Merge pull request #338 from TonyKim9401/main

[TONY] WEEK 02 Solutions

Author: TonyKim9401

construct-binary-tree-from-preorder-and-inorder-traversal/TomyKim9401.java

@@ -0,0 +1,23 @@
+class Solution {
+    private int i, p;
+    public TreeNode buildTree(int[] preorder, int[] inorder) {
+        // Time complexity: O(n)
+        // Space complexity: O(n)
+        return builder(preorder, inorder, Integer.MIN_VALUE);
+    }
+
+    private TreeNode builder(int[] preorder, int[] inorder, int stop) {
+        if (p >= preorder.length) return null;
+        if (inorder[i] == stop) {
+            i += 1;
+            return null;
+        }
+
+        TreeNode node = new TreeNode(preorder[p]);
+        p += 1;
+
+        node.left = builder(preorder, inorder, node.val);
+        node.right = builder(preorder, inorder, stop);
+        return node;
+    }
+}

counting-bits/TonyKim9401.java

@@ -0,0 +1,10 @@
+class Solution {
+    public int[] countBits(int n) {
+        // time complexity: O(n)
+        // space complexity: O(n)
+        int[] output = new int[n+1];
+        int num = 0;
+        while (num <= n) output[num] = Integer.bitCount(num++);
+        return output;
+    }
+}

decode-ways/TonyKim9401.java

@@ -0,0 +1,21 @@
+class Solution {
+    public int numDecodings(String s) {
+        // time complexity: O(n)
+        // space complexity: O(n)
+        if (s.charAt(0) == '0') return 0;
+
+        int[] dp = new int[s.length() + 1];
+        dp[0] = 1;
+        dp[1] = 1;
+
+        for (int i = 2; i <= s.length(); i++) {
+            int oneDigit = Integer.parseInt(s.substring(i-1, i));
+            int twoDigits = Integer.parseInt(s.substring(i-2, i));
+
+            if (oneDigit > 0 && oneDigit < 10) dp[i] += dp[i-1];
+            if (twoDigits >= 10 && twoDigits <= 26) dp[i] += dp[i-2];
+        }
+
+        return dp[s.length()];
+    }
+}

encode-and-decode-strings/TonyKim9401.java

@@ -0,0 +1,32 @@
+public class Solution {
+    /*
+     * @param strs: a list of strings
+     * @return: encodes a list of strings to a single string.
+     */
+    public String encode(List<String> strs) {
+        // write your code here
+        StringBuilder sb = new StringBuilder();
+        for (String str : strs) {
+            sb.append(str.length()).append("#").append(str);
+        }
+        return sb.toString();
+    }
+
+    /*
+     * @param str: A string
+     * @return: decodes a single string to a list of strings
+     */
+    public List<String> decode(String str) {
+        // write your code here
+        List<String> output = new ArrayList<>();
+        int i = 0;
+        while (i < str.length()) {
+            int idx = str.indexOf('#', i);
+            int length = Integer.parseInt(str.substring(i, idx));
+            String s = str.substring(idx + 1, idx + 1 + length);
+            output.add(s);
+            i = idx + 1 + length;
+        }
+        return output;
+    }
+}

valid-anagram/TonyKim9401.java

@@ -0,0 +1,15 @@
+class Solution {
+    public boolean isAnagram(String s, String t) {
+        // time complexity: O(n log n)
+        // space complexity: O(n)
+        if (s.length() != t.length()) return false;
+
+        char[] sArr = s.toCharArray();
+        char[] tArr = t.toCharArray();
+
+        Arrays.sort(sArr);
+        Arrays.sort(tArr);
+
+        return Arrays.equals(sArr, tArr);
+    }
+}