Merge pull request #346 from obzva/main

[Flynn] week2

Author: obzva

construct-binary-tree-from-preorder-and-inorder-traversal/flynn.cpp

@@ -0,0 +1,49 @@
+/**
+ * For the number of given nodes N,
+ * 
+ * Time complexity: O(N)
+ * 
+ * Space complexity: O(N) at worst
+ */
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
+        unordered_map<int, int> inorder_index_map;
+        stack<TreeNode*> tree_stack;
+
+        for (int i = 0; i < inorder.size(); i++) inorder_index_map[inorder[i]] = i;
+
+        TreeNode* root = new TreeNode(preorder[0]);
+        tree_stack.push(root);
+
+        for (int i = 1; i < preorder.size(); i++) {
+            TreeNode* curr = new TreeNode(preorder[i]);
+
+            if (inorder_index_map[curr->val] < inorder_index_map[tree_stack.top()->val]) {
+                tree_stack.top()->left = curr;
+            } else {
+                TreeNode* parent;
+                while (!tree_stack.empty() && inorder_index_map[curr->val] > inorder_index_map[tree_stack.top()->val]) {
+                    parent = tree_stack.top();
+                    tree_stack.pop();
+                }
+                parent->right = curr;
+            }
+            tree_stack.push(curr);
+        }
+
+        return root;
+    }
+};

counting-bits/flynn.cpp

@@ -0,0 +1,22 @@
+/**
+ * Time complexity: O(N)
+ * 
+ * Space complexity: O(1)
+ */
+
+class Solution {
+public:
+    vector<int> countBits(int n) {
+        vector<int> res;
+        res.push_back(0);
+
+        int i = 1, j = 1;
+        while (i <= n) {
+            res.push_back(res[i - j] + 1);
+            i++;
+            if (i == j * 2) j *= 2;
+        }
+
+        return res;
+    }
+};

decode-ways/flynn.cpp

@@ -0,0 +1,53 @@
+/**
+ * For the length of the given string N,
+ * 
+ * Time complexity: O(N)
+ * 
+ * Space complexity: O(N)
+ */
+
+class Solution {
+public:
+    int numDecodings(string s) {
+        if (s[0] == '0') return 0;
+
+        int memo[s.size() + 1];
+
+        fill(memo, memo + s.size() + 1, 0);
+        memo[0] = 1;
+        memo[1] = 1;
+
+        for (int i = 2; i <= s.size(); i++) {
+            int s_i = i - 1;
+            if (s[s_i] != '0') memo[i] = memo[i - 1];
+            int two_digits = stoi(s.substr(s_i - 1, 2));
+            if (10 <= two_digits && two_digits <= 26) memo[i] += memo[i - 2];
+        } 
+
+        return memo[s.size()];
+    }
+};
+
+/**
+ * Space complexity O(1) solution
+ */
+
+// class Solution {
+// public:
+//     int numDecodings(string s) {
+//         if (s[0] == '0') return 0;
+
+//         int one_digit_memo = 1, two_digit_memo = 1;
+
+//         for (int i = 1; i < s.size(); i++) {
+//             int tmp = 0;
+//             if (s[i] != '0') tmp = one_digit_memo;
+//             int two_digits = stoi(s.substr(i - 1, 2));
+//             if (10 <= two_digits && two_digits <= 26) tmp += two_digit_memo;
+//             two_digit_memo = one_digit_memo;
+//             one_digit_memo = tmp;
+//         } 
+
+//         return one_digit_memo;
+//     }
+// };

encode-and-decode-strings/flynn.cpp

@@ -0,0 +1,57 @@
+/**
+ * For the number of given strings N, and the length of the longest string M,
+ * 
+ * Encode
+ * - Time complexity: O(N)
+ * - Space complexity: O(1)
+ * 
+ * Decode
+ * - Time complexity: O(NM)
+ * - Space complexity: O(M)
+ */
+
+class Codec {
+public:
+
+    // Encodes a list of strings to a single string.
+    string encode(vector<string>& strs) {
+        string res = "";
+        for (auto str : strs) {
+            res += to_string(str.size());
+            res.push_back('.');
+            res += str;
+        }
+        return res;
+    }
+
+    // Decodes a single string to a list of strings.
+    vector<string> decode(string s) {
+        vector<string> res;
+
+        auto it = s.begin();
+        while (it != s.end()) {
+            int str_size = 0;
+            string tmp = "";
+
+            while (*it != '.') {
+                str_size = str_size * 10 + (*it - '0');
+                it++;
+            } 
+
+            it++;
+
+            for (int i = 0; i < str_size; i++) {
+                tmp.push_back(*it);
+                it++;
+            }
+
+            res.push_back(tmp);
+        }
+
+        return res;
+    }
+};
+
+// Your Codec object will be instantiated and called as such:
+// Codec codec;
+// codec.decode(codec.encode(strs));

valid-anagram/flynn.cpp

@@ -0,0 +1,22 @@
+/**
+ * For length of given strings N,
+ * 
+ * Time complexity: O(N)
+ * - iteration for given strings
+ * 
+ * Space complexity: O(1)
+ * - the size of the container `count` is constant
+ */
+
+class Solution {
+public:
+    bool isAnagram(string s, string t) {
+        int count[26] = {0};
+
+        for (auto c : s) count[c - 'a']++;
+        for (auto c : t) count[c - 'a']--;
+
+        for (int i = 0; i < 26; i++) if (count[i]) return false;
+        return true;
+    }
+};