weekly contest 115

958. Check Completeness of a Binary Tree

Given a binary tree, determine if it is a complete binary tree.

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

Example 1:

Input: [1,2,3,4,5,6]
Output: true
Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.

Example 2:

Input: [1,2,3,4,5,null,7]
Output: false
Explanation: The node with value 7 isn't as far left as possible.

 

Note:

1. The tree will have between 1 and 100 nodes.

Approach #1: C++.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isCompleteTree(TreeNode* root) {
        queue<TreeNode*> nodes;
        nodes.push(root);
        while (true) {
            TreeNode* cur = nodes.front();
            if (cur == NULL) break;
            nodes.push(cur->left);
            nodes.push(cur->right);
            nodes.pop();
        }
        while (!nodes.empty()) {
            TreeNode* cur = nodes.front(); 
            if (cur != NULL) return false;
            nodes.pop();
        }
        return true;
    }
};

　　

957. Prison Cells After N Days

There are 8 prison cells in a row, and each cell is either occupied or vacant.

Each day, whether the cell is occupied or vacant changes according to the following rules:

• If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
• Otherwise, it becomes vacant.

(Note that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors.)

We describe the current state of the prison in the following way: cells[i] == 1 if the i-th cell is occupied, else cells[i] == 0.

Given the initial state of the prison, return the state of the prison after N days (and N such changes described above.)

 

Example 1:

Input: cells = [0,1,0,1,1,0,0,1], N = 7
Output: [0,0,1,1,0,0,0,0]
Explanation: 
The following table summarizes the state of the prison on each day:
Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
Day 7: [0, 0, 1, 1, 0, 0, 0, 0]

Example 2:

Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]

Note:

1. cells.length == 8
2. cells[i] is in {0, 1}
3. 1 <= N <= 10^9

Appraoch #1: C++.

class Solution {
public:
    vector<int> prisonAfterNDays(vector<int>& cells, int N) {
        unordered_map<string, int> seen;
        while (N > 0) {
            string temp(cells.begin(), cells.end());
            seen[temp] = N--;
            vector<int> dummy(8, 0);
            for (int j = 1; j < 7; ++j) 
                dummy[j] = cells[j-1] == cells[j+1] ? 1 : 0;
            cells = dummy;
            string ant(cells.begin(), cells.end());
            if (seen.count(ant)) 
                N %= seen[ant] - N;
        }
        return cells;
    }
};

　　

Analysis:

There are only six binary numbers,  so all of the difference possibilities is 2^6. 

so we can use a map to store the case, if N is very large we can reduce the number of calculation by 

N %= seen[ant] - N;