Minimum Path Cost in a Grid

This problem asks to find the minimum cost path from the top row to the bottom row of a grid, where the cost is the sum of the cell values and the movement costs between cells. We can efficiently solve this using dynamic programming.

Approach:

The dynamic programming approach builds a solution from subproblems. We'll use a DP table (or array in the optimized version) to store the minimum cost to reach each cell.

1. Initialization: The minimum cost to reach any cell in the first row is simply the value of that cell.

2. Iteration: For each subsequent row, we calculate the minimum cost to reach each cell in that row. To reach a cell (i, j), we consider all possible cells (i-1, k) in the previous row (where 0 <= k < n) and calculate the cost of moving from (i-1, k) to (i, j). This cost consists of:

   • The minimum cost to reach (i-1, k) (obtained from our DP table).
   • The movement cost from grid[i-1][k] to column j (obtained from moveCost).
   • The value of the current cell grid[i][j].

   We choose the minimum cost among all possible k values.

3. Result: The minimum cost to reach the bottom row is the minimum value in the last row of our DP table.

Optimization:

Since we only need the information from the previous row to compute the current row's minimum costs, we can optimize space complexity by using only two arrays to store the minimum cost for the current row and previous row instead of a full DP table.

Time and Space Complexity:

• Time Complexity: O(m * n^2), where 'm' is the number of rows and 'n' is the number of columns in the grid. This is because for each cell in rows 1 to m-1, we iterate through all n cells in the previous row to find the minimum cost.

• Space Complexity: O(n) in the optimized version (using two arrays to represent rows), as opposed to O(m*n) for the non-optimized version which uses a complete DP table.

Code Implementation:

Here are implementations in several programming languages:

Python:

import sys
 
def minPathCost(grid, moveCost):
    m, n = len(grid), len(grid[0])
    prev_row = grid[0]  # Initialize with the first row
 
    for i in range(1, m):
        curr_row = [sys.maxsize] * n # Initialize current row with maximum possible value
        for j in range(n):
            for k in range(n):
                curr_row[j] = min(curr_row[j], prev_row[k] + moveCost[grid[i - 1][k]][j] + grid[i][j])
        prev_row = curr_row
 
    return min(prev_row)
 
 

Java:

import java.util.Arrays;
 
class Solution {
    public int minPathCost(int[][] grid, int[][] moveCost) {
        int m = grid.length;
        int n = grid[0].length;
        int[] prevRow = grid[0];
 
        for (int i = 1; i < m; i++) {
            int[] currRow = new int[n];
            Arrays.fill(currRow, Integer.MAX_VALUE);
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    currRow[j] = Math.min(currRow[j], prevRow[k] + moveCost[grid[i - 1][k]][j] + grid[i][j]);
                }
            }
            prevRow = currRow;
        }
 
        int minCost = Integer.MAX_VALUE;
        for (int cost : prevRow) {
            minCost = Math.min(minCost, cost);
        }
        return minCost;
    }
}

C++:

#include <vector>
#include <limits> // for numeric_limits
 
class Solution {
public:
    int minPathCost(vector<vector<int>>& grid, vector<vector<int>>& moveCost) {
        int m = grid.size();
        int n = grid[0].size();
        vector<int> prevRow = grid[0];
 
        for (int i = 1; i < m; i++) {
            vector<int> currRow(n, numeric_limits<int>::max()); 
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    currRow[j] = min(currRow[j], prevRow[k] + moveCost[grid[i - 1][k]][j] + grid[i][j]);
                }
            }
            prevRow = currRow;
        }
 
        int minCost = numeric_limits<int>::max();
        for (int cost : prevRow) {
            minCost = min(minCost, cost);
        }
        return minCost;
    }
};

These solutions demonstrate the optimized dynamic programming approach, offering a clear and efficient way to solve the "Minimum Path Cost in a Grid" problem. Remember to handle potential integer overflow issues depending on the input constraints.