Posts tagged as “hard”

花花酱 LeetCode 927. Three Equal Parts

By zxi on October 20, 2018

Problem

Given an array A of 0s and 1s, divide the array into 3 non-empty parts such that all of these parts represent the same binary value.

If it is possible, return any [i, j] with i+1 < j, such that:

A[0], A[1], ..., A[i] is the first part;
A[i+1], A[i+2], ..., A[j-1] is the second part, and
A[j], A[j+1], ..., A[A.length - 1] is the third part.
All three parts have equal binary value.

If it is not possible, return [-1, -1].

Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.

Example 1:

Input: [1,0,1,0,1]
Output: [0,3]

Example 2:

Input: [1,1,0,1,1]
Output: [-1,-1]

Note:

3 <= A.length <= 30000
A[i] == 0 or A[i] == 1

Solution:

each part should have the same number of 1 s.

Find the suffix (without leading os) of the last part which should have 1/3 of the total ones.

Time complexity: O(n^2) in theory but close to O(n) in practice

Space complexity: O(n)

C++

// Author: Huahua, 44 ms

class Solution {

public:

vector<int> threeEqualParts(vector<int>& A) {

string s(begin(A), end(A));

int ones = accumulate(begin(A), end(A), 0);

if (ones % 3 != 0) return {-1, -1};

if (ones == 0) return {0, A.size() - 1};

ones /= 3;

int right = A.size() - 1;

while (ones) if (A[right--]) --ones;

string suffix(begin(s) + right + 1, end(s));

size_t l = suffix.length();

size_t left = s.find(suffix);

if (left == std::string::npos) return {-1, -1};

size_t mid = s.find(suffix, left + l);

if (mid == std::string::npos

|| mid + 2 * l > s.length()) return {-1, -1};

return {left + l - 1, mid + l};

}

};

花花酱 LeetCode 924. Minimize Malware Spread

By zxi on October 13, 2018

Problem

In a network of nodes, each node i is directly connected to another node j if and only if graph[i][j] = 1.

Some nodes initial are initially infected by malware. Whenever two nodes are directly connected and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.

Suppose M(initial) is the final number of nodes infected with malware in the entire network, after the spread of malware stops.

We will remove one node from the initial list. Return the node that if removed, would minimize M(initial). If multiple nodes could be removed to minimize M(initial), return such a node with the smallest index.

Note that if a node was removed from the initial list of infected nodes, it may still be infected later as a result of the malware spread.

Example 1:

Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
Output: 0

Example 2:

Input: graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2]
Output: 0

Example 3:

Input: graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2]
Output: 1

Note:

1 < graph.length = graph[0].length <= 300
0 <= graph[i][j] == graph[j][i] <= 1
graph[i][i] = 1
1 <= initial.length < graph.length
0 <= initial[i] < graph.length

Solution: BFS

Time complexity: O(n^3)

Space complexity: O(n^2)

C++

// Author: Huahua

class Solution {

public:

int minMalwareSpread(vector<vector<int>>& graph, vector<int>& initial) {

int min_affected = INT_MAX;

int min_node = -1;

sort(begin(initial), end(initial));

for (int t : initial) {

vector<int> bad(graph.size(), 0);

queue<int> q;

for (int n : initial)

if (n != t) {

bad[n] = 1;

q.push(n);

}

int affected = initial.size() - 1;

while (!q.empty()) {

int size = q.size();

while (size--) {

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

for (int i = 0; i < graph[n].size(); ++i) {

if (graph[n][i] == 0 || bad[i]) continue;

++affected;

bad[i] = 1;

q.push(i);

}

if (affected < min_affected) {

min_affected = affected;

min_node = t;

}

return min_node;

}

};

花花酱 LeetCode 920. Number of Music Playlists

By zxi on October 6, 2018

Problem

Your music player contains N different songs and she wants to listen to L (not necessarily different) songs during your trip. You create a playlist so that:

Every song is played at least once
A song can only be played again only if K other songs have been played

Return the number of possible playlists. As the answer can be very large, return it modulo 10^9 + 7.

Example 1:

Input: N = 3, L = 3, K = 1
Output: 6
Explanation: There are 6 possible playlists. [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].

Example 2:

Input: N = 2, L = 3, K = 0
Output: 6
Explanation: There are 6 possible playlists. [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], [1, 2, 2]

Example 3:

Input: N = 2, L = 3, K = 1
Output: 2
Explanation: There are 2 possible playlists. [1, 2, 1], [2, 1, 2]

Note:

0 <= K < N <= L <= 100

Solution: DP

dp[i][j] := # of playlists of length i using j different songs.

dp[i][j] = dp[i – 1][j – 1] * (N – (j – 1)) + // Adding a new song. j – 1 used, choose any one from (N – (j – 1)) unused.
dp[i -1][j] * max(j – K, 0) // Reuse an existing song.

Time complexity: O(LN)

Space complexity: O(LN) -> O(N)

C++/O(LN)

// Author: Huahua

class Solution {

public:

int numMusicPlaylists(int N, int L, int K) {

constexpr long kMod = 1e9 + 7;

vector<vector<long>> dp(L + 1, vector<long>(N + 1, 0));

dp[0][0] = 1;

for (int i = 1; i <= L; ++i)

for (int j = 1; j <= min(i, N); ++j)

dp[i][j] = (dp[i - 1][j - 1] * (N - (j - 1)) +

dp[i - 1][j] * max(j - K, 0)) % kMod;

return dp[L][N];

}

};

C++/O(N)

// Author: Huahua, 0 ms

class Solution {

public:

int numMusicPlaylists(int N, int L, int K) {

constexpr long kMod = 1e9 + 7;

vector<long> dp(N + 1, 0);

for (int i = 1; i <= L; ++i) {

dp[0] = i == 1;

for (int j = min(i, N); j >= 1; --j)

dp[j] = (dp[j - 1] * (N - (j - 1)) +

dp[j] * max(j - K, 0)) % kMod;

}

return dp[N];

}

};

花花酱 LeetCode 32. Longest Valid Parentheses

By zxi on October 3, 2018

Problem

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.

Example 1:

Input: "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()"

Example 2:

Input: ")()())" Output: 4 Explanation: The longest valid parentheses substring is "()()"

Solution: Stack

Use a stack to track the index of all unmatched open parentheses.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int longestValidParentheses(string s) {

stack<int> q;

int start = 0;

int ans = 0;

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

if(s[i] == '(') {

q.push(i);

} else {

if (q.empty()) {

start = i + 1;

} else {

int index = q.top(); q.pop();

ans = max(ans, q.empty() ? i - start + 1 : i - q.top());

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def longestValidParentheses(self, s):

q = []

start, ans = 0, 0

for i in range(len(s)):

if s[i] == '(':

q.append(i)

continue

if not q:

start = i + 1

else:

q.pop()

ans = max(ans, i - q[-1] if q else i - start + 1)

return ans

Posts tagged as “hard”

花花酱 LeetCode 927. Three Equal Parts

Problem

Solution:

C++

花花酱 LeetCode 924. Minimize Malware Spread

Problem

Solution: BFS

C++

花花酱 LeetCode 920. Number of Music Playlists

Problem

Solution: DP

C++/O(LN)

C++/O(N)

花花酱 LeetCode 32. Longest Valid Parentheses

Problem

Solution: Stack

C++

Python3

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花花酱 LeetCode 25. Reverse Nodes in k-Group

Problem

Solution

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