[crispy] week 2 solution

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

@@ -0,0 +1,37 @@
+// Time Complexity: O(n^2)
+// Spatial Complexity: O(n)
+
+class Solution {
+private:
+    int findIndex(int targetVal, vector<int>& inorder) {
+        auto pos = find(inorder.begin(), inorder.end(), targetVal);
+        if (pos == inorder.end()) {
+            return -1;
+        }
+
+        return pos - inorder.begin();
+    }
+
+    TreeNode* dfs(vector<int>& preorder, vector<int>& inorder, int preorderIndex, int startIndex, int endIndex) {
+        if (preorder.size() <= preorderIndex || startIndex > endIndex) {
+            return nullptr;
+        }
+
+        int targetValue = preorder[preorderIndex];
+        int rootIndex = this->findIndex(targetValue, inorder);
+        if(rootIndex < 0) {
+            return nullptr;
+        }
+
+        int leftSubtreeLength = rootIndex - startIndex;
+
+        TreeNode* left = dfs(preorder, inorder, preorderIndex + 1, startIndex, rootIndex - 1); 
+        TreeNode* right = dfs(preorder, inorder, preorderIndex + 1 + leftSubtreeLength, rootIndex + 1, endIndex);
+
+        return new TreeNode(targetValue, left, right);
+    }
+public:
+    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
+        return this->dfs(preorder, inorder, 0, 0, preorder.size() - 1);
+    }
+};

counting-bits/heozeop.cpp

@@ -0,0 +1,31 @@
+// Time Complexity: O(n^2)
+// Spatial Complexity: O(n)
+
+class Solution {
+private:
+    int count1(int n) {
+        int ans = 0;
+
+        while(n) {
+            if (n % 2) {
+                ++ans;
+            }
+
+            n /= 2;
+        }
+
+        return ans;
+    }
+
+public:
+    vector<int> countBits(int n) {
+        vector<int> ans(n + 1);
+
+        for(int i = 0; i <= n; ++i) {
+            ans[i] = this->count1(i);
+        }
+
+        return ans;
+    }
+};
+

decode-ways/heozeop.cpp

@@ -0,0 +1,31 @@
+// Time Complexity: O(n)
+// Spatial Complexity: O(n)
+
+class Solution {
+public:
+  int numDecodings(string s) {
+    if(s.length() < 1 || s[0] == '0') {
+      return 0;
+    }
+
+    vector<int> dp(s.length() + 1, 0);
+    dp[0] = dp[1] = 1;
+    if(s[1] == '0') {
+      dp[1] = 0;
+    }
+
+    int prev,pprev;
+    for(int i = 2; i <= s.length(); ++i) {
+      prev = s[i - 1] - '0';
+      if (prev <= 9 && prev > 0) {
+        dp[i] += dp[i-1];
+      }
+      pprev = (s[i - 2] - '0') * 10 + prev;
+      if(pprev <= 26 && pprev > 9) {
+        dp[i] += dp[i-2];
+      }
+    }
+
+    return dp[s.length()];
+  }
+};

encode-and-decode-strings/heozeop.cpp

@@ -0,0 +1,55 @@
+// Time Complexity: O(n)
+// Spatial Complexity: O(n)
+
+class Codec {
+private:
+    string genereateRandomString(size_t size) {
+        string randomString = "";
+        for(size_t i = 0; i < size; ++i) {
+            randomString += static_cast<char>(rand() % 256);
+        }
+
+        return randomString;
+    }
+
+    vector<string> split(string target, string delimiter) {
+        vector<string> ans;
+
+        int delimiterLength = delimiter.size();
+        size_t pos = target.find(delimiter);
+        while(pos != string::npos) {
+            ans.push_back(target.substr(0, pos));
+
+            target = target.substr(pos + delimiterLength, target.size());
+            pos = target.find(delimiter);
+        }
+        ans.push_back(target);
+
+        return ans;
+    }
+
+    string delimiter = this->genereateRandomString(10);
+
+public:
+
+    // Encodes a list of strings to a single string.
+    string encode(vector<string>& strs) {
+        string encodedString = strs[0];
+
+        for(int i = 1; i < strs.size(); ++i) {
+            encodedString += this->delimiter + strs[i];
+        }
+
+        return encodedString;
+    }
+
+    // Decodes a single string to a list of strings.
+    vector<string> decode(string s) {
+
+        return split(s, this->delimiter);
+    }
+};
+
+// Your Codec object will be instantiated and called as such:
+// Codec codec;
+// codec.decode(codec.encode(strs));

valid-anagram/heozeop.cpp

@@ -0,0 +1,26 @@
+// Time Complexity: O(n)
+// Spatial Complexity: O(1)
+
+class Solution {
+public:
+    bool isAnagram(string s, string t) {
+        int numberOfAlphabet[26];
+
+        for(char character : s) {
+            numberOfAlphabet[character - 'a']++;
+        }
+
+        for(char character : t) {
+            numberOfAlphabet[character - 'a']--;
+        }
+
+        for(int i = 0; i < 26; ++i) {
+            if (numberOfAlphabet[i] != 0) {
+                return false;
+            }
+        }
+
+        return true;
+    }
+};
+