[Flynn] WEEK 01 Solutions

contains-duplicate/flynn.cpp

@@ -0,0 +1,24 @@
+/**
+ * for given size of input nums N,
+ * 
+ * Time complexity: O(N)
+ *   - iteration: O(N)
+ *   - unorderd_set find method: O(1) on average
+ *   - unorderd_set insert method: O(1) on average
+ * 
+ * Space complexity: O(N)
+ */
+
+class Solution {
+public:
+    bool containsDuplicate(vector<int>& nums) {
+        unordered_set<int> unique_numbers;
+
+        auto it = nums.begin();
+        while (it != nums.end()) {
+            if (unique_numbers.find(*it) == unique_numbers.end()) unique_numbers.insert(*(it++));
+            else return true;
+        }
+        return false;
+    }
+};

kth-smallest-element-in-a-bst/flynn.cpp

@@ -0,0 +1,41 @@
+/**
+ * For the height H of the given BST,
+ * 
+ * Time complexity: O(max(H, K))
+ *   - if H > K, O(H) at worst
+ *   - else, O(K)
+ * 
+ * Space complexity: O(H > K ? H + K : K)
+ *   - additional vector to save nums O(K)
+ *   - if H > K, call stack O(H)
+ *   - else, O(K)
+ */
+
+/**
+ * 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:
+    void inorder(TreeNode* node, vector<int>& v, int max_size) {
+        if (!node || v.size() == max_size) return;
+
+        if (node->left) inorder(node->left, v, max_size);
+        v.push_back(node->val);
+        if (node->right) inorder(node->right, v, max_size);
+    }
+
+    int kthSmallest(TreeNode* root, int k) {
+        vector<int> nums;
+        inorder(root, nums, k);
+
+        return nums[k - 1];
+    }
+};

number-of-1-bits/flynn.cpp

@@ -0,0 +1,21 @@
+/**
+ * For given integer N,
+ * 
+ * Time complexity: O(logN)
+ * 
+ * Space complexity: O(1)
+ */
+
+class Solution {
+public:
+    int hammingWeight(int n) {
+        int res = 0;
+
+        while (n) {
+            if (n & 1) res++;
+            n >>= 1;
+        }
+
+        return res;
+    }
+};

palindromic-substrings/flynn.cpp

@@ -0,0 +1,34 @@
+/**
+ * For the length N of given string s,
+ * 
+ * Time complexity: O(N^3)
+ * 
+ * Space complexity: O(1)
+ */
+
+class Solution {
+public:
+    int countSubstrings(string s) {
+        int res = 0;
+
+        for (int i = 0; i < s.size(); i++) {
+            for (int j = i; j < s.size(); j++) {
+                int start = i, end = j;
+                bool is_palindrome = true;
+
+                while (start <= end) {
+                    if (s[start] != s[end]) {
+                        is_palindrome = false;
+                        break;
+                    }
+                    start++;
+                    end--;
+                }
+
+                if (is_palindrome) res++;
+            }
+        }
+
+        return res;
+    }
+};

top-k-frequent-elements/flynn.cpp

@@ -0,0 +1,46 @@
+/**
+ * For given size of nums N and K,
+ * 
+ * Time complexity: O(N + MlogM + KlogM) <= O(NlogN)
+ *   - the first iteration: O(N)
+ *     - N times of 
+ *     - unordered_map find: O(1) on average
+ *     - unordered_map insert: O(1) on average
+ *   - the second iteration: O(MlogM) such that M is the number of unique numbers
+ *     - M times of
+ *     - priority_queue push: O(logM)
+ *   - the last iteration of making result: O(KlogM)
+ *     - K times of
+ *     - priority_queue pop: O(logM)
+ * 
+ *  Space complexity: O(N) at worst
+ */
+
+class Solution {
+public:
+    vector<int> topKFrequent(vector<int>& nums, int k) {
+        vector<int> res;
+        priority_queue<pair<int, int>> pq;
+        unordered_map<int, int> m;
+
+        auto nums_it = nums.begin();
+        while (nums_it != nums.end()) {
+            if (m.find(*nums_it) == m.end()) m.insert({*nums_it, 1});
+            else m[*nums_it]++;
+            nums_it++;
+        }
+
+        auto m_it = m.begin();
+        while (m_it != m.end()) {
+            pq.push({(*m_it).second, (*m_it).first});
+            m_it++;
+        }
+
+        while (k--) {
+            res.push_back(pq.top().second);
+            pq.pop();
+        }
+
+        return res;
+    }
+};