花花酱 LeetCode 916. Word Subsets

By zxi on September 30, 2018

Problem

We are given two arrays A and B of words. Each word is a string of lowercase letters.

Now, say that word b is a subset of word a if every letter in b occurs in a, including multiplicity. For example, "wrr" is a subset of "warrior", but is not a subset of "world".

Now say a word a from A is universal if for every b in B, b is a subset of a.

Return a list of all universal words in A. You can return the words in any order.

Example 1:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
Output: ["facebook","google","leetcode"]

Example 2:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
Output: ["apple","google","leetcode"]

Example 3:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
Output: ["facebook","google"]

Example 4:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
Output: ["google","leetcode"]

Example 5:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
Output: ["facebook","leetcode"]

Note:

1 <= A.length, B.length <= 10000
1 <= A[i].length, B[i].length <= 10
A[i] and B[i] consist only of lowercase letters.
All words in A[i] are unique: there isn’t i != j with A[i] == A[j].

Solution: Hashtable

Find the max requirement for each letter from B.

Time complexity: O(|A|+|B|)

Space complexity: O(26)

C++

// Author: Huahua, 104 ms

class Solution {

public:

vector<string> wordSubsets(vector<string>& A, vector<string>& B) {

vector<int> req(26);

for (const string& b : B) {

vector<int> cur(getCount(b));

for (int i = 0; i < 26; ++i)

req[i] = max(req[i], cur[i]);

}

vector<string> ans;

for (const string& a : A) {

vector<int> cur(getCount(a));

bool universal = true;

for (int i = 0; i < 26; ++i)

if (cur[i] < req[i]) {

universal = false;

break;

}

if (universal) ans.push_back(a);

}

return ans;

}

private:

vector<int> getCount(const string& a) {

vector<int> count(26);

for (char c : a)

++count[c - 'a'];

return count;

}

};

花花酱 LeetCode 914. X of a Kind in a Deck of Cards

By zxi on September 30, 2018

Problem

n a deck of cards, each card has an integer written on it.

Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where:

Each group has exactly X cards.
All the cards in each group have the same integer.

Example 1:

Input: [1,2,3,4,4,3,2,1]
Output: true
Explanation: Possible partition [1,1],[2,2],[3,3],[4,4]

Example 2:

Input: [1,1,1,2,2,2,3,3]
Output: false
Explanation: No possible partition.

Example 3:

Input: [1]
Output: false
Explanation: No possible partition.

Example 4:

Input: [1,1]
Output: true
Explanation: Possible partition [1,1]

Example 5:

Input: [1,1,2,2,2,2]
Output: true
Explanation: Possible partition [1,1],[2,2],[2,2]

Note:

1 <= deck.length <= 10000
0 <= deck[i] < 10000

Solution 1: HashTable + Brute Force

Try all possible Xs. 2 ~ deck.size()

Time complexity: ~O(n)

Space complexity: O(1)

C++

// Author: Huahua, 16 ms

class Solution {

public:

bool hasGroupsSizeX(vector<int>& deck) {

unordered_map<int, int> counts;

for (int card : deck) ++counts[card];

for (int i = 2; i <= deck.size(); ++i) {

if (deck.size() % i) continue;

bool ok = true;

for (const auto& pair : counts)

if (pair.second % i) {

ok = false;

break;

}

if (ok) return true;

}

return false;

}

};

Solution 2: HashTable + GCD

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua, 16 ms

class Solution {

public:

bool hasGroupsSizeX(vector<int>& deck) {

unordered_map<int, int> counts;

for (int card : deck) ++counts[card];

int X = deck.size();

for (const auto& p : counts)

X = __gcd(X, p.second);

return X >= 2;

}

};

Java

// Author: Somebody

class Solution {

public boolean hasGroupsSizeX(int[] deck) {

HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

for (int i=0; i<deck.length; i++) {

int curr = deck[i];

if (map.containsKey(curr)) {

map.put(curr, map.get(curr) + 1);

} else {

map.put(curr, 1);

}

// O(nLogn)

int gcd = 0;

for(Integer key: map.keySet()) {

gcd = findGCDOfTwoNumbers(map.get(key), gcd);

}

return gcd >= 2;

}

// O(Log n)

public static int findGCDOfTwoNumbers (int a, int b) {

if (b == 0) {

return a;

}

return findGCDOfTwoNumbers(b, a%b);

}

花花酱 LeetCode 88. Merge Sorted Array

By zxi on September 27, 2018

Problem

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

The number of elements initialized in nums1 and nums2 are m and n respectively.
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.

Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]

Solution:

Fill nums1 from back to front

Time complexity: O(m + n)

Space complexity: O(1) in-place

C++

// Author: Huahua

class Solution {

public:

void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {

int i = m - 1;

int j = n - 1;

int tail = m + n - 1;

while (j >= 0)

nums1[tail--] = (i >= 0 && nums1[i] > nums2[j]) ? nums1[i--] : nums2[j--];

}

};

花花酱 LeetCode 910. Smallest Range II

By zxi on September 26, 2018

Problem

Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add x to A[i] (only once).

