Posts tagged as “grid”

Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.

Example 1:

Input: matrix =
[
[0,1,1,1],
[1,1,1,1],
[0,1,1,1]
]
Output: 15
Explanation:
There are 10 squares of side 1.
There are 4 squares of side 2.
There is  1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.


Example 2:

Input: matrix =
[
[1,0,1],
[1,1,0],
[1,1,0]
]
Output: 7
Explanation:
There are 6 squares of side 1.
There is 1 square of side 2.
Total number of squares = 6 + 1 = 7.


Constraints:

• 1 <= arr.length <= 300
• 1 <= arr[0].length <= 300
• 0 <= arr[i][j] <= 1

Solution: DP

dp[i][j] := edge of largest square with bottom right corner at (i, j)
dp[i][j] = min(dp[i – 1][j], dp[i – 1][j – 1], dp[i][j – 1]) if m[i][j] == 1 else 0
ans = sum(dp)

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

C++

You are given a map of a server center, represented as a m * n integer matrix grid, where 1 means that on that cell there is a server and 0 means that it is no server. Two servers are said to communicate if they are on the same row or on the same column.

Return the number of servers that communicate with any other server.

Example 1:

Input: grid = [[1,0],[0,1]]
Output: 0
Explanation: No servers can communicate with others.

Example 2:

Input: grid = [[1,0],[1,1]]
Output: 3
Explanation: All three servers can communicate with at least one other server.


Example 3:

Input: grid = [[1,1,0,0],[0,0,1,0],[0,0,1,0],[0,0,0,1]]
Output: 4
Explanation: The two servers in the first row can communicate with each other. The two servers in the third column can communicate with each other. The server at right bottom corner can't communicate with any other server.


Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m <= 250
• 1 <= n <= 250
• grid[i][j] == 0 or 1

Solution: Counting

Two passes:
First pass, count number of computers for each row and each column.
Second pass, count grid[i][j] if rows[i] or cols[j] has more than 1 computer.

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

C++

In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.

Return the maximum amount of gold you can collect under the conditions:

• Every time you are located in a cell you will collect all the gold in that cell.
• From your position you can walk one step to the left, right, up or down.
• You can’t visit the same cell more than once.
• Never visit a cell with 0 gold.
• You can start and stop collecting gold from any position in the grid that has some gold.

Example 1:

Input: grid = [[0,6,0],[5,8,7],[0,9,0]]
Output: 24
Explanation:
[[0,6,0],
[5,8,7],
[0,9,0]]
Path to get the maximum gold, 9 -> 8 -> 7.


Example 2:

Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
Output: 28
Explanation:
[[1,0,7],
[2,0,6],
[3,4,5],
[0,3,0],
[9,0,20]]
Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.


Constraints:

• 1 <= grid.length, grid[i].length <= 15
• 0 <= grid[i][j] <= 100
• There are at most 25 cells containing gold.

Solution: DFS

Time compleixty: O(4^25) ???
Space complexity: O(25)

C++

Given a 2D grid of 0s and 1s, return the number of elements in the largest square subgrid that has all 1s on its border, or 0 if such a subgrid doesn’t exist in the grid.

Example 1:

Input: grid = [[1,1,1],[1,0,1],[1,1,1]]
Output: 9


Example 2:

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


Constraints:

• 1 <= grid.length <= 100
• 1 <= grid[0].length <= 100
• grid[i][j] is 0 or 1

Solution: DP

Compute the sums of all rectangles that has left-top corner at (0, 0) in O(m*n) time.
For each square and check whether its borders are all ones in O(1) time.

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

C++

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++

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