Posts tagged as “hard”

花花酱 LeetCode 2642. Design Graph With Shortest Path Calculator

By zxi on April 29, 2023

There is a directed weighted graph that consists of n nodes numbered from 0 to n - 1. The edges of the graph are initially represented by the given array edges where edges[i] = [from_i, to_i, edgeCost_i] meaning that there is an edge from from_i to to_i with the cost edgeCost_i.

Implement the Graph class:

Graph(int n, int[][] edges) initializes the object with n nodes and the given edges.
addEdge(int[] edge) adds an edge to the list of edges where edge = [from, to, edgeCost]. It is guaranteed that there is no edge between the two nodes before adding this one.
int shortestPath(int node1, int node2) returns the minimum cost of a path from node1 to node2. If no path exists, return -1. The cost of a path is the sum of the costs of the edges in the path.

Example 1:

Input
["Graph", "shortestPath", "shortestPath", "addEdge", "shortestPath"]
[[4, [[0, 2, 5], [0, 1, 2], [1, 2, 1], [3, 0, 3]]], [3, 2], [0, 3], [[1, 3, 4]], [0, 3]]
Output
[null, 6, -1, null, 6]

Explanation

Graph g = new Graph(4, [[0, 2, 5], [0, 1, 2], [1, 2, 1], [3, 0, 3]]); g.shortestPath(3, 2); // return 6. The shortest path from 3 to 2 in the first diagram above is 3 -> 0 -> 1 -> 2 with a total cost of 3 + 2 + 1 = 6.
g.shortestPath(0, 3); // return -1. There is no path from 0 to 3. 
g.addEdge([1, 3, 4]); // We add an edge from node 1 to node 3, and we get the second diagram above. 
g.shortestPath(0, 3); // return 6. The shortest path from 0 to 3 now is 0 -> 1 -> 3 with a total cost of 2 + 4 = 6.

Constraints:

1 <= n <= 100
0 <= edges.length <= n * (n - 1)
edges[i].length == edge.length == 3
0 <= from_i, to_i, from, to, node1, node2 <= n - 1
1 <= edgeCost_i, edgeCost <= 10⁶
There are no repeated edges and no self-loops in the graph at any point.
At most 100 calls will be made for addEdge.
At most 100 calls will be made for shortestPath.

Solution 1: Floyd-Washall

Time complexity:
Init O(n³)
addEdge O(n²)
shortestPath O(1)

Space complexity: O(1)

C++

// Author: Huahua

class Graph {

public:

Graph(int n, vector<vector<int>>& edges):

n_(n),

d_(n, vector<long>(n, 1e18)) {

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

d_[i][i] = 0;

for (const vector<int>& e : edges)

d_[e[0]][e[1]] = e[2];

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

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

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

d_[i][j] = min(d_[i][j], d_[i][k] + d_[k][j]);

}

void addEdge(vector<int> edge) {

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

for (int j = 0; j < n_; ++j)

d_[i][j] = min(d_[i][j], d_[i][edge[0]] + edge[2] + d_[edge[1]][j]);

}

int shortestPath(int node1, int node2) {

return d_[node1][node2] >= 1e18 ? -1 : d_[node1][node2];

}

private:

int n_;

vector<vector<long>> d_;

};

Solution 2: Dijkstra

Time complexity:
Init: O(|E|) ~ O(n²)
AddEdge: O(1)
ShortestPath: O(|V|*log(|E|)) ~ O(n*logn)

Space complexity: O(E|) ~ O(n²)

C++

// Author: Huahua

class Graph {

public:

Graph(int n, vector<vector<int>>& edges): n_(n), g_(n) {

for (const auto& e : edges)

g_[e[0]].emplace_back(e[1], e[2]);

}

void addEdge(vector<int> edge) {

g_[edge[0]].emplace_back(edge[1], edge[2]);

}

int shortestPath(int node1, int node2) {

vector<int> dist(n_, INT_MAX);

dist[node1] = 0;

priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> q;

q.emplace(0, node1);

while (!q.empty()) {

const auto [d, u] = q.top(); q.pop();

if (u == node2) return d;

for (const auto& [v, c] : g_[u])

if (dist[u] + c < dist[v]) {

dist[v] = dist[u] + c;

q.emplace(dist[v], v);

}

return -1;

}

private:

int n_;

vector<vector<pair<int, int>>> g_;

};

花花酱 LeetCode 2435. Paths in Matrix Whose Sum Is Divisible by K

By zxi on October 10, 2022

You are given a 0-indexed m x n integer matrix grid and an integer k. You are currently at position (0, 0) and you want to reach position (m - 1, n - 1) moving only down or right.

