week8 mission done

Author: dev-jonghoonpark

combination-sum/dev-jonghoonpark.md

@@ -0,0 +1,78 @@
+- https://leetcode.com/problems/combination-sum/
+- time complexity : O(2^n)
+- space complexity : O(2^n)
+- https://algorithm.jonghoonpark.com/2024/06/23/leetcode-39
+
+## bfs 로 풀기
+
+```java
+class Solution {
+    public List<List<Integer>> combinationSum(int[] candidates, int target) {
+        Arrays.sort(candidates);
+        List<List<Integer>> result = new ArrayList<>();
+
+        Queue<Holder> queue = new LinkedList<>();
+        queue.offer(new Holder(target));
+
+        while (!queue.isEmpty()) {
+            Holder holder = queue.poll();
+            int lastInput = !holder.combination.isEmpty() ? holder.combination.get(holder.combination.size() - 1) : 0;
+            int left = holder.left;
+            if (left == 0) {
+                result.add(holder.combination);
+                continue;
+            }
+
+            for (int candidate : candidates) {
+                if (candidate < lastInput) {
+                    continue;
+                }
+
+                if (left - candidate >= 0) {
+                    queue.add(holder.next(candidate));
+                } else {
+                    break;
+                }
+            }
+        }
+
+        return result;
+    }
+}
+
+class Holder {
+    int left;
+    List<Integer> combination;
+
+    public Holder(int left) {
+        this.left = left;
+        this.combination = new ArrayList<>();
+    }
+
+    private Holder(int left, List<Integer> combination) {
+        this.left = left;
+        this.combination = combination;
+    }
+
+    public Holder next(int minus) {
+        List<Integer> combinedList = new ArrayList<>(this.combination.size() + 1);
+        combinedList.addAll(this.combination);
+        combinedList.add(minus);
+        return new Holder(this.left - minus, combinedList);
+    }
+
+    @Override
+    public String toString() {
+        return "{" +
+                "left=" + left +
+                ", combination=" + combination +
+                '}';
+    }
+}
+```
+
+## backtracking 으로 풀기
+
+```java
+// TODO
+```

construct-binary-tree-from-preorder-and-inorder-traversal/dev-jonghoonpark.md

@@ -0,0 +1,110 @@
+- https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
+- time complexity : O(n)
+- space complexity : O(n)
+- https://algorithm.jonghoonpark.com/2024/06/23/leetcode-105
+
+```java
+class Solution {
+    public TreeNode buildTree(int[] preorder, int[] inorder) {
+        Map<Integer, Integer> inorderIndexMap = new HashMap<>();
+        for (int i = 0; i < inorder.length; i++) {
+            inorderIndexMap.put(inorder[i], i);
+        }
+
+        return buildTree(inorderIndexMap, new Traversal(preorder), new Traversal(inorder));
+    }
+
+    public TreeNode buildTree(Map<Integer, Integer> inorderIndexMap, Traversal preorderTraversal, Traversal inorderTraversal) {
+        if(preorderTraversal.start > preorderTraversal.end) {
+            return null;
+        }
+
+        TreeNode treeNode = new TreeNode(preorderTraversal.getFirst());
+        if(preorderTraversal.start == preorderTraversal.end) {
+            return treeNode;
+        }
+
+        int rootIndex = inorderIndexMap.get(preorderTraversal.getFirst());
+        int leftSize = rootIndex - inorderTraversal.start;
+        treeNode.left = buildTree(
+                inorderIndexMap,
+                preorderTraversal.subIterator(preorderTraversal.start + 1, preorderTraversal.start + leftSize),
+                inorderTraversal.subIterator(inorderTraversal.start, rootIndex - 1)
+        );
+        treeNode.right = buildTree(
+                inorderIndexMap,
+                preorderTraversal.subIterator(preorderTraversal.start + leftSize + 1, preorderTraversal.end),
+                inorderTraversal.subIterator(rootIndex + 1, inorderTraversal.end)
+        );
+
+        return treeNode;
+    }
+}
+
+class TreeNode {
+    int val;
+    TreeNode left;
+    TreeNode right;
+
+    TreeNode() {
+    }
+
+    TreeNode(int val) {
+        this.val = val;
+    }
+
+    TreeNode(int val, TreeNode left, TreeNode right) {
+        this.val = val;
+        this.left = left;
+        this.right = right;
+    }
+
+    @Override
+    public String toString() {
+        return "{" +
+                "val=" + val +
+                ", left=" + left +
+                ", right=" + right +
+                '}';
+    }
+}
+
+class Traversal {
+    int[] array;
+    int start;
+    int end;
+
+    public Traversal(int[] array) {
+        this.array = array;
+        this.start = 0;
+        this.end = array.length - 1;
+    }
+
+    private Traversal(int[] array, int start, int end) {
+        this.array = array;
+        this.start = start;
+        this.end = end;
+    }
+
+    public Traversal subIterator(int start, int end) {
+        return new Traversal(array, start, end);
+    }
+
+    public int getFirst() {
+        return array[start];
+    }
+
+    public Traversal(int start, int end) {
+        this.start = start;
+        this.end = end;
+    }
+
+    @Override
+    public String toString() {
+        return "{" +
+                "start=" + start +
+                ", end=" + end +
+                '}';
+    }
+}
+```

