花花酱 LeetCode 936. Stamping The Sequence

By zxi on December 8, 2018

Problem

You want to form a target string of lowercase letters.

At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.

On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.

For example, if the initial sequence is “?????”, and your stamp is "abc", then you may make “abc??”, “?abc?”, “??abc” in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)

If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.

For example, if the sequence is “ababc”, and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves “?????” -> “abc??” -> “ababc”.

Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.

Example 1:

Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)

Example 2:

Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]

Note:

1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters.

Solution: Greedy + Reverse Simulation

Reverse the stamping process. Each time find a full or partial match. Replace the matched char to ‘?’.

Don’t forget the reverse the answer as well.

T = “ababc”, S = “abc”

T = “ab???”, index = 2

T = “?????”, index = 0

ans = [0, 2]

Time complexity: O((T – S)*S)

Space complexity: O(T)

C++

// Author: Huahua, running time: 12 ms

class Solution {

public:

vector<int> movesToStamp(string stamp, string target) {

vector<int> ans;

vector<int> seen(target.length());

int total = 0;

while (total < target.length()) {

bool found = false;

for (int i = 0; i <= target.length() - stamp.length(); ++i) {

if (seen[i]) continue;

int l = unStamp(stamp, target, i);

if (l == 0) continue;

seen[i] = 1;

total += l;

ans.push_back(i);

found = true;

}

if (!found) return {};

}

reverse(begin(ans), end(ans));

return ans;

}

private:

int unStamp(const string& stamp, string& target, int s) {

int l = stamp.size();

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

if (target[s + i] == '?')

--l;

else if (target[s + i] != stamp[i])

return 0;

}

if (l != 0)

std::fill(begin(target) + s, begin(target) + s + stamp.length(), '?');

return l;

}

};

花花酱 LeetCode 952. Largest Component Size by Common Factor

By zxi on December 1, 2018

Problem

Given a non-empty array of unique positive integers A, consider the following graph:

There are A.length nodes, labelled A[0] to A[A.length - 1];
There is an edge between A[i] and A[j] if and only if A[i] and A[j] share a common factor greater than 1.

Return the size of the largest connected component in the graph.

Example 1:

Input: [4,6,15,35]
Output: 4

Example 2:

Input: [20,50,9,63]
Output: 2

Example 3:

Input: [2,3,6,7,4,12,21,39]
Output: 8

Note:

1 <= A.length <= 20000
1 <= A[i] <= 100000

Solution: Union Find

For each number, union itself with all its factors.

E.g. 6, union(6,2), union(6,3)

Time complexity: \( O(\Sigma{sqrt(A[i])}) \)

Space complexity: \( O(max(A)) \)

C++

// Author: Huahua, running time: 148 ms

class DSU {

public:

DSU(int n): p_(n) {

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

p_[i] = i;

}

void Union(int x, int y) {

p_[Find(x)] = p_[Find(y)];

}

int Find(int x) {

if (p_[x] != x) p_[x] = Find(p_[x]);

return p_[x];

}

private:

vector<int> p_;

};

class Solution {

public:

int largestComponentSize(vector<int>& A) {

int n = *max_element(begin(A), end(A));

DSU dsu(n + 1);

for (int a : A) {

int t = sqrt(a);

for (int k = 2; k <= t; ++k)

if (a % k == 0) {

dsu.Union(a, k);

dsu.Union(a, a / k);

}

unordered_map<int, int> c;

int ans = 1;

for (int a : A)

ans = max(ans, ++c[dsu.Find(a)]);

return ans;

}

};

Python3

# Author: Huahua, running time: 3432 ms

class Solution:

def largestComponentSize(self, A):

p = list(range(max(A) + 1))

def find(x):

while p[x] != x:

p[x] = p[p[x]]

x = p[x]

return x

def union(x, y):

p[find(x)] = p[find(y)]

for a in A:

for k in range(2, int(math.sqrt(a) + 1)):

if a % k == 0:

union(a, k)

union(a, a // k)

return collections.Counter([find(a) for a in A]).most_common(1)[0][1]

LeetCode Weekly Contest 113

By zxi on December 1, 2018

花花酱 LeetCode 948. Bag of Tokens

By zxi on November 29, 2018

Problem

You have an initial power P, an initial score of 0 points, and a bag of tokens.

Each token can be used at most once, has a value token[i], and has potentially two ways to use it.

If we have at least token[i] power, we may play the token face up, losing token[i] power, and gaining 1 point.
If we have at least 1 point, we may play the token face down, gaining token[i]power, and losing 1 point.

Return the largest number of points we can have after playing any number of tokens.

Example 1:

Input: tokens = [100], P = 50
Output: 0

Example 2:

Input: tokens = [100,200], P = 150
Output: 1

Example 3:

Input: tokens = [100,200,300,400], P = 200
Output: 2

Note:

tokens.length <= 1000
0 <= tokens[i] < 10000
0 <= P < 10000

Solution: Greedy + Two Pointers

Sort the tokens, gain points from the low end gain power from the high end.

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua, Running time: 8 ms

class Solution {

public:

int bagOfTokensScore(vector<int>& tokens, int P) {

sort(begin(tokens), end(tokens));

int points = 0;

int ans = 0;

int i = 0;

int j = tokens.size() - 1;

while (i <= j)

if (P >= tokens[i]) {

P -= tokens[i++];

ans = max(ans, ++points);

} else if (points > 0) {

P += tokens[j--];

--points;

} else {

break;

}

return ans;

}

};

花花酱 LeetCode 947. Most Stones Removed with Same Row or Column

By zxi on November 29, 2018

Problem

On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at most one stone.

Now, a move consists of removing a stone that shares a column or row with another stone on the grid.

What is the largest possible number of moves we can make?

Example 1:

Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5

Example 2:

Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3

Example 3:

Input: stones = [[0,0]]
Output: 0

Note:

1 <= stones.length <= 1000
0 <= stones[i][j] < 10000

Solution 2: Union Find

Find all connected components (islands)

Ans = # of stones – # of islands

C++

// Author: Huahua, running time: 16 ms

class Solution {

public:

int removeStones(vector<vector<int>>& stones) {

const int kSize = 10000;

DSU dsu(kSize * 2);

for (const auto& stone : stones)

dsu.Union(stone[0], stone[1] + kSize);

unordered_set<int> seen;

for (const auto& stone : stones)

seen.insert(dsu.Find(stone[0]));

return stones.size() - seen.size();

}

private:

class DSU {

public:

DSU(int n): p_(n) {

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

p_[i] = i;

}

void Union(int i, int j) {

p_[Find(i)] = Find(j);

}

int Find(int i) {

if (p_[i] != i) p_[i] = Find(p_[i]);

return p_[i];

}

private:

vector<int> p_;

};

Huahua's Tech Road

花花酱 LeetCode 936. Stamping The Sequence

Problem

Solution: Greedy + Reverse Simulation

C++

花花酱 LeetCode 952. Largest Component Size by Common Factor

Problem

Solution: Union Find

C++

Python3

LeetCode Weekly Contest 113

花花酱 LeetCode 948. Bag of Tokens

Problem

Solution: Greedy + Two Pointers

C++

花花酱 LeetCode 947. Most Stones Removed with Same Row or Column

Problem

Solution 2: Union Find

C++

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