Return the number of paths where the sum of the elements on the path is divisible by k. Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: grid = [[5,2,4],[3,0,5],[0,7,2]], k = 3
Output: 2
Explanation: There are two paths where the sum of the elements on the path is divisible by k.
The first path highlighted in red has a sum of 5 + 2 + 4 + 5 + 2 = 18 which is divisible by 3.
The second path highlighted in blue has a sum of 5 + 3 + 0 + 5 + 2 = 15 which is divisible by 3.

Example 2:

Input: grid = [[0,0]], k = 5
Output: 1
Explanation: The path highlighted in red has a sum of 0 + 0 = 0 which is divisible by 5.

Example 3:

Input: grid = [[7,3,4,9],[2,3,6,2],[2,3,7,0]], k = 1
Output: 10
Explanation: Every integer is divisible by 1 so the sum of the elements on every possible path is divisible by k.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 5 * 10⁴
1 <= m * n <= 5 * 10⁴
0 <= grid[i][j] <= 100
1 <= k <= 50

Solution: DP

Let dp[i][j][r] := # of paths from (0,0) to (i,j) with path sum % k == r.

init: dp[0][0][grid[0][0] % k] = 1

dp[i][j][(r + grid[i][j]) % k] = dp[i-1][j][r] + dp[i][j-1][r]

ans = dp[m-1][n-1][0]

Time complexity: O(m*n*k)
Space complexity: O(m*n*k) -> O(n*k)

C++

// Author: Huahua

class Solution {

public:

int numberOfPaths(vector<vector<int>>& grid, int k) {

const int kMod = 1e9 + 7;

const int m = grid.size();

const int n = grid[0].size();

vector<vector<vector<int>>> dp(m, vector<vector<int>>(n, vector<int>(k)));

dp[0][0][grid[0][0] % k] = 1;

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

for (int j = 0; j < n; ++j) {

if (i == 0 && j == 0) continue;

for (int r = 0; r < k; ++r)

dp[i][j][(r + grid[i][j]) % k] =

((j ? dp[i][j - 1][r] : 0) + (i ? dp[i - 1][j][r] : 0)) % kMod;

}

return dp[m - 1][n - 1][0];

}

};

花花酱 LeetCode 2416. Sum of Prefix Scores of Strings

By zxi on September 18, 2022

You are given an array words of size n consisting of non-empty strings.

We define the score of a string word as the number of strings words[i] such that word is a prefix of words[i].

For example, if words = ["a", "ab", "abc", "cab"], then the score of "ab" is 2, since "ab" is a prefix of both "ab" and "abc".

Return an array answer of size n where answer[i] is the sum of scores of every non-empty prefix of words[i].

Note that a string is considered as a prefix of itself.

Example 1:

Input: words = ["abc","ab","bc","b"]
Output: [5,4,3,2]
Explanation: The answer for each string is the following:
- "abc" has 3 prefixes: "a", "ab", and "abc".
- There are 2 strings with the prefix "a", 2 strings with the prefix "ab", and 1 string with the prefix "abc".
The total is answer[0] = 2 + 2 + 1 = 5.
- "ab" has 2 prefixes: "a" and "ab".
- There are 2 strings with the prefix "a", and 2 strings with the prefix "ab".
The total is answer[1] = 2 + 2 = 4.
- "bc" has 2 prefixes: "b" and "bc".
- There are 2 strings with the prefix "b", and 1 string with the prefix "bc".
The total is answer[2] = 2 + 1 = 3.
- "b" has 1 prefix: "b".
- There are 2 strings with the prefix "b".
The total is answer[3] = 2.

Example 2:

Input: words = ["abcd"]
Output: [4]
Explanation:
"abcd" has 4 prefixes: "a", "ab", "abc", and "abcd".
Each prefix has a score of one, so the total is answer[0] = 1 + 1 + 1 + 1 = 4.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 1000
words[i] consists of lowercase English letters.

Solution: Trie

Insert all the words into a tire whose node val is the number of substrings that have the current prefix.

During query time, sum up the values along the prefix path.

Time complexity: O(sum(len(word))
Space complexity: O(sum(len(word))

C++

// Author: Huahua

struct Trie {

Trie* ch[26] = {};

int cnt = 0;

void insert(string_view s) {

Trie* cur = this;

for (char c : s) {

if (!cur->ch[c - 'a'])

cur->ch[c - 'a'] = new Trie();

cur = cur->ch[c - 'a'];

++cur->cnt;

}

int query(string_view s) {

Trie* cur = this;

int ans = 0;

for (char c : s) {

cur = cur->ch[c - 'a'];

ans += cur->cnt;

}

return ans;

}

};

class Solution {

public:

vector<int> sumPrefixScores(vector<string>& words) {

Trie* root = new Trie();

for (const string& w : words)

root->insert(w);

vector<int> ans;

for (const string& w : words)

ans.push_back(root->query(w));

return ans;

}

};

花花酱 LeetCode 2407. Longest Increasing Subsequence II

By zxi on September 11, 2022

You are given an integer array nums and an integer k.