After this process, we have some array B.

Return the smallest possible difference between the maximum value of B and the minimum value of B.

Example 1:

Input: A = [1], K = 0
Output: 0
Explanation: B = [1]

Example 2:

Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]

Example 3:

Input: A = [1,3,6], K = 3
Output: 3
Explanation: B = [4,6,3]

Note:

1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000

Solution: Greedy

Sort the array and compare adjacent numbers.

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int smallestRangeII(vector<int>& A, int K) {

sort(begin(A), end(A));

int ans = A.back() - A.front();

for (int i = 1; i < A.size(); ++i) {

int l = min(A.front() + K, A[i] - K);

int h = max(A.back() - K, A[i - 1] + K);

ans = min(ans, h - l);

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def smallestRangeII(self, A, K):

A.sort()

ans = A[-1] - A[0]

for a, b in zip(A[0:-1], A[1:]):

l = min(A[0] + K, b - K)

h = max(A[-1] - K, a + K)

ans = min(ans, h - l)

return ans

花花酱 LeetCode 909. Snakes and Ladders

By zxi on September 25, 2018

Problem

On an N x N board, the numbers from 1 to N*N are written boustrophedonically starting from the bottom left of the board, and alternating direction each row. For example, for a 6 x 6 board, the numbers are written as follows:

You start on square 1 of the board (which is always in the last row and first column). Each move, starting from square x, consists of the following:

You choose a destination square S with number x+1, x+2, x+3, x+4, x+5, or x+6, provided this number is <= N*N.
- (This choice simulates the result of a standard 6-sided die roll: ie., there are always at most 6 destinations.)
If S has a snake or ladder, you move to the destination of that snake or ladder. Otherwise, you move to S.

A board square on row r and column c has a “snake or ladder” if board[r][c] != -1. The destination of that snake or ladder is board[r][c].

Note that you only take a snake or ladder at most once per move: if the destination to a snake or ladder is the start of another snake or ladder, you do not continue moving. (For example, if the board is [[4,-1],[-1,3]], and on the first move your destination square is 2, then you finish your first move at 3, because you do notcontinue moving to 4.)

Return the least number of moves required to reach square N*N. If it is not possible, return -1.

Example 1:

Input: [
[-1,-1,-1,-1,-1,-1],
[-1,-1,-1,-1,-1,-1],
[-1,-1,-1,-1,-1,-1],
[-1,35,-1,-1,13,-1],
[-1,-1,-1,-1,-1,-1],
[-1,15,-1,-1,-1,-1]]
Output: 4
Explanation: 
At the beginning, you start at square 1 [at row 5, column 0].
You decide to move to square 2, and must take the ladder to square 15.
You then decide to move to square 17 (row 3, column 5), and must take the snake to square 13.
You then decide to move to square 14, and must take the ladder to square 35.
You then decide to move to square 36, ending the game.
It can be shown that you need at least 4 moves to reach the N*N-th square, so the answer is 4.

Note:

2 <= board.length = board[0].length <= 20
board[i][j] is between 1 and N*N or is equal to -1.
The board square with number 1 has no snake or ladder.
The board square with number N*N has no snake or ladder.

Solution: BFS

Time complexity: O(n*n)

Space complexity: O(n*n)

C++

// Author: Huahua, 8 ms (beats 100%)

class Solution {

public:

int snakesAndLadders(vector<vector<int>>& board) {

const int N = board.size();

int steps = 0;

vector<int> seen(N * N + 1, 0);

queue<int> q;

q.push(1);

seen[1] = 1;

while (!q.empty()) {

int size = q.size();

++steps;

while (size--) {

int s = q.front(); q.pop();

for (int x = s + 1; x <= min(s + 6, N * N); ++x) {

int r, c;

getRC(x, N, &r, &c);

int nx = board[r][c] == -1 ? x : board[r][c];

if (seen[nx]) continue;

if (nx == N * N) return steps;

q.push(nx);

seen[nx] = 1;

}

return -1;

}

private:

void getRC(int s, int N, int* r, int* c) {

int y = (s - 1) / N;

int x = (s - 1) % N;

*r = N - 1 - y;

*c = (y % 2) ? N - 1 - x : x;

}

};

Huahua's Tech Road

花花酱 LeetCode 916. Word Subsets

Problem

Solution: Hashtable

C++

花花酱 LeetCode 914. X of a Kind in a Deck of Cards

Problem

Solution 1: HashTable + Brute Force

C++

Solution 2: HashTable + GCD

C++

Java

花花酱 LeetCode 88. Merge Sorted Array

Problem

Solution:

C++

花花酱 LeetCode 910. Smallest Range II

Problem

Solution: Greedy

C++

Python3

花花酱 LeetCode 909. Snakes and Ladders

Problem

Solution: BFS

C++