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花花酱 LeetCode 1391. Check if There is a Valid Path in a Grid

Given a m x ngrid. Each cell of the grid represents a street. The street of grid[i][j] can be:

  • 1 which means a street connecting the left cell and the right cell.
  • 2 which means a street connecting the upper cell and the lower cell.
  • 3 which means a street connecting the left cell and the lower cell.
  • 4 which means a street connecting the right cell and the lower cell.
  • 5 which means a street connecting the left cell and the upper cell.
  • 6 which means a street connecting the right cell and the upper cell.

You will initially start at the street of the upper-left cell (0,0). A valid path in the grid is a path which starts from the upper left cell (0,0) and ends at the bottom-right cell (m - 1, n - 1)The path should only follow the streets.

Notice that you are not allowed to change any street.

Return true if there is a valid path in the grid or false otherwise.

Example 1:

Input: grid = [[2,4,3],[6,5,2]]
Output: true
Explanation: As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).

Example 2:

Input: grid = [[1,2,1],[1,2,1]]
Output: false
Explanation: As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)

Example 3:

Input: grid = [[1,1,2]]
Output: false
Explanation: You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).

Example 4:

Input: grid = [[1,1,1,1,1,1,3]]
Output: true

Example 5:

Input: grid = [[2],[2],[2],[2],[2],[2],[6]]
Output: true

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 300
  • 1 <= grid[i][j] <= 6

Solution: BFS

Need to check both sides (x, y) -> (tx, ty) and (tx, ty) -> (x, y) to make sure a path exist.

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

C++

花花酱 LeetCode 87. Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great
   /    \
  gr    eat
 / \    /  \
g   r  e   at
           / \
          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat
   /    \
  rg    eat
 / \    /  \
r   g  e   at
           / \
          a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae
   /    \
  rg    tae
 / \    /  \
r   g  ta  e
       / \
      t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

Solution: Recursion

isScramble(s1, s2)
if s1 == s2: return true
if sorted(s1) != sroted(s2): return false
We try all possible partitions:

  1. s1[0:l] v.s s2[0:l] && s1[l:] vs s2[l:]
  2. s1[0:l] vs s2[L-l:l] && s1[l:] vs s2[0:L-l]

Time complexity: O(n^5)
Space complexity: O(n^4)

C++

Python3

花花酱 LeetCode 306. Additive Number

Additive number is a string whose digits can form additive sequence.

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

Given a string containing only digits '0'-'9', write a function to determine if it’s an additive number.

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Example 1:

Input: "112358"
Output: true
Explanation: The digits can form an additive sequence: 1, 1, 2, 3, 5, 8. 
             1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8

Example 2:

Input: "199100199"
Output: true
Explanation: The additive sequence is: 1, 99, 100, 199. 
             1 + 99 = 100, 99 + 100 = 199

Constraints:

  • num consists only of digits '0'-'9'.
  • 1 <= num.length <= 35

Solution: DFS

Time complexity: O(n^2)
Space complexity: O(n)

C++

Python3

花花酱 LeetCode 529. Minesweeper

Let’s play the minesweeper game (Wikipediaonline game)!

You are given a 2D char matrix representing the game board. ‘M’ represents an unrevealed mine, ‘E’ represents an unrevealed empty square, ‘B’ represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit (‘1’ to ‘8’) represents how many mines are adjacent to this revealed square, and finally ‘X’ represents a revealed mine.

Now given the next click position (row and column indices) among all the unrevealed squares (‘M’ or ‘E’), return the board after revealing this position according to the following rules:

  1. If a mine (‘M’) is revealed, then the game is over – change it to ‘X’.
  2. If an empty square (‘E’) with no adjacent mines is revealed, then change it to revealed blank (‘B’) and all of its adjacent unrevealed squares should be revealed recursively.
  3. If an empty square (‘E’) with at least one adjacent mine is revealed, then change it to a digit (‘1’ to ‘8’) representing the number of adjacent mines.
  4. Return the board when no more squares will be revealed.

Example 1:

Input: 

[['E', 'E', 'E', 'E', 'E'],
 ['E', 'E', 'M', 'E', 'E'],
 ['E', 'E', 'E', 'E', 'E'],
 ['E', 'E', 'E', 'E', 'E']]

Click : [3,0]

Output: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'M', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Explanation:

Example 2:

Input: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'M', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Click : [1,2]

Output: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'X', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Explanation:

Note:

  1. The range of the input matrix’s height and width is [1,50].
  2. The click position will only be an unrevealed square (‘M’ or ‘E’), which also means the input board contains at least one clickable square.
  3. The input board won’t be a stage when game is over (some mines have been revealed).
  4. For simplicity, not mentioned rules should be ignored in this problem. For example, you don’t need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.

Solution: DFS

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

C++

Solution 2: BFS

Python3

花花酱 LeetCode 1377. Frog Position After T Seconds

Given an undirected tree consisting of n vertices numbered from 1 to n. A frog starts jumping from the vertex 1. In one second, the frog jumps from its current vertex to another unvisited vertex if they are directly connected. The frog can not jump back to a visited vertex. In case the frog can jump to several vertices it jumps randomly to one of them with the same probability, otherwise, when the frog can not jump to any unvisited vertex it jumps forever on the same vertex. 

The edges of the undirected tree are given in the array edges, where edges[i] = [fromi, toi] means that exists an edge connecting directly the vertices fromi and toi.

Return the probability that after t seconds the frog is on the vertex target.

Example 1:

Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4
Output: 0.16666666666666666 
Explanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 probability to the vertex 2 after second 1 and then jumping with 1/2 probability to vertex 4 after second 2. Thus the probability for the frog is on the vertex 4 after 2 seconds is 1/3 * 1/2 = 1/6 = 0.16666666666666666. 

Example 2:

Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 1, target = 7
Output: 0.3333333333333333
Explanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 = 0.3333333333333333 probability to the vertex 7 after second 1. 

Example 3:

Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 20, target = 6
Output: 0.16666666666666666

Constraints:

  • 1 <= n <= 100
  • edges.length == n-1
  • edges[i].length == 2
  • 1 <= edges[i][0], edges[i][1] <= n
  • 1 <= t <= 50
  • 1 <= target <= n
  • Answers within 10^-5 of the actual value will be accepted as correct.

Solution: BFS

key: if a node has children, the fog jumps to to children so the probability at current node will become 0.

Time complexity: O(n)
Space complexity: O(n)

C++

python3