# Posts tagged as “BFS”

A certain bug’s home is on the x-axis at position x. Help them get there from position 0.

The bug jumps according to the following rules:

• It can jump exactly a positions forward (to the right).
• It can jump exactly b positions backward (to the left).
• It cannot jump backward twice in a row.
• It cannot jump to any forbidden positions.

The bug may jump forward beyond its home, but it cannot jump to positions numbered with negative integers.

Given an array of integers forbidden, where forbidden[i] means that the bug cannot jump to the position forbidden[i], and integers ab, and x, return the minimum number of jumps needed for the bug to reach its home. If there is no possible sequence of jumps that lands the bug on position x, return -1.

Example 1:

Input: forbidden = [14,4,18,1,15], a = 3, b = 15, x = 9
Output: 3
Explanation: 3 jumps forward (0 -> 3 -> 6 -> 9) will get the bug home.


Example 2:

Input: forbidden = [8,3,16,6,12,20], a = 15, b = 13, x = 11
Output: -1


Example 3:

Input: forbidden = [1,6,2,14,5,17,4], a = 16, b = 9, x = 7
Output: 2
Explanation: One jump forward (0 -> 16) then one jump backward (16 -> 7) will get the bug home.


Constraints:

• 1 <= forbidden.length <= 1000
• 1 <= a, b, forbidden[i] <= 2000
• 0 <= x <= 2000
• All the elements in forbidden are distinct.
• Position x is not forbidden.

## Solution: BFS

Normal BFS with two tricks:
1. For each position, we need to track whether it’s reached via a forward jump or backward jump
2. How far should we go? If we don’t limit, it can go forever which leads to TLE/MLE. We can limit the distance to 2*max_jump, e.g. 4000, that’s maximum distance we can jump back to home in one shot.

Time complexity: O(max_distance * 2)
Space complexity: O(max_distance * 2)

## C++

You are a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move updownleft, or right, and you wish to find a route that requires the minimum effort.

A route’s effort is the maximum absolute differencein heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

Example 1:

Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
Output: 2
Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.


Example 2:

Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
Output: 1
Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].


Example 3:

Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
Output: 0
Explanation: This route does not require any effort.


Constraints:

• rows == heights.length
• columns == heights[i].length
• 1 <= rows, columns <= 100
• 1 <= heights[i][j] <= 106

## Solution: “Lazy BFS / DP”

dp[y][x] = min(max(dp[ty][tx], abs(h[ty][tx] – h[y][x]))) (x, y) and (tx, ty) are neighbors
repeat this process for at most rows * cols times.
if dp does not change after one round which means we found the optimal solution and can break earlier.

Time complexity: O(n^2*m^2))
Space complexity: O(nm)

## Solution 2: Binary Search + BFS

Use binary search to guess a cost and then check whether there is path that is under the cost.

Time complexity: O(mn*log(max(h) – min(h)))
Space complexity: O(mn)

## Solution 3: Dijkstra

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

## C++

There are n cities numbered from 0 to n-1 and n-1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.

Roads are represented by connections where connections[i] = [a, b] represents a road from city a to b.

This year, there will be a big event in the capital (city 0), and many people want to travel to this city.

Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed.

It’s guaranteed that each city can reach the city 0 after reorder.

Example 1:

Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
Output: 3
Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 2:

Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
Output: 2
Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 3:

Input: n = 3, connections = [[1,0],[2,0]]
Output: 0


Constraints:

• 2 <= n <= 5 * 10^4
• connections.length == n-1
• connections[i].length == 2
• 0 <= connections[i][0], connections[i][1] <= n-1
• connections[i][0] != connections[i][1]

## Solution: BFS

Augment the graph
g[u][v] = 1, g[v][u] = 0, u->v is an edge in the original graph.

BFS from 0, sum up all the edge costs to visit all the nodes.

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

## C++

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++

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)

## python3

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