# Problem

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1:
Input:

0 0 0
0 1 0
0 0 0


Output:

0 0 0
0 1 0
0 0 0


Example 2:
Input:

0 0 0
0 1 0
1 1 1


Output:

0 0 0
0 1 0
1 2 1


Note:

1. The number of elements of the given matrix will not exceed 10,000.
2. There are at least one 0 in the given matrix.
3. The cells are adjacent in only four directions: up, down, left and right.

# Solution 1: DP

Two passes:

1. down, right
2. up, left

Time complexity: O(mn)

Space complexity: O(mn)

# Solution 2: BFS

Start from all 0 cells and find shortest paths to rest of the cells.

Time complexity: O(mn)

Space complexity: O(mn)

# Problem

We are given a 2-dimensional grid"." is an empty cell, "#" is a wall, "@" is the starting point, ("a""b", …) are keys, and ("A""B", …) are locks.

We start at the starting point, and one move consists of walking one space in one of the 4 cardinal directions.  We cannot walk outside the grid, or walk into a wall.  If we walk over a key, we pick it up.  We can’t walk over a lock unless we have the corresponding key.

For some 1 <= K <= 6, there is exactly one lowercase and one uppercase letter of the first K letters of the English alphabet in the grid.  This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.

Return the lowest number of moves to acquire all keys.  If it’s impossible, return -1.

Example 1:

Input: ["@.a.#","###.#","b.A.B"]
Output: 8


Example 2:

Input: ["@..aA","..B#.","....b"]
Output: 6


Note:

1. 1 <= grid.length <= 30
2. 1 <= grid[0].length <= 30
3. grid[i][j] contains only '.''#''@''a'-'f' and 'A'-'F'
4. The number of keys is in [1, 6].  Each key has a different letter and opens exactly one lock.

# Solution: BFS

Time complexity: O(m*n*64)

Space complexity: O(m*n*64)

# Problem

https://leetcode.com/problems/shortest-path-visiting-all-nodes/description/

An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1) is given as graph.

graph.length = N, and j != i is in the list graph[i] exactly once, if and only if nodes i and j are connected.

Return the length of the shortest path that visits every node. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.

Example 1:

Input: [[1,2,3],[0],[0],[0]]
Output: 4
Explanation: One possible path is [1,0,2,0,3]

Example 2:

Input: [[1],[0,2,4],[1,3,4],[2],[1,2]]
Output: 4
Explanation: One possible path is [0,1,4,2,3]


# Solution: BFS

Time complexity: O(n*2^n)

Space complexity: O(n*2^n)

C++

C++ / vector

# Problem

https://leetcode.com/problems/bus-routes/description/

We have a list of bus routes. Each routes[i] is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->… forever.

We start at bus stop S (initially not on a bus), and we want to go to bus stop T. Travelling by buses only, what is the least number of buses we must take to reach our destination? Return -1 if it is not possible.

Example:
Input:
routes = [[1, 2, 7], [3, 6, 7]]
S = 1
T = 6
Output: 2
Explanation:
The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.


Note:

• 1 <= routes.length <= 500.
• 1 <= routes[i].length <= 500.
• 0 <= routes[i][j] < 10 ^ 6.

# Solution: BFS

Time Complexity: O(m*n) m: # of buses, n: # of routes

Space complexity: O(m*n + m)

C++

# Problem

https://leetcode.com/problems/minimum-genetic-mutation/description/

A gene string can be represented by an 8-character long string, with choices from "A""C""G""T".

Suppose we need to investigate about a mutation (mutation from “start” to “end”), where ONE mutation is defined as ONE single character changed in the gene string.

For example, "AACCGGTT" -> "AACCGGTA" is 1 mutation.

Also, there is a given gene “bank”, which records all the valid gene mutations. A gene must be in the bank to make it a valid gene string.

Now, given 3 things – start, end, bank, your task is to determine what is the minimum number of mutations needed to mutate from “start” to “end”. If there is no such a mutation, return -1.

Note:

1. Starting point is assumed to be valid, so it might not be included in the bank.
2. If multiple mutations are needed, all mutations during in the sequence must be valid.
3. You may assume start and end string is not the same.

Example 1:

start: "AACCGGTT"
end:   "AACCGGTA"
bank: ["AACCGGTA"]

return: 1


Example 2:

start: "AACCGGTT"
end:   "AAACGGTA"
bank: ["AACCGGTA", "AACCGCTA", "AAACGGTA"]

return: 2


Example 3:

start: "AAAAACCC"
end:   "AACCCCCC"
bank: ["AAAACCCC", "AAACCCCC", "AACCCCCC"]

return: 3

# Solution: BFS Shortest Path

Time complexity: O(n^2)

Space complexity: O(n)

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