Minimum Path Cost in a Grid

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Dynamic Programming

You are given a 0-indexed m x n integer matrix grid consisting of distinct integers from 0 to m * n - 1. You can move in this matrix from a cell to any other cell in the next row. That is, if you are in cell (x, y) such that x < m - 1, you can move to any of the cells (x + 1, 0), (x + 1, 1), ..., (x + 1, n - 1). Note that it is not possible to move from cells in the last row.

Each possible move has a cost given by a 0-indexed 2D array moveCost of size (m * n) x n, where moveCost[i][j] is the cost of moving from a cell with value i to a cell in column j of the next row. The cost of moving from cells in the last row of grid can be ignored.

The cost of a path in grid is the sum of all values of cells visited plus the sum of costs of all the moves made. Return the minimum cost of a path that starts from any cell in the first row and ends at any cell in the last row.

Example 1:

Input: grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]]
Output: 17
Explanation: The path with the minimum possible cost is the path 5 -> 0 -> 1.
- The sum of the values of cells visited is 5 + 0 + 1 = 6.
- The cost of moving from 5 to 0 is 3.
- The cost of moving from 0 to 1 is 8.
So the total cost of the path is 6 + 3 + 8 = 17.

Example 2:

Input: grid = [[5,1,2],[4,0,3]], moveCost = [[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]]
Output: 6
Explanation: The path with the minimum possible cost is the path 2 -> 3.
- The sum of the values of cells visited is 2 + 3 = 5.
- The cost of moving from 2 to 3 is 1.
So the total cost of this path is 5 + 1 = 6.

Constraints:

m == grid.length
n == grid[i].length
2 <= m, n <= 50
grid consists of distinct integers from 0 to m * n - 1.
moveCost.length == m * n
moveCost[i].length == n
1 <= moveCost[i][j] <= 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the grid and what is the range of values for both the grid cells and the move costs?
Can the values in the grid or the move costs be negative, zero, or non-integer?
Is it guaranteed that a path exists from the top row to the bottom row?
If multiple minimum-cost paths exist, is any one of them acceptable, or is there a specific path I should aim to find?
What is the expected output if the input grid is empty or null?

Brute Force Solution

Approach

The brute force approach for finding the minimum path cost in a grid involves exploring every possible path from the top row to the bottom row. We calculate the cost of each path and then identify the path with the lowest total cost. This method guarantees finding the optimal solution, but it can be inefficient for larger grids.

Here's how the algorithm would work step-by-step:

Start at each possible entry point in the top row of the grid.
From each starting point, explore all possible moves to the next row below.
For each move, add the current cell's cost and the transition cost to the running total cost of that path.
Continue this process, row by row, until you reach the bottom row.
Once you reach the bottom row, you have a complete path from top to bottom and its total cost.
Repeat this entire process for every possible starting point in the top row, generating all possible paths.
After generating all paths and their costs, compare the total costs of all paths.
Select the path with the minimum total cost as the optimal solution.

Code Implementation

def minimum_path_cost_brute_force(grid, move_cost):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])

    min_path_cost = float('inf')

    def calculate_path_cost(row, column, current_path_cost):
        # If we've reached the bottom row, update the minimum cost
        if row == number_of_rows - 1:
            return current_path_cost

        min_cost_from_here = float('inf')

        # Explore all possible moves to the next row
        for next_column in range(number_of_columns):
            new_path_cost = current_path_cost + move_cost[grid[row][column]][next_column] + grid[row + 1][next_column]
            min_cost_from_here = min(min_cost_from_here, calculate_path_cost(row + 1, next_column, new_path_cost))

        return min_cost_from_here

    # Iterate through each starting column in the top row.
    for start_column in range(number_of_columns):
        starting_cost = grid[0][start_column]
        path_cost = calculate_path_cost(0, start_column, starting_cost)
        min_path_cost = min(min_path_cost, path_cost)

    return min_path_cost

Big(O) Analysis

Time Complexity

O(m * n^m) – The algorithm explores all possible paths from the top row to the bottom row of the grid. Let 'm' be the number of rows and 'n' be the number of columns. From each of the 'n' starting points in the top row, we have 'n' choices for each of the remaining 'm-1' rows. Thus, the total number of possible paths is n * n^(m-1), which simplifies to n^m. Since we need to traverse and calculate the cost of each of these paths, the overall time complexity becomes O(m * n^m). The multiplication by m accounts for the work done to sum the cost of each path.

Space Complexity

O(m*n) – The brute force approach explores every possible path using recursion. In the worst-case scenario, the recursion depth can reach m (the number of rows) for each of the n (number of columns in the first row) starting positions. For each recursive call, we store the current path's cost. Therefore, the space used by the recursion stack and to maintain the current path costs will be proportional to m*n in the worst case. Thus the auxiliary space complexity is O(m*n) where m is the number of rows, and n is the number of columns.

Optimal Solution

Approach

The best way to find the minimum path cost is to build the solution step-by-step, always making the cheapest choice at each step. We remember the best (lowest cost) way to reach each point in the grid as we go. This avoids recomputing costs and efficiently leads us to the overall minimum cost path.

