From 858b7402425aedcb02942478ac03ba227b35c720 Mon Sep 17 00:00:00 2001
From: BaeKwangho <jason@monymony.co>
Date: Mon, 15 Dec 2025 23:00:55 +0900
Subject: [PATCH 1/2] week6

---
 container-with-most-water/Baekwangho.ts       |  43 ++++++++
 .../Baekwangho.ts                             | 102 ++++++++++++++++++
 valid-parentheses/Baekwangho.ts               |  35 ++++++
 3 files changed, 180 insertions(+)
 create mode 100644 container-with-most-water/Baekwangho.ts
 create mode 100644 design-add-and-search-words-data-structure/Baekwangho.ts
 create mode 100644 valid-parentheses/Baekwangho.ts

diff --git a/container-with-most-water/Baekwangho.ts b/container-with-most-water/Baekwangho.ts
new file mode 100644
index 000000000..5fd8851a6
--- /dev/null
+++ b/container-with-most-water/Baekwangho.ts
@@ -0,0 +1,43 @@
+/**
+f((a,ax), (b,bx)) => min(ax, bx) * (b - a) (단, b > a)
+라고 하자
+(1,1) (2,8) => 1
+(1,1) (3,6) => 2
+(2,8) (3,6) => 6
+(1,1) (4,2) => 3
+(2,8) (4,2) => 4
+(3,6) (4,2) => 2
+...
+(2,8) (5,5) => 15
+
+근데 생각해보니, 순회하면서 가장 큰 왼쪽 봉을 구하고, 한칸씩 움직이며 그 봉으로부터의 너비를 계산하면 되는 것 같음.
+그러다가 왼쪽 봉보다 더 큰 봉이 있다면 거기서부터 시작하되, 최댓값은 그대로 두기
+
+를 생각했다가 보니, 투 포인터를 사용해서 가장 큰 수를 구하는 방법이 있을 것 같음.
+힌트를 보니, 낮은 바를 항상 움직이라고 함.
+
+시간 복잡도: o(n) / 공간 복잡도: o(1)
+*/
+function maxArea(height: number[]): number {
+  let left = 0;
+  let right = height.length - 1;
+  let max = 0;
+
+  while (left < right) {
+    const ltemp = height[left];
+    const rtemp = height[right];
+
+    if (ltemp < rtemp) {
+      left++;
+    } else {
+      right--;
+    }
+
+    const result = (right - left + 1) * Math.min(ltemp, rtemp);
+    if (result > max) {
+      max = result;
+    }
+  }
+
+  return max;
+}
diff --git a/design-add-and-search-words-data-structure/Baekwangho.ts b/design-add-and-search-words-data-structure/Baekwangho.ts
new file mode 100644
index 000000000..66e3157b6
--- /dev/null
+++ b/design-add-and-search-words-data-structure/Baekwangho.ts
@@ -0,0 +1,102 @@
+/**
+search 의 역할을 보았을 때, 각 글자의 길이별로 set 을 두는 형태로 구현해볼 수 있을 듯 하다.
+class WordDictionary {
+    setArray: Array<Set<string>>
+    wordSet: Set<string>
+    constructor() {
+        this.setArray = Array.from({length:25}, () => new Set())
+        this.wordSet = new Set()
+    }
+
+    addWord(word: string): void {
+        for(let i=0; i<word.length; i++) {
+            this.setArray[i].add(word[i])
+        }
+        this.wordSet.add(word)
+    }
+
+    search(word: string): boolean {
+        if (word.includes('.')) {
+            for(let i=0; i<word.length; i++) {
+                console.log(word[i], this.setArray[i])
+                if (word[i] === '.') {
+                    if (this.setArray[i].size) continue
+                    return false
+                } else {
+                    if (this.setArray[i].has(word[i])) continue
+                    return false
+                }
+            }
+            return true
+        } else {
+            if (this.wordSet.has(word)) return true
+            return false
+        }
+    }
+}
+ */
+
+/**
+실패했다. 한글자의 '.' 케이스를 충족시킬 수 없었다. 
+실제 word 기반한 데이터가 아니기 때문에 발생하는 오류인 듯 하다.
+힌트에서 trie 를 얘기하니, 그것을 구현해보자.
+...
+근데 결국 search 에서 탐색이 최악의 경우 '..............n' 등에서 최악일 듯 하다.
+...
+하지만 힌트를 보니, dfs 기반으로 구현하라고 하는구나.
+
+시간 복잡도: O(L) ~ O(n * L) / 공간 복잡도: O(n * L)
+O(L) 은 search 에 wildcard 가 없을 때, L은 글자 길이
+ */
+class WordDictionary {
+  headMap: any;
+
+  constructor() {
+    this.headMap = {};
+  }
+
+  addWord(word: string): void {
+    let curMap = this.headMap;
+
+    for (let i = 0; i < word.length; i++) {
+      const cur = curMap[word[i]];
+      if (!cur) {
+        const temp = {};
+        curMap[word[i]] = temp;
+        curMap = temp;
+      } else {
+        curMap = cur;
+      }
+    }
+    curMap["count"] ? curMap["count"]++ : (curMap["count"] = 1);
+  }
+
+  search(word: string): boolean {
+    function call(dict: any, subset?: string): boolean {
+      if (dict["count"] && !subset) {
+        return true;
+      }
+
+      const firstChar = subset![0];
+      if (firstChar === ".") {
+        return Object.values(dict).some((el) =>
+          call(el, subset!.slice(1, subset!.length))
+        );
+      } else {
+        const firstCharDict = dict[firstChar];
+        if (!firstCharDict) return false;
+
+        return call(firstCharDict, subset!.slice(1, subset!.length));
+      }
+    }
+
+    return call(this.headMap, word);
+  }
+}
+
+/**
+ * Your WordDictionary object will be instantiated and called as such:
+ * var obj = new WordDictionary()
+ * obj.addWord(word)
+ * var param_2 = obj.search(word)
+ */
diff --git a/valid-parentheses/Baekwangho.ts b/valid-parentheses/Baekwangho.ts
new file mode 100644
index 000000000..600b1e700
--- /dev/null
+++ b/valid-parentheses/Baekwangho.ts
@@ -0,0 +1,35 @@
+/**
+ * 공간 복잡도: o(n) 시간 복잡도: o(n)
+ */
+function isValid(s: string): boolean {
+  const stack = [];
+
+  const start = ["(", "{", "["];
+
+  for (let i = 0; i < s.length; i++) {
+    if (start.includes(s[i])) {
+      stack.push(s[i]);
+    } else {
+      const last = stack.pop();
+      switch (last) {
+        case "(": {
+          if (s[i] !== ")") return false;
+          break;
+        }
+        case "{": {
+          if (s[i] !== "}") return false;
+          break;
+        }
+        case "[": {
+          if (s[i] !== "]") return false;
+          break;
+        }
+        default: {
+          return false;
+        }
+      }
+    }
+  }
+
+  return stack.length > 0 ? false : true;
+}

