[minji-go] week 14 solutions

binary-tree-level-order-traversal/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/binary-tree-level-order-traversal/">week14-2. binary-tree-level-order-traversal</a>
+ * <li>Description: Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level)</li>
+ * <li>Topics: Tree, Breadth-First Search, Binary Tree</li>
+ * <li>Time Complexity: O(N), Runtime 1ms       </li>
+ * <li>Space Complexity: O(N), Memory 45.3MB    </li>
+ */
+
+class Solution {
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        if (root == null) {
+            return Collections.emptyList();
+        }
+
+        Queue<TreeNode> queue = new LinkedList<>();
+        queue.offer(root);
+
+        List<List<Integer>> results = new ArrayList<>();
+        while (!queue.isEmpty()) {
+            List<Integer> result = new ArrayList<>();
+
+            int size = queue.size();
+            for (int i = 0; i < size; i++) {
+                TreeNode node = queue.poll();
+                result.add(node.val);
+                if (node.left != null) queue.offer(node.left);
+                if (node.right != null) queue.offer(node.right);
+            }
+            results.add(result);
+        }
+
+        return results;
+    }
+}

counting-bits/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/counting-bits/">week14-1. counting-bits</a>
+ * <li>Description: Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i</li>
+ * <li>Topics: Dynamic Programming, Bit Manipulation</li>
+ * <li>Time Complexity: O(N), Runtime 2ms       </li>
+ * <li>Space Complexity: O(N), Memory 46.64MB   </li>
+ */
+class Solution {
+    public int[] countBits(int n) {
+        int[] count = new int[n + 1];
+
+        int offset = 1;
+        for (int i = 1; i <= n; i++) {
+            if (i == offset * 2) offset *= 2;
+            count[i] = count[i - offset] + 1;
+        }
+
+        return count;
+    }
+}

house-robber-ii/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/house-robber-ii/">week14-3. house-robber-ii</a>
+ * <li>Description: Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police</li>
+ * <li>Topics: Array, Dynamic Programming   </li>
+ * <li>Time Complexity: O(N), Runtime 0ms   </li>
+ * <li>Space Complexity: O(1), Memory 40.6MB</li>
+ */
+
+class Solution {
+    public int rob(int[] nums) {
+        int n = nums.length;
+        if (n == 1) return nums[0];
+        if (n == 2) return Math.max(nums[0], nums[1]);
+
+        return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1));
+    }
+
+    public int rob(int[] nums, int start, int end) {
+        int prev1 = 0, prev2 = 0;
+        for (int i = start; i <= end; i++) {
+            int temp = Math.max(prev2, prev1 + nums[i]);
+            prev1 = prev2;
+            prev2 = temp;
+        }
+        return prev2;
+    }
+}

linked-list-cycle/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/linked-list-cycle/">week9-1. linked-list-cycle</a>
+ * <li>Description: Return true if there is a cycle in the linked list. </li>
+ * <li>Topics: Hash Table, Linked List, Two Pointers</li>
+ * <li>Time Complexity: O(N), Runtime 0ms   </li>
+ * <li>Space Complexity: O(1), Memory 44.37MB</li>
+ */
+public class Solution {
+    public boolean hasCycle(ListNode head) {
+        if (head == null) {
+            return false;
+        }
+
+        ListNode slow = head;
+        ListNode fast = head.next;
+
+        while (slow != fast) {
+            if (fast == null || fast.next == null) {
+                return false;
+            }
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+
+        return true;
+    }
+}

