[HoonDongKang] Week 8 solutions

clone-graph/HoonDongKang.ts

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+/**
+ * [Problem]: [133] Clone Graph
+ * (https://leetcode.com/problems/longest-repeating-character-replacement/description/)
+ */
+
+//  Definition for _Node.
+class _Node {
+    val: number;
+    neighbors: _Node[];
+
+    constructor(val?: number, neighbors?: _Node[]) {
+        this.val = val === undefined ? 0 : val;
+        this.neighbors = neighbors === undefined ? [] : neighbors;
+    }
+}
+
+function cloneGraph(node: _Node | null): _Node | null {
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    function dfsFunc(node: _Node | null): _Node | null {
+        if (!node) return null;
+        const map = new Map<_Node, _Node>();
+
+        function dfs(node: _Node): _Node {
+            if (map.has(node)) return map.get(node)!;
+
+            const copy = new _Node(node.val);
+            map.set(node, copy);
+
+            for (const neighbor of node.neighbors) {
+                copy.neighbors.push(dfs(neighbor));
+            }
+
+            return copy;
+        }
+
+        return dfs(node);
+    }
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    function bfsFunc(node: _Node | null): _Node | null {
+        if (!node) return null;
+
+        const map = new Map<_Node, _Node>();
+        const queue: _Node[] = [];
+
+        const clone = new _Node(node.val);
+        map.set(node, clone);
+        queue.push(node);
+
+        while (queue.length > 0) {
+            const cur = queue.shift()!;
+
+            for (const neighbor of cur.neighbors) {
+                if (!map.has(neighbor)) {
+                    map.set(neighbor, new _Node(neighbor.val));
+                    queue.push(neighbor);
+                }
+
+                map.get(cur)!.neighbors.push(map.get(neighbor)!);
+            }
+        }
+        return clone;
+    }
+}

longest-common-subsequence/HoonDongKang.ts

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+/**
+ * [Problem]: [1143] Longest Common Subsequence
+ * (https://leetcode.com/problems/longest-common-subsequence/description/)
+ */
+function longestCommonSubsequence(text1: string, text2: string): number {
+    //시간복잡도 O(2^n)
+    //공간복잡도 O(n)
+    // Time Limit Exceeded
+    function dfsFunc(text1: string, text2: string): number {
+        function dfs(i: number, j: number): number {
+            if (i === text1.length || j === text2.length) return 0;
+
+            if (text1[i] === text2[j]) {
+                return 1 + dfs(i + 1, j + 1);
+            } else {
+                return Math.max(dfs(i, j + 1), dfs(i + 1, j));
+            }
+        }
+
+        return dfs(0, 0);
+    }
+
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    function memoizationFunc(text1: string, text2: string): number {
+        const memo = new Map<string, number>();
+
+        function dfs(i: number, j: number) {
+            if (i === text1.length || j === text2.length) return 0;
+
+            const key = `${i}, ${j}`;
+            if (memo.has(key)) return memo.get(key)!;
+
+            let result = 0;
+            if (text1[i] === text2[j]) {
+                result = 1 + dfs(i + 1, j + 1);
+            } else {
+                result = Math.max(dfs(i, j + 1), dfs(i + 1, j));
+            }
+
+            memo.set(key, result);
+            return result;
+        }
+
+        return dfs(0, 0);
+    }
+
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    function dpFunc(tex1: string, text2: string): number {
+        const length1 = text1.length;
+        const length2 = text2.length;
+        const dp: number[][] = Array.from({ length: length1 + 1 }, () =>
+            Array(length2 + 1).fill(0)
+        );
+
+        for (let i = 1; i <= length1; i++) {
+            for (let j = 1; j <= length2; j++) {
+                if (text1[i - 1] === text2[j - 1]) {
+                    dp[i][j] = 1 + dp[i - 1][j - 1];
+                } else {
+                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
+                }
+            }
+        }
+
+        return dp[length1][length2];
+    }
+}

longest-repeating-character-replacement/HoonDongKang.ts

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+/**
+ * [Problem]: [424] Longest Repeating Character Replacement
+ * (https://leetcode.com/problems/longest-repeating-character-replacement/description/)
+ */
+function characterReplacement(s: string, k: number): number {
+    //시간복잡도 O(n)
+    //공간복잡도 O(n)
+    let left = 0;
+    let right = 0;
+    let result = 0;
+    let map = new Map<string, number>();
+
+    while (right < s.length) {
+        map.set(s[right], (map.get(s[right]) || 0) + 1);
+
+        const maxFreq = Math.max(...Array.from(map.values()));
+        if (right - left + 1 - maxFreq > k) {
+            map.set(s[left], map.get(s[left])! - 1);
+            left++;
+        }
+
+        result = Math.max(result, right - left + 1);
+        right++;
+    }
+
+    return result;
+}

palindromic-substrings/HoonDongKang.ts

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+/**
+ * [Problem]: [647] Palindromic Substrings
+ * (https://leetcode.com/problems/palindromic-substrings/description/)
+ */
+
+function countSubstrings(s: string): number {
+    //시간복잡도 O(n^3)
+    //공간복잡도 O(1)
+    function bruteForceFunc(s: string): number {
+        let count = 0;
+
+        for (let i = 0; i < s.length; i++) {
+            for (let j = i; j < s.length; j++) {
+                if (isPanlindrome(i, j)) count++;
+            }
+        }
+
+        function isPanlindrome(left: number, right: number): boolean {
+            while (left < right) {
+                if (s[left] !== s[right]) return false;
+                left++;
+                right--;
+            }
+
+            return true;
+        }
+
+        return count;
+    }
+    //시간복잡도 O(n^2)
+    //공간복잡도 O(n^2)
+    function dpFunc(s: string): number {
+        const dp = new Map<string, boolean>();
+        let count = 0;
+
+        for (let end = 0; end < s.length; end++) {
+            for (let start = end; start >= 0; start--) {
+                const key = `${start},${end}`;
+                if (start === end) {
+                    dp.set(key, true);
+                } else if (start + 1 === end) {
+                    dp.set(key, s[start] === s[end]);
+                } else {
+                    const innerKey = `${start + 1},${end - 1}`;
+                    const isInnerPalindrome = dp.get(innerKey) || false;
+                    dp.set(key, s[start] === s[end] && isInnerPalindrome);
+                }
+
+                if (dp.get(key)) {
+                    count++;
+                }
+            }
+        }
+
+        return count;
+    }
+}

reverse-bits/HoonDongKang.ts

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+/**
+ * [Problem]: [190] Reverse Bits
+ * (https://leetcode.com/problems/reverse-bits/description/)
+ */
+function reverseBits(n: number): number {
+    //시간복잡도 O(1)
+    //공간복잡도 O(1)
+    function stackFunc(n: number): number {
+        const stack: number[] = [];
+        let output = 0;
+        let scale = 1;
+
+        while (stack.length < 32) {
+            stack.push(n % 2);
+            n = Math.floor(n / 2);
+        }
+
+        while (stack.length) {
+            output += stack.pop()! * scale;
+            scale *= 2;
+        }
+
+        return output;
+    }
+    //시간복잡도 O(1)
+    //공간복잡도 O(1)
+    function bitFunc(n: number): number {
+        let result = 0;
+        for (let i = 0; i < 32; i++) {
+            result <<= 1;
+            result |= n & 1;
+            n >>>= 1;
+        }
+
+        return result >>> 0;
+    }
+}