[hyunjung-choi] WEEK 06 solutions

container-with-most-water/hyunjung-choi.kt

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+/**
+ * 시간 복잡도: O(n)
+ * 공간 복잡도: O(1)
+ */
+
+class Solution {
+    fun maxArea(height: IntArray): Int {
+        var i = 0
+        var j = height.size - 1
+        var max = 0
+
+        while (i < j) {
+            val h = minOf(height[i], height[j])
+            max = maxOf(max, (j - i) * h)
+
+            if (height[i] <= height[j]) i++
+            else j--
+        }
+
+        return max
+    }
+}

design-add-and-search-words-data-structure/hyunjung-choi.kt

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+class WordDictionary() {
+
+    class TrieNode {
+        val children = mutableMapOf<Char, TrieNode>()
+        var isEndOfWord = false
+    }
+
+    private val root = TrieNode()
+
+    fun addWord(word: String) {
+        var current = root
+
+        for (char in word) {
+            if (char !in current.children) {
+                current.children[char] = TrieNode()
+            }
+            current = current.children[char]!!
+        }
+
+        current.isEndOfWord = true
+    }
+
+    fun search(word: String): Boolean {
+        return searchHelper(word, 0, root)
+    }
+
+    private fun searchHelper(word: String, index: Int, node: TrieNode): Boolean {
+        if (index == word.length) {
+            return node.isEndOfWord
+        }
+
+        val char = word[index]
+
+        return if (char == '.') {
+            for (child in node.children.values) {
+                if (searchHelper(word, index + 1, child)) {
+                    return true
+                }
+            }
+            false
+        } else {
+            val child = node.children[char] ?: return false
+            searchHelper(word, index + 1, child)
+        }
+    }
+}

valid-parentheses/hyunjung-choi.kt

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+/**
+ * 시간 복잡도: O(n)
+ * 공간 복잡도: O(n)
+ */
+
+class Solution {
+    fun isValid(s: String): Boolean {
+        if (s.length % 2 == 1) return false
+
+        val stack = Stack<Char>()
+        val pairs = mapOf(')' to '(', ']' to '[', '}' to '{')
+
+        for (char in s) {
+            when {
+                char in pairs -> {
+                    if (stack.isEmpty() || stack.pop() != pairs[char]) {
+                        return false
+                    }
+                }
+
+                else -> stack.push(char)
+            }
+        }
+
+        return stack.isEmpty()
+    }
+}