pacific-atlantic-water-flow/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/pacific-atlantic-water-flow/">week9-2. pacific-atlantic-water-flow</a>
+ * <li>Description: Return a 2D list of grid coordinates result where result[i] = [ri, ci] denotes that rain water can flow from cell (ri, ci) to both the Pacific and Atlantic oceans. </li>
+ * <li>Topics: Array, Depth-First Search, Breadth-First Search, Matrix</li>
+ * <li>Time Complexity: O(MN), Runtime 9ms   </li>
+ * <li>Space Complexity: O(MN), Memory 45.18MB</li>
+ */
+class Solution {
+    private int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
+
+    public List<List<Integer>> pacificAtlantic(int[][] heights) {
+        int m = heights.length;
+        int n = heights[0].length;
+
+        Queue<int[]> pacificQueue = new LinkedList<>();
+        boolean[][] pacific = new boolean[m][n];
+        for (int i = 0; i < m; i++) {
+            pacific[i][0] = true;
+            pacificQueue.offer(new int[]{i, 0});
+        }
+        for (int i = 0; i < n; i++) {
+            pacific[0][i] = true;
+            pacificQueue.offer(new int[]{0, i});
+        }
+        bfs(heights, pacificQueue, pacific);
+
+        Queue<int[]> atlanticQueue = new LinkedList<>();
+        boolean[][] atlantic = new boolean[m][n];
+        for (int i = 0; i < m; i++) {
+            atlantic[i][n - 1] = true;
+            atlanticQueue.offer(new int[]{i, n - 1});
+        }
+        for (int i = 0; i < n; i++) {
+            atlantic[m - 1][i] = true;
+            atlanticQueue.offer(new int[]{m - 1, i});
+        }
+        bfs(heights, atlanticQueue, atlantic);
+
+        List<List<Integer>> result = new ArrayList<>();
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if (pacific[i][j] && atlantic[i][j]) {
+                    result.add(List.of(i, j));
+                }
+            }
+        }
+        return result;
+    }
+
+
+    private void bfs(int[][] heights, Queue<int[]> queue, boolean[][] visit) {
+        while (!queue.isEmpty()) {
+            int[] curr = queue.poll();
+            int cr = curr[0];
+            int cc = curr[1];
+
+            for (int[] dir : directions) {
+                int nr = cr + dir[0];
+                int nc = cc + dir[1];
+
+                if (nr < 0 || nr > heights.length - 1 || nc < 0 || nc > heights[0].length - 1 || visit[nr][nc]) {
+                    continue;
+                }
+                if (heights[nr][nc] >= heights[cr][cc]) {
+                    visit[nr][nc] = true;
+                    queue.offer(new int[]{nr, nc});
+                }
+            }
+        }
+    }
+}

word-search-ii/minji-go.java

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+/**
+ * <a href="https://leetcode.com/problems/word-search-ii/">week14-5. word-search-ii</a>
+ * <li>Description: Given an m x n board of characters and a list of strings words, return all words on the board.</li>
+ * <li>Topics: Array, String, Backtracking, Trie, Matrix</li>
+ * <li>Time Complexity: O(M*N*4^L), Runtime 446ms   </li>
+ * <li>Space Complexity: O(K*L), Memory 44.81MB</li>
+ * <li>Note: Refer to answer </li>
+ */
+class Solution {
+
+    class TrieNode {
+        Map<Character, TrieNode> children = new HashMap<>();
+        String word = null;
+    }
+
+    public List<String> findWords(char[][] board, String[] words) {
+        TrieNode root = buildTrie(words);
+        List<String> result = new ArrayList<>();
+
+        for (int i = 0; i < board.length; i++) {
+            for (int j = 0; j < board[0].length; j++) {
+                if (root.children.containsKey(board[i][j])) {
+                    TrieNode node = root.children.get(board[i][j]);
+                    findWord(board, i, j, node, result);
+                }
+            }
+        }
+
+        return result;
+    }
+
+    private TrieNode buildTrie(String[] words) {
+        TrieNode root = new TrieNode();
+
+        for (String word : words) {
+            TrieNode node = root;
+            for (char c : word.toCharArray()) {
+                node = node.children.computeIfAbsent(c, k -> new TrieNode());
+            }
+            node.word = word;
+        }
+        return root;
+    }
+
+    private int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
+
+    private void findWord(char[][] board, int r, int c, TrieNode node, List<String> result) {
+        if (node.word != null) {
+            result.add(node.word);
+            node.word = null;
+        }
+
+        char letter = board[r][c];
+        board[r][c] = '#';
+
+        for (int[] direction : directions) {
+            int nr = r + direction[0];
+            int nc = c + direction[1];
+
+            if (nr < 0 || nr > board.length - 1 || nc < 0 || nc > board[0].length - 1) {
+                continue;
+            }
+            if (node.children.containsKey(board[nr][nc])) {
+                TrieNode nextNode = node.children.get(board[nr][nc]);
+                findWord(board, nr, nc, nextNode, result);
+            }
+        }
+
+        board[r][c] = letter;
+    }
+
+}
+