Here's how the algorithm would work step-by-step:

Start at the top row of the grid. The cost to reach any cell in the top row is just its own value.
Move down one row at a time. For each cell in the current row, consider all possible cells we could have come from in the row above, using the given move costs.
Calculate the total cost to reach the current cell from each of those possible previous cells. This is the cost to reach the previous cell, plus the move cost from that previous cell to the current cell, plus the value of the current cell itself.
Choose the path (previous cell) that gives the minimum total cost to reach the current cell. Record this minimum cost as the cost to reach this cell.
Repeat steps 2-4 for each row until you reach the bottom row.
Once you've reached the bottom row, find the cell in the bottom row with the minimum cost. This is the minimum cost to reach any cell in the bottom row from the top row.
That minimum cost is the answer.

Code Implementation

def min_path_cost(grid, move_cost):
    number_of_rows = len(grid)
    number_of_cols = len(grid[0])

    # cost_to_reach[row][col] is the min cost to reach cell (row, col).
    cost_to_reach = [[0] * number_of_cols for _ in range(number_of_rows)]

    # Initialize the first row's cost.
    for col_index in range(number_of_cols):
        cost_to_reach[0][col_index] = grid[0][col_index]

    # Iterate through the rows, starting from the second row.
    for row_index in range(1, number_of_rows):
        for col_index in range(number_of_cols):
            cost_to_reach[row_index][col_index] = float('inf')

            # Iterate through the possible previous cells in the row above.
            for previous_col_index in range(number_of_cols):

                #Find minimum cost to reach the current cell.
                current_cost = cost_to_reach[row_index - 1][previous_col_index] \
                    + move_cost[grid[row_index - 1][previous_col_index]][col_index] \
                    + grid[row_index][col_index]

                cost_to_reach[row_index][col_index] = min(cost_to_reach[row_index][col_index], current_cost)

    # Find the minimum cost in the last row.
    minimum_final_cost = float('inf')
    for col_index in range(number_of_cols):
        # Find the minimum cost to reach any cell in the bottom row.
        minimum_final_cost = min(minimum_final_cost, cost_to_reach[number_of_rows - 1][col_index])

    return minimum_final_cost

Big(O) Analysis

Time Complexity

O(m * n * n) – Let m be the number of rows and n be the number of columns in the grid. The algorithm iterates through each cell in the grid, which takes O(m * n) time. For each cell, it considers all possible cells in the previous row as potential predecessors. Since there are n columns in the previous row, this takes O(n) time. Therefore, the overall time complexity is O(m * n * n) because for each cell in the grid, we iterate through all possible previous cells.

Space Complexity

O(rows * cols) – The algorithm stores the minimum cost to reach each cell in the grid. Since we remember the best (lowest cost) way to reach each point in the grid, we are essentially creating a table (or matrix) with the same dimensions as the input grid to store these intermediate minimum costs. Therefore, the space used is proportional to the number of cells in the grid, which is rows multiplied by cols, where rows is the number of rows and cols is the number of columns in the input grid. This gives us a space complexity of O(rows * cols).

Edge Cases

Case	How to Handle
Null or empty grid	Return 0 as the minimum cost, indicating no path exists or the grid is invalid.
Grid with only one row and one column	Return the value of the single cell as the minimum path cost.
Grid with only one row	Sum the values in the first row, and then add the cost of moving from each cell to the next in the 'moveCost' array.
Grid with only one column	Sum the values in the first column assuming moves down each row are always to the same column at cost of 0.
Large grid dimensions leading to potential integer overflow	Use 'long' data type for path costs to accommodate large sums, preventing integer overflow.
moveCost values are very large	Use 'long' data type for moveCost values to avoid overflow when adding it to the path.
All values in the grid are identical	The algorithm should still correctly find the minimum path using the moveCost to differentiate possible paths.
Negative values in grid or moveCost	The algorithm should correctly handle the negative values to find the true minimum (if allowed by the problem statement).

Null or empty grid

How to Handle:

Return 0 as the minimum cost, indicating no path exists or the grid is invalid.

Grid with only one row and one column

How to Handle:

Return the value of the single cell as the minimum path cost.

Grid with only one row

How to Handle:

Sum the values in the first row, and then add the cost of moving from each cell to the next in the 'moveCost' array.

Grid with only one column

How to Handle:

Sum the values in the first column assuming moves down each row are always to the same column at cost of 0.

Large grid dimensions leading to potential integer overflow

How to Handle:

Use 'long' data type for path costs to accommodate large sums, preventing integer overflow.

moveCost values are very large

How to Handle:

Use 'long' data type for moveCost values to avoid overflow when adding it to the path.

All values in the grid are identical

How to Handle:

The algorithm should still correctly find the minimum path using the moveCost to differentiate possible paths.

Negative values in grid or moveCost

How to Handle:

The algorithm should correctly handle the negative values to find the true minimum (if allowed by the problem statement).