From 1204a0a222b6f2a0fbada2b6dc7d45674b9991ba Mon Sep 17 00:00:00 2001
From: BaeKwangho <jason@monymony.co>
Date: Thu, 18 Dec 2025 00:00:32 +0900
Subject: [PATCH 2/2] week 06

---
 longest-increasing-subsequence/Baekwangho.ts |  0
 spiral-matrix/Baekwangho.ts                  | 86 ++++++++++++++++++++
 2 files changed, 86 insertions(+)
 create mode 100644 longest-increasing-subsequence/Baekwangho.ts
 create mode 100644 spiral-matrix/Baekwangho.ts

diff --git a/longest-increasing-subsequence/Baekwangho.ts b/longest-increasing-subsequence/Baekwangho.ts
new file mode 100644
index 000000000..e69de29bb
diff --git a/spiral-matrix/Baekwangho.ts b/spiral-matrix/Baekwangho.ts
new file mode 100644
index 000000000..15a1578a9
--- /dev/null
+++ b/spiral-matrix/Baekwangho.ts
@@ -0,0 +1,86 @@
+/**
+(0,0) 에서 시작해서, 위 아래 끝값 및 왼쪽 오른쪽의 끝값을 갱신해가며
+현재 위치가 이보다 커질 경우 각각을 -1 하고 방향을 바꿔준다.
+ */
+function spiralOrder(matrix: number[][]): number[] {
+  let hmin = 0,
+    hmax = matrix[0].length - 1,
+    vmin = 0,
+    vmax = matrix.length - 1;
+  let x = 0,
+    y = 0;
+  const result = [];
+  let currentStep = [1, 0];
+
+  while (hmax >= hmin && vmax >= vmin) {
+    result.push(matrix[y][x]);
+
+    const next = applyCurrentStep(currentStep, x, y, hmin, hmax, vmin, vmax);
+    currentStep = next.nextStep;
+    x = next.x;
+    y = next.y;
+    hmin = next.hmin;
+    hmax = next.hmax;
+    vmin = next.vmin;
+    vmax = next.vmax;
+
+    // console.log(currentStep, x, y, hmin, hmax, vmin, vmax)
+  }
+  return result;
+}
+
+function applyCurrentStep(
+  currentStep: number[],
+  x: number,
+  y: number,
+  hmin: number,
+  hmax: number,
+  vmin: number,
+  vmax: number
+) {
+  let nextStep = currentStep;
+
+  switch (true) {
+    case currentStep[0] === 1 && currentStep[1] === 0: {
+      if (x >= hmax) {
+        nextStep = [0, 1];
+        y += 1;
+        vmin += 1;
+      } else {
+        x += 1;
+      }
+      break;
+    }
+    case currentStep[0] === 0 && currentStep[1] === 1: {
+      if (y >= vmax) {
+        nextStep = [-1, 0];
+        x -= 1;
+        hmax -= 1;
+      } else {
+        y += 1;
+      }
+      break;
+    }
+    case currentStep[0] === -1 && currentStep[1] === 0: {
+      if (hmin >= x) {
+        nextStep = [0, -1];
+        y -= 1;
+        vmax -= 1;
+      } else {
+        x -= 1;
+      }
+      break;
+    }
+    case currentStep[0] === 0 && currentStep[1] === -1: {
+      if (vmin >= y) {
+        nextStep = [1, 0];
+        x += 1;
+        hmin += 1;
+      } else {
+        y -= 1;
+      }
+      break;
+    }
+  }
+  return { nextStep, x, y, hmin, hmax, vmin, vmax };
+}