+        if (!curNode.children[char]) {
+            curNode.children[char] = new TrieNode();
+        }
+        curNode = curNode.children[char];
+    }
+    curNode.endOfWord = true;
+};
+
+/**
+ * @param {string} word
+ * @return {boolean}
+ */
+Trie.prototype.search = function(word) {
+    let curNode = this.root;
+    for (let i = 0; i < word.length; i++) {
+        const char = word[i];
+        if (!curNode.children[char]) {
+            return false;
+        }
+        curNode = curNode.children[char];
+    }
+    return curNode.endOfWord;
+};
+
+/**
+ * @param {string} prefix
+ * @return {boolean}
+ */
+Trie.prototype.startsWith = function(prefix) {
+    let curNode = this.root;
+    for (let i = 0; i < prefix.length; i++) {
+        const char = prefix[i];
+        if (!curNode.children[char]) {
+            return false;
+        }
+        curNode = curNode.children[char];
+    }
+    return true;
+};

word-break/hwanmini.js

@@ -0,0 +1,27 @@
+// 시간복잡도: O(n * m * n)
+// 공간복잡도: O(n)
+/**
+ * @param {string} s
+ * @param {string[]} wordDict
+ * @return {boolean}
+ */
+var wordBreak = function(s, wordDict) {
+    const wordChecks = Array.from({length: s.length}).fill(false)
+    const dp = [true, ...wordChecks]
+
+    for (let i = 0 ; i <= s.length; i++) {
+        for (let j = 0; j < wordDict.length; j++) {
+            const dict = wordDict[j]
+            const dictLen  = dict.length
+
+            const checkWord = s.slice(i - dictLen, i)
+            if (dp[i - dictLen] && checkWord === dict) {
+                dp[i] = true
+            }
+        }
+    }
+
+    return dp[dp.length - 1]
+};
+
+console.log(wordBreak("leetcode", ["leet","code"]))