Today is Saturday, I gonna review the tasks I've done this week, and finish today's leetcode's January LeetCoding Challenge with cpp.

LeetCode Review

Merge Two Sorted Lists

too easy to review

Sort Array By Parity II

too easy to review

Construct Quad Tree

too easy to review

Check Array Formation Through Concatenation

too easy to review

Task Scheduler

no need to review

Reach a Number

no need to review

Pseudo-Palindromic Paths in a Binary Tree

no need to review

Largest Rectangle in Histogram

use a stack just need one iteration

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        vector<int> st{-1};
        heights.push_back(0);
        int len = heights.size();
        int answer = 0;
        for(int i = 0; i < len; ++i) {
            while(st.size() && st.back() != -1 && heights[i] < heights[st.back()]) {
                int index = st.back();
                st.pop_back();
                int h = heights[index];
                int w = i - st.back() - 1;
                answer = max(answer, h*w);
            }
            st.push_back(i);
        }
        return answer;
    }
};

Game of Life

another way to enumerate

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int rows = board.size();
        int cols = board[0].size();
        for(int i = 0; i < rows; ++i) {
            for(int j = 0; j < cols; ++j) {
                int count = 0;
                for(int row = max(0, i-1); row <= min(rows-1, i+1); ++row) {
                    for(int col = max(0, j-1); col <= min(cols-1, j+1); ++col) {
                        count += (board[row][col]&1);
                    }
                }
                count -= board[i][j];
                if(count == 3 || (count == 2 && board[i][j] == 1)) {
                    board[i][j] |= 2;
                }
            }
        }
        for(auto &row : board) {
            for(auto &cell : row) {
                cell >>= 1;
            }
        }
    }
};

Lowest Common Ancestor of a Binary Tree

using parent pointers

class Solution {
    unordered_map<TreeNode*, int> level;
    unordered_map<TreeNode*, TreeNode*> parent;
    void build(TreeNode *root) {
        if(!root) return;
        if(root->left) {
            level[root->left] = level[root] + 1;
            parent[root->left] = root;
            build(root->left);
        }
        if(root->right) {
            level[root->right] = level[root] + 1;
            parent[root->right] = root;
            build(root->right);
        }
    }
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        build(root);
        while(level[p] > level[q]) p = parent[p];
        while(level[q] > level[p]) q = parent[q];
        while(q != p) {
            p = parent[p];
            q = parent[q];
        }
        return q;
    }
};

January LeetCoding Challenge 2

Description

Find a Corresponding Node of a Binary Tree in a Clone of That Tree

Given two binary trees original and cloned and given a reference to a node target in the original tree.

The cloned tree is a copy of the original tree.

Return a reference to the same node in the cloned tree.

Note that you are not allowed to change any of the two trees or the target node and the answer must be a reference to a node in the cloned tree.

Follow up: Solve the problem if repeated values on the tree are allowed.

Example 1:

Input: tree = [7,4,3,null,null,6,19], target = 3
Output: 3
Explanation: In all examples the original and cloned trees are shown. The target node is a green node from the original tree. The answer is the yellow node from the cloned tree.

Example 2:

Input: tree = [7], target =  7
Output: 7

Example 3:

Input: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4
Output: 4

Example 4:

Input: tree = [1,2,3,4,5,6,7,8,9,10], target = 5
Output: 5

Example 5:

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

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• The values of the nodes of the tree are unique.
• target node is a node from the original tree and is not null.

Solution

class Solution {
public:
    TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) {
        if(!original) return nullptr;
        if(original == target) return cloned;
        TreeNode *left = getTargetCopy(original->left, cloned->left, target);
        if(left) return left;
        TreeNode *right = getTargetCopy(original->right, cloned->right, target);
        if(right) return right;
        return nullptr;
    }
};