Weekly Contest 133

1030. Matrix Cells in Distance Order

We are given a matrix with R rows and C columns has cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.

Additionally, we are given a cell in that matrix with coordinates (r0, c0).

Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from smallest distance to largest distance.  Here, the distance between two cells (r1, c1) and (r2, c2) is the Manhattan distance, |r1 - r2| + |c1 - c2|.  (You may return the answer in any order that satisfies this condition.)

 

Example 1:

Input: R = 1, C = 2, r0 = 0, c0 = 0
Output: [[0,0],[0,1]]
Explanation: The distances from (r0, c0) to other cells are: [0,1]

Example 2:

Input: R = 2, C = 2, r0 = 0, c0 = 1
Output: [[0,1],[0,0],[1,1],[1,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]
The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.

Example 3:

Input: R = 2, C = 3, r0 = 1, c0 = 2
Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]
There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].

 

Note:

1. 1 <= R <= 100
2. 1 <= C <= 100
3. 0 <= r0 < R
4. 0 <= c0 < C

 

Approach #1: 

class Solution {
public:
    vector<vector<int>> allCellsDistOrder(int R, int C, int r0, int c0) {
        map<int, vector<pair<int, int>>> m;
        for (int i = 0; i < R; ++i) {
            for (int j = 0; j < C; ++j) {
                int dis = abs(i - r0) + abs(j - c0);
                m[dis].push_back({i, j});
            }
        }
        
        int index = 0;
        vector<vector<int>> ret = vector(R*C, vector<int>(2));
        map<int, vector<pair<int, int>>>::iterator it;
        
        for (it = m.begin(); it != m.end(); ++it) {
            vector<pair<int, int>> temp = it->second;
            for (int i = 0; i < temp.size(); ++i) {
                ret[index][0] = temp[i].first;
                ret[index][1] = temp[i].second;
                index++;
            }
        }
        
        return ret;
    }
};

　　

1029. Two City Scheduling

There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.

 

Example 1:

Input: [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation: 
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

 

Note:

1. 1 <= costs.length <= 100
2. It is guaranteed that costs.length is even.
3. 1 <= costs[i][0], costs[i][1] <= 1000

 

Approach #1: 

class Solution {
    public int twoCitySchedCost(int[][] costs) {
        int N = costs.length / 2;
        int[][] dp = new int[N+1][N+1];
        for (int i = 1; i <= N; ++i) {
            dp[i][0] = dp[i-1][0] + costs[i-1][0];
        }
        for (int i = 1; i <= N; ++i) {
            dp[0][i] = dp[0][i-1] + costs[i-1][1];
        }
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j <= N; ++j) {
                dp[i][j] = Math.min(dp[i-1][j] + costs[i+j-1][0], dp[i][j-1] + costs[i+j-1][1]);
            }
        }
        return dp[N][N];
    }
}

　　

1031. Maximum Sum of Two Non-Overlapping Subarrays

Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M.  (For clarification, the L-length subarray could occur before or after the M-length subarray.)

Formally, return the largest V for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) and either:

• 0 <= i < i + L - 1 < j < j + M - 1 < A.length, or
• 0 <= j < j + M - 1 < i < i + L - 1 < A.length.

 

Example 1:

Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.

Example 2:

Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.

Example 3:

Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.

 

Note:

1. L >= 1
2. M >= 1
3. L + M <= A.length <= 1000
4. 0 <= A[i] <= 1000

 

Approach #1:

class Solution {
    public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        for (int i = 1; i < A.length; ++i) {
            A[i] += A[i-1];
        }
        int res = A[L+M-1], Lmax = A[L-1], Mmax = A[M-1];
        for (int i = L + M; i < A.length; ++i) {
            Lmax = Math.max(Lmax, A[i-M] - A[i-M-L]);
            Mmax = Math.max(Mmax, A[i-L] - A[i-M-L]);
            res = Math.max(res, Math.max(Lmax + A[i] - A[i-M], Mmax + A[i] - A[i-L]));
        }
        
        return res;
    }
}

　　

1032. Stream of Characters

Implement the StreamChecker class as follows:

• StreamChecker(words): Constructor, init the data structure with the given words.
• query(letter): returns true if and only if for some k >= 1, the last k characters queried (in order from oldest to newest, including this letter just queried) spell one of the words in the given list.

 

Example:

StreamChecker streamChecker = new StreamChecker(["cd","f","kl"]); // init the dictionary.
streamChecker.query('a');          // return false
streamChecker.query('b');          // return false
streamChecker.query('c');          // return false
streamChecker.query('d');          // return true, because 'cd' is in the wordlist
streamChecker.query('e');          // return false
streamChecker.query('f');          // return true, because 'f' is in the wordlist
streamChecker.query('g');          // return false
streamChecker.query('h');          // return false
streamChecker.query('i');          // return false
streamChecker.query('j');          // return false
streamChecker.query('k');          // return false
streamChecker.query('l');          // return true, because 'kl' is in the wordlist

 

Note:

• 1 <= words.length <= 2000
• 1 <= words[i].length <= 2000
• Words will only consist of lowercase English letters.
• Queries will only consist of lowercase English letters.
• The number of queries is at most 40000.

 

Approach #1:

class StreamChecker {
    public class TriNode {
        boolean isEnd = false;
        TriNode[] next = new TriNode[26];
    }
    
    TriNode root = new TriNode();
    StringBuilder buf = new StringBuilder();
    
    void insert(String word) {
        TriNode temp = root;
        for (int i = 0; i < word.length(); ++i) {
            char ch = word.charAt(word.length()-i-1);
            if (temp.next[ch-'a'] == null) temp.next[ch-'a'] = new TriNode();
            temp = temp.next[ch-'a'];
        }
        temp.isEnd = true;
    }
    
    public StreamChecker(String[] words) {
        for (String word : words) {
            insert(word);
        }
    }
    
    public boolean query(char letter) {
        buf.append(letter);
        // System.out.println(buf.toString());
        TriNode p = root;
        for (int i = buf.length() - 1; i >= 0; --i) {
            char ch = buf.charAt(i);
            p = p.next[ch-'a'];
            if (p == null) return false;
            if (p.isEnd) return true;
        }
        return false;
    }
}

/**
 * Your StreamChecker object will be instantiated and called as such:
 * StreamChecker obj = new StreamChecker(words);
 * boolean param_1 = obj.query(letter);
 */