2021-12-05 Daily-Challenge

Posted on:
Posted modified:
In: DailyChallenge
|

Today I have done leetcode's December LeetCoding Challenge with cpp.

December LeetCoding Challenge 5

Description

House Robber III

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called root.

Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if two directly-linked houses were broken into on the same night.

Given the root of the binary tree, return the maximum amount of money the thief can rob without alerting the police.

Example 1:

img

Input: root = [3,2,3,null,3,null,1]
Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

img

Input: root = [3,4,5,1,3,null,1]
Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • 0 <= Node.val <= 104

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
  pair<int, int> solve(TreeNode *root) {
    if(!root) {
      return {0, 0};
    }
    auto [maxLeft, maxNotRobLeft] = solve(root->left);
    auto [maxRight, maxNotRobRight] = solve(root->right);
    int maxRob = root->val + maxNotRobLeft + maxNotRobRight;
    int maxNotRob = maxLeft + maxRight;
    return {max(maxRob, maxNotRob), maxNotRob};
  }
public:
  int rob(TreeNode* root) {
    return solve(root).first;
  }
};

// Accepted
// 124/124 cases passed (4 ms)
// Your runtime beats 99.42 % of cpp submissions
// Your memory usage beats 70.99 % of cpp submissions (22.2 MB)
Algorithm