implement-trie-prefix-tree/dev-jonghoonpark.md

@@ -0,0 +1,58 @@
+- https://leetcode.com/problems/implement-trie-prefix-tree/
+- https://algorithm.jonghoonpark.com/2024/06/23/leetcode-208
+
+## TC, SC
+
+insert, search, startsWith 메소드의 경우 입력된 문자열의 길이를 n 이라 하였을 때 시간 복잡도는 `O(n)`이다. 공간 복잡도는 `insert된 문자열의 갯수` 를 `N` 이라 하고 `insert된 문자열의 길이의 평균` 를 `L`이라고 하였을 때 `O(N * L * 26)`이다. 26은 계수이기 때문에 생략할 수 있다.
+
+## 풀이
+
+```java
+class Trie {
+
+    Node root = new Node();
+
+    public Trie() {
+
+    }
+
+    public void insert(String word) {
+        Node currentNode = root;
+        for(char c : word.toCharArray()) {
+            if(currentNode.nodes[c - 97] == null) {
+                currentNode.nodes[c - 97] = new Node();
+            }
+            currentNode = currentNode.nodes[c - 97];
+        }
+        currentNode.val = word;
+    }
+
+    public boolean search(String word) {
+        Node currentNode = root;
+        for(char c : word.toCharArray()) {
+            if(currentNode.nodes[c - 97] == null) {
+                return false;
+            }
+            currentNode = currentNode.nodes[c - 97];
+        }
+
+        return currentNode.val != null && currentNode.val.equals(word);
+    }
+
+    public boolean startsWith(String prefix) {
+        Node currentNode = root;
+        for(char c : prefix.toCharArray()) {
+            if(currentNode.nodes[c - 97] == null) {
+                return false;
+            }
+            currentNode = currentNode.nodes[c - 97];
+        }
+        return true;
+    }
+}
+
+class Node {
+    String val;
+    Node[] nodes = new Node[26];
+}
+```

kth-smallest-element-in-a-bst/dev-jonghoonpark.md

@@ -0,0 +1,44 @@
+- https://leetcode.com/problems/kth-smallest-element-in-a-bst/
+- time complexity : O(n)
+- space complexity : O(n), 트리가 균등할 경우 O(logn)에 가까워진다.
+- https://algorithm.jonghoonpark.com/2024/06/23/leetcode-230
+
+```java
+class Solution {
+    public int kthSmallest(TreeNode root, int k) {
+        return dfs(root, new Holder(k));
+    }
+
+    public int dfs(TreeNode root, Holder holder) {
+        if(root.left != null) {
+            int left = dfs(root.left, holder);
+            if (left != -1) {
+                return left;
+            }
+        }
+        holder.decrease();
+        if (holder.k == 0) {
+            return root.val;
+        }
+        if(root.right != null) {
+            int right = dfs(root.right, holder);
+            if (right != -1) {
+                return right;
+            }
+        }
+        return -1;
+    }
+}
+
+class Holder {
+    int k;
+
+    public Holder(int k) {
+        this.k = k;
+    }
+
+    public void decrease() {
+        this.k = this.k - 1;
+    }
+}
+```

word-search/dev-jonghoonpark.md

@@ -0,0 +1,48 @@
+- https://leetcode.com/problems/word-search/
+- https://algorithm.jonghoonpark.com/2024/06/23/leetcode-79
+
+## TC, SC
+
+`board` 의 길이를 `w`, `board[i]`의 길이를 `h`, 단어의 길이를 `n` 이라고 하였을 때.
+시간 복잡도는 `O(w * h * n)` 이다. 공간 복잡도는 `O(n)`이다.
+
+## 풀이
+
+```java
+class Solution {
+    public boolean exist(char[][] board, String word) {
+        char[] chars = word.toCharArray();
+        for (int i = 0; i < board.length; i++) {
+            for (int j = 0; j < board[0].length; j++) {
+                if (dfs(board, chars, i, j, 0, word.length() - 1)) {
+                    return true;
+                }
+            }
+        }
+
+        return false;
+    }
+
+    public boolean dfs(char[][] board, char[] chars, int i, int j, int pointer, int end) {
+        if (i < 0 || i > board.length - 1 || j < 0 || j > board[0].length - 1 || board[i][j] != chars[pointer]) {
+            return false;
+        }
+
+        if(pointer == end) {
+            return true;
+        }
+
+        int next = pointer + 1;
+        char temp = board[i][j];
+        board[i][j] = ' ';
+        boolean result = dfs(board, chars, i + 1, j, next, end)
+                || dfs(board, chars, i - 1, j, next, end)
+                || dfs(board, chars, i, j + 1, next, end)
+                || dfs(board, chars, i, j - 1, next, end);
+        if(!result) {
+            board[i][j] = temp;
+        }
+        return result;
+    }
+}
+```