Given a cost matrix cost[][] and a position (m, n) in cost[][], write a function that returns cost of minimum cost path to reach (m, n) from (0, 0). Each cell of the matrix represents a cost to traverse through that cell. Total cost of a path to reach (m, n) is sum of all the costs on that path (including both source and destination). You can only traverse down, right and diagonally lower cells from a given cell, i.e., from a given cell (i, j), cells (i+1, j), (i, j+1) and (i+1, j+1) can be traversed. You may assume that all costs are positive integers.
For example, in the following figure, what is the minimum cost path to (2, 2)?
The path with minimum cost is highlighted in the following figure. The path is (0, 0) –> (0, 1) –> (1, 2) –> (2, 2). The cost of the path is 8 (1 + 2 + 2 + 3).
Python
# Dynamic Programming Python implementation of Min Cost Path# problemR = 3C = 3def minCost(cost, m, n): # Instead of following line, we can use int tc[m + 1][n + 1] or # dynamically allocate memory to save space. The following # line is used to keep te program simple and make it working # on all compilers. tc = [[0 for x in range(C)] for x in range(R)] tc[0][0] = cost[0][0] # Initialize first column of total cost(tc) array for i in range(1, m + 1): tc[i][0] = tc[i-1][0] + cost[i][0] # Initialize first row of tc array for j in range(1, n + 1): tc[0][j] = tc[0][j-1] + cost[0][j] # Construct rest of the tc array for i in range(1, m + 1): for j in range(1, n + 1): tc[i][j] = min(tc[i-1][j-1], tc[i-1][j], tc[i][j-1]) + cost[i][j] return tc[m][n]# Driver program to test above functionscost = [[1, 2, 3], [4, 8, 2], [1, 5, 3]]print(minCost(cost, 2, 2))# This code is contributed by Bhavya Jain |
8
Time Complexity: O(m*n)
Auxiliary Space: O(m*n)
Please refer complete article on Dynamic Programming | Set 6 (Min Cost Path) for more details!
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