Leetcode 1257 @ Smallest Common Region

1257. Smallest Common Region

Medium

You are given some lists of regions where the first region of each list includes all other regions in that list.

Naturally, if a region X contains another region Y then X is bigger than Y. Also by definition a region X contains itself.

Given two regions region1, region2, find out the smallest region that contains both of them.

If you are given regions r1, r2 and r3 such that r1 includes r3, it is guaranteed there is no r2 such that r2 includes r3.

It's guaranteed the smallest region exists.

Example 1:

Input:
regions = [["Earth","North America","South America"],
["North America","United States","Canada"],
["United States","New York","Boston"],
["Canada","Ontario","Quebec"],
["South America","Brazil"]],
region1 = "Quebec",
region2 = "New York"
Output: "North America"

Constraints:

• 2 <= regions.length <= 10^4
• region1 != region2
• All strings consist of English letters and spaces with at most 20 letters.

这道题就是变相的lowest common ancestor，整个2维数组给的就是一颗树，从earth出发，延伸出north america, south america。思路是这样，可是关键，我们需要构造出一棵树，然后再做lowest common ancestor么？未免有点繁琐，尤其它给的数据结构是2维数组。

那么非常直观的思路是，我们把region1, region2从顶点到末端的路线找出来，然后再一一比对，找到最低的ancestor就可以了。

那么找路线，是bottom up or top down呢？top down，可以stack, DFS。又或者bottom up，用一个hashmap来保存每一个region的parent region，这样按图索骥，就可以把region1 to root的路线找出来了。

注意一点，push back parent node时，要从i = 1开始，因为若从0开始，那么就会有parent[region] = region，后面做while find时容易不小心陷入死循环。

想一想， 如果top down, DFS该怎么做呢？

class Solution
{
    public:
    string findSmallestRegion(vector<vector<string>>& regions, string region1, string region2) 
    {
        // lowest common ancestor
        // use parents hashmap to print out the paths
        unordered_map<string, string> parents;
        vector<string> p1, p2;
        for (int i = 0; i<regions.size(); i++)
        {
            for (int j = 1; j<regions[i].size(); j++)
            {
                parents[regions[i][j]] = regions[i][0];
            }
        }
        while(!region1.empty())
        {
            p1.push_back(region1);
            region1 = parents[region1];
        }
        while(!region2.empty())
        {
            p2.push_back(region2);
            region2 = parents[region2];
        }
        for (auto &r1 : p1)
        {
            for (auto &r2 : p2)
            {
                if (r2 == r1)
                    return r1;
            }
        }
        return NULL;
    }
};

Leetcode 251 @ Flattern 2D Vector

Design and implement an iterator to flatten a 2d vector. It should support the following operations: next and hasNext.

Example:

Vector2D iterator = new Vector2D([[1,2],[3],[4]]);

iterator.next(); // return 1
iterator.next(); // return 2
iterator.next(); // return 3
iterator.hasNext(); // return true
iterator.hasNext(); // return true
iterator.next(); // return 4
iterator.hasNext(); // return false

Notes:

1. Please remember to RESET your class variables declared in Vector2D, as static/class variables are persisted across multiple test cases. Please see here for more details.
2. You may assume that next() call will always be valid, that is, there will be at least a next element in the 2d vector when next() is called.

这道题很简单，没什么复杂。唯一要注意的一点是，2D vector的每一行的大小不一定一样，所以用 v.size() * v[0].size() 去计算总体的大小是不对的。以及v传进来后，要把它放入自己的数据结构，就是把它flattern，一维化。

class Vector2D {
    int index;
    vector<int> data;
public:
    Vector2D(vector<vector<int>>& v) {
        for (auto &row : v)
        {
            for (int &it : row)
            {
                data.push_back(it);
            }
        }
        index = 0;
    }
    
    int next() {
        return data[index++];
    }
    
    bool hasNext() {
         return index < data.size();
    }
};

/**
 * Your Vector2D object will be instantiated and called as such:
 * Vector2D* obj = new Vector2D(v);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */

Leetcode @ Bulls and Cows

299. Bulls and Cows

Easy

You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called "bulls") and how many digits match the secret number but locate in the wrong position (called "cows"). Your friend will use successive guesses and hints to eventually derive the secret number.

Write a function to return a hint according to the secret number and friend's guess, use A to indicate the bulls and B to indicate the cows. 

Please note that both secret number and friend's guess may contain duplicate digits.

Example 1:

Input: secret = "1807", guess = "7810"

Output: "1A3B"

Explanation: 1 bull and 3 cows. The bull is 8, the cows are 0, 1 and 7.

Example 2:

Input: secret = "1123", guess = "0111"

Output: "1A1B"

Explanation: The 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow.

Note: You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.

理解题意很重要，里面只算digit，而不是两个连在一起的数字，比如11 vs 110，算2bulls，而不是1 bull，并不是把"11"作为一个数字去看，最开始我是这样想的就想复杂了。

理解清楚题意后，就变得很简单，用两个数组来分别存当secret 和 guess某一位 数字不想等时的count，如果相等，就bull++，最后在iterate through 两个数组，大小为10，取每位上两者中的最小的那个，因为要找重合的个数么，所以找交集。

class Solution {
public:
    string getHint(string secret, string guess) {
        int size = secret.size();
        vector<int> rec1(10, 0), rec2(10, 0);
        int bulls = 0, cows = 0;
        for (int i = 0; i<size; i++)
        {
            if (secret[i] == guess[i])
            {
                bulls++;
            }
            else
            {
                rec1[secret[i]-'0']++;
                rec2[guess[i]-'0']++;
            }
        }
        for (int i = 0; i<=9; i++)
        {
            cows += min(rec1[i], rec2[i]);
        }
        return to_string(bulls) + "A" + to_string(cows) + "B";
    }
};