Find the longest subsequence of nums that meets the following requirements:

The subsequence is strictly increasing and
The difference between adjacent elements in the subsequence is at most k.

Return the length of the longest subsequence that meets the requirements.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [4,2,1,4,3,4,5,8,15], k = 3
Output: 5
Explanation:
The longest subsequence that meets the requirements is [1,3,4,5,8].
The subsequence has a length of 5, so we return 5.
Note that the subsequence [1,3,4,5,8,15] does not meet the requirements because 15 - 8 = 7 is larger than 3.

Example 2:

Input: nums = [7,4,5,1,8,12,4,7], k = 5
Output: 4
Explanation:
The longest subsequence that meets the requirements is [4,5,8,12].
The subsequence has a length of 4, so we return 4.

Example 3:

Input: nums = [1,5], k = 1
Output: 1
Explanation:
The longest subsequence that meets the requirements is [1].
The subsequence has a length of 1, so we return 1.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i], k <= 10⁵

Solution: DP + Segment Tree | Max range query

Let dp[i] := length of LIS end with number i.
dp[i] = 1 + max(dp[i-k:i])

Naive dp takes O(n*k) time which will cause TLE.

We can use segment tree to speed up the max range query to log(m), where m is the max value of the array.

Time complexity: O(n*logm)
Space complexity: O(m)

C++

// Author: Huahua

class Solution {

public:

int lengthOfLIS(vector<int>& nums, int k) {

const int n = *max_element(begin(nums), end(nums));

vector<int> dp(2 * (n + 1));

auto query = [&](int l, int r) -> int {

int ans = 0;

for (l += n, r += n; l < r; l >>= 1, r >>= 1) {

if (l & 1) ans = max(ans, dp[l++]);

if (r & 1) ans = max(ans, dp[--r]);

}

return ans;

};

auto update = [&](int i, int val) -> void {

dp[i += n] = val;

while (i > 1) {

i >>= 1;

dp[i] = max(dp[i * 2], dp[i * 2 + 1]);

}

};

int ans = 0;

for (int x : nums) {

int cur = 1 + query(max(1, x - k), x);

update(x, cur);

ans = max(ans, cur);

}

return ans;

}

};

花花酱 LeetCode 2267. Check if There Is a Valid Parentheses String Path

By zxi on May 10, 2022

A parentheses string is a non-empty string consisting only of '(' and ')'. It is valid if any of the following conditions is true:

It is ().
It can be written as AB (A concatenated with B), where A and B are valid parentheses strings.
It can be written as (A), where A is a valid parentheses string.

You are given an m x n matrix of parentheses grid. A valid parentheses string path in the grid is a path satisfying all of the following conditions:

The path starts from the upper left cell (0, 0).
The path ends at the bottom-right cell (m - 1, n - 1).
The path only ever moves down or right.
The resulting parentheses string formed by the path is valid.

Return true if there exists a valid parentheses string path in the grid. Otherwise, return false.

Example 1:

Input: grid = [["(","(","("],[")","(",")"],["(","(",")"],["(","(",")"]]
Output: true
Explanation: The above diagram shows two possible paths that form valid parentheses strings.
The first path shown results in the valid parentheses string "()(())".
The second path shown results in the valid parentheses string "((()))".
Note that there may be other valid parentheses string paths.

Example 2:

Input: grid = [[")",")"],["(","("]]
Output: false
Explanation: The two possible paths form the parentheses strings "))(" and ")((". Since neither of them are valid parentheses strings, we return false.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 100
grid[i][j] is either '(' or ')'.

Solution: DP

Let dp(i, j, b) denote whether there is a path from (i,j) to (m-1, n-1) given b open parentheses.
if we are at (m – 1, n – 1) and b == 0 then we found a valid path.
dp(i, j, b) = dp(i + 1, j, b’) or dp(i, j + 1, b’) where b’ = b + 1 if grid[i][j] == ‘(‘ else -1

Time complexity: O(m*n*(m + n))
Space complexity: O(m*n*(m + n))

Python3

# Author: Huahua

class Solution:

def hasValidPath(self, grid: List[List[str]]) -> bool:

m, n = len(grid), len(grid[0])

@cache

def dp(i: int, j: int, b: int) -> bool:

if b < 0 or i == m or j == n: return False

b += 1 if grid[i][j] == '(' else -1

if i == m - 1 and j == n - 1 and b == 0: return True

return dp(i + 1, j, b) or dp(i, j + 1, b)

return dp(0, 0, 0)