Minimum initial vertices to traverse whole matrix with given conditions
We are given a matrix that contains different values in each cell. Our aim is to find the minimal set of positions in the matrix such that the entire matrix can be traversed starting from the positions in the set.
We can traverse the matrix under the below conditions:
- We can move only to those neighbors that contain values less than or equal to the current cell’s value. A neighbour of the cell is defined as the cell that shares a side with the given cell.
Input : 1 2 3 2 3 1 1 1 1 Output : 1 1 0 2 If we start from 1, 1 we can cover 6 vertices in the order (1, 1) -> (1, 0) -> (2, 0) -> (2, 1) -> (2, 2) -> (1, 2). We cannot cover the entire matrix with this vertex. Remaining vertices can be covered (0, 2) -> (0, 1) -> (0, 0). Input : 3 3 1 1 Output : 0 1 If we start from 0, 1, we can traverse the entire matrix from this single vertex in this order (0, 0) -> (0, 1) -> (1, 1) -> (1, 0). Traversing the matrix in this order satisfies all the conditions stated above.
From the above examples, we can easily identify that in order to use a minimum number of positions, we have to start from the positions having the highest cell value. Therefore, we pick the positions that contain the highest value in the matrix. We take the vertices having the highest value in a separate array. We perform DFS at every vertex starting from the
highest value. If we encounter any unvisited vertex during dfs then we have to include this vertex in our set. When all the cells have been processed, then the set contains the required vertices.
How does this work?
We need to visit all vertices and to reach the largest values we must start with them. If the two largest values are not adjacent, then both of them must be picked. If the two largest values are adjacent, then any of them can be picked as moving to equal value neighbors is allowed.
Time Complexity : O(N*M *log(N*M))
Space Complexity : O(N*M)
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