Suppose we have a matrix A of integers with R rows and C columns, we have to find the maximum score of a path starting from [0,0] and ending at [R-1,C-1]. Here the scoring technique will be the minimum value in that path. For example, the value of the path 8 → 4 → 5 → 9 is 4. A path moves some number of times from one visited cell to any neighboring unvisited cell in one of the 4 cardinal directions (north, east, west, south).
For example, if the grid is like −
5 | 4 | 5 |
1 | 2 | 6 |
7 | 4 | 6 |
The orange cells will be the path. The output is 4
To solve this, we will follow these steps −
- r := number of rows and c := number of columns
- ans := minimum of A[0, 0] and A[r – 1, c – 1]
- make one matrix called visited of the order same as A, and fill this with FALSE
- h := a list, where we store a tuple (-A[0, 0], 0, 0)
- Make heap from h
- while h is not empty
- v, x, y := delete the h from heap, and store three values
- if x = r – 1 and y := c – 1, then come out from loop
- ans := min of ans, A[x, y]
- visited[x, y] := true
- for dy, dx in the list [(-1, 0), (1, 0), (0, 1), (0, -1)], do
- a := x + dx and b := y + dy
- if a in range 0 to r – 1 and b in range 0 to c – 1 and visited[a, b] is false,
- insert (-A[a, b], a, b) into heap with h
- return ans
Let us see the following implementation to get better understanding −
Example
import heapq class Solution(object): def maximumMinimumPath(self, A): """ :type A: List[List[int]] :rtype: int """ r,c = len(A),len(A[0]) ans = min(A[0][0],A[-1][-1]) visited = [[False for i in range(c)] for j in range(r)] h = [(-A[0][0],0,0)] heapq.heapify(h) while h: # print(h) v,x,y = heapq.heappop(h) if x== r-1 and y == c-1: break ans = min(ans,A[x][y]) visited[x][y]= True for dx,dy in {(-1,0),(1,0),(0,1),(0,-1)}: a,b = x+dx,y+dy if a>=0 and a<r and b>=0 and b<c and not visited[a][b]: heapq.heappush(h,(-A[a][b],a,b)) return ans
Input
[[5,4,5],[1,2,6],[7,4,6]]
Output
4