Today I have done Delete Leaves With a Given Value and leetcode's August LeetCoding Challenge with cpp.

Delete Leaves With a Given Value

Description

Given a binary tree root and an integer target, delete all the leaf nodes with value target.

Note that once you delete a leaf node with value target, if it's parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you can't).

Example 1:

Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left). 
After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).

Example 2:

Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]

Example 3:

Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.

Example 4:

Input: root = [1,1,1], target = 1
Output: []

Example 5:

Input: root = [1,2,3], target = 1
Output: [1,2,3]

Constraints:

• 1 <= target <= 1000
• The given binary tree will have between 1 and 3000 nodes.
• Each node's value is between [1, 1000].

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  TreeNode* removeLeafNodes(TreeNode* root, int target) {
    if(!root) return nullptr;
    root->left = removeLeafNodes(root->left, target);
    root->right = removeLeafNodes(root->right, target);
    if(root->val == target && !root->left && !root->right) return nullptr;
    return root;
  }
};

// Accepted
// 50/50 cases passed (12 ms)
// Your runtime beats 88.29 % of cpp submissions
// Your memory usage beats 14.7 % of cpp submissions (22 MB)

August LeetCoding Challenge 12

Description

Group Anagrams

Given an array of strings strs, group the anagrams together. You can return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

Input: strs = ["eat","tea","tan","ate","nat","bat"]
Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Example 2:

Input: strs = [""]
Output: [[""]]

Example 3:

Input: strs = ["a"]
Output: [["a"]]

Constraints:

• 1 <= strs.length <= 10^4
• 0 <= strs[i].length <= 100
• strs[i] consists of lower-case English letters.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  vector<vector<string>> groupAnagrams(vector<string>& strs) {
    unordered_map<string, vector<string>> groups;
    for(auto s : strs) {
      string key = s;
      sort(key.begin(), key.end());
      groups[key].push_back(s);
    }
    vector<vector<string>> answer;
    for(auto [key, group] : groups) {
      answer.emplace_back(group);
    }
    return answer;
  }
};

// Accepted
// 114/114 cases passed (28 ms)
// Your runtime beats 89.97 % of cpp submissions
// Your memory usage beats 35.7 % of cpp submissions (20.8 MB)