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Posts tagged as “BFS”

花花酱 LeetCode 838. Push Dominoes

here are N dominoes in a line, and we place each domino vertically upright.

In the beginning, we simultaneously push some of the dominoes either to the left or to the right.

After each second, each domino that is falling to the left pushes the adjacent domino on the left.

Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right.

When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces.

For the purposes of this question, we will consider that a falling domino expends no additional force to a falling or already fallen domino.

Given a string “S” representing the initial state. S[i] = 'L', if the i-th domino has been pushed to the left; S[i] = 'R', if the i-th domino has been pushed to the right; S[i] = '.', if the i-th domino has not been pushed.

Return a string representing the final state. 

Example 1:

Input: ".L.R...LR..L.."
Output: "LL.RR.LLRRLL.."

Example 2:

Input: "RR.L"
Output: "RR.L"
Explanation: The first domino expends no additional force on the second domino.

Note:

  1. 0 <= N <= 10^5
  2. String dominoes contains only 'L‘, 'R' and '.'

Solution: Simulation

Simulate the push process, record the steps from L and R for each domino.
steps(L) == steps(R) => “.”
steps(L) < steps(R) => “L”
steps(L) > steps(R) => “R”

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

C++

花花酱 8 Puzzles – Bidirectional A* vs Bidirectional BFS

8 Puzzles # nodes expended of 1000 solvable instances

Conclusion:

Nodes expended: BiDirectional A* << A* (Manhattan) <= Bidirectional BFS < A* Hamming << BFS
Running time: BiDirectional A* < Bidirectional BFS <= A* (Manhattan) < A* Hamming << BFS

Code:

C++ Version

花花酱 LeetCode 994. Rotting Oranges

In a given grid, each cell can have one of three values:

  • the value 0 representing an empty cell;
  • the value 1 representing a fresh orange;
  • the value 2 representing a rotten orange.

Every minute, any fresh orange that is adjacent (4-directionally) to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange.  If this is impossible, return -1instead.

Example 1:

Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4

Example 2:

Input: [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation:  The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.

Example 3:

Input: [[0,2]]
Output: 0
Explanation:  Since there are already no fresh oranges at minute 0, the answer is just 0.

Note:

  1. 1 <= grid.length <= 10
  2. 1 <= grid[0].length <= 10
  3. grid[i][j] is only 01, or 2.

Solution: BFS

Time complexity: O(mn)
Space complexity: O(mn)

C++

花花酱 LeetCode 934. Shortest Bridge

Problem

https://leetcode.com/problems/shortest-bridge/description/

In a given 2D binary array A, there are two islands.  (An island is a 4-directionally connected group of 1s not connected to any other 1s.)

Now, we may change 0s to 1s so as to connect the two islands together to form 1 island.

Return the smallest number of 0s that must be flipped.  (It is guaranteed that the answer is at least 1.)

Example 1:

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

Example 2:

Input: [[0,1,0],[0,0,0],[0,0,1]]
Output: 2

Example 3:

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

Note:

  1. 1 <= A.length = A[0].length <= 100
  2. A[i][j] == 0 or A[i][j] == 1

Solution: DFS + BFS

  1. Use DFS to find one island and color all the nodes as 2 (BLUE).
  2. Use BFS to find the shortest path from any nodes with color 2 (BLUE) to any nodes with color 1 (RED).

Time complexity: O(mn)

Space complexity: O(mn)

C++

Related Problems

花花酱 LeetCode 924. Minimize Malware Spread

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. 1 < graph.length = graph[0].length <= 300
  2. 0 <= graph[i][j] == graph[j][i] <= 1
  3. graph[i][i] = 1
  4. 1 <= initial.length < graph.length
  5. 0 <= initial[i] < graph.length

Solution: BFS

Time complexity: O(n^3)

Space complexity: O(n^2)

C++