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Path With Maximum Minimum Value

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Topics:

ArraysBinary SearchGraphsDynamic Programming

Given a matrix of integers A with R rows and C columns, find the largest minimum value of any path from [0, 0] to [R-1, C-1].

A path is a sequence of cells in the matrix such that we can move from one cell to any adjacent cell in the 4 cardinal directions (up, down, left, right). The score of a path is the minimum value in that path.

For example, the path [8, 4, 5, 9] has a value of 4.

Return the largest minimum value of any path from [0, 0] to [R-1, C-1].

Example 1:

Input: A = [[5,4,5],[1,2,6],[7,4,6]]
Output: 4
Explanation: 
The path with the maximum minimum value is highlighted in yellow.

Example 2:

Input: A = [[2,2,1,2,2],[1,2,2,2,1],[1,1,1,2,2]]
Output: 2

Example 3:

Input: A = [[3,4,6,3,4],[0,2,1,1,7],[8,8,3,2,7],[4,9,6,1,1],[5,6,5,2,6]]
Output: 3

Constraints:

1 <= R, C <= 50
0 <= A[i][j] <= 10^9

Try coding it in LeetCode

Path With Maximum Minimum Value

Medium

Asked by:

40 views

Topics:

ArraysBinary SearchGraphsDynamic Programming

Given a matrix of integers A with R rows and C columns, find the largest minimum value of any path from [0, 0] to [R-1, C-1].

For example, the path [8, 4, 5, 9] has a value of 4.

Return the largest minimum value of any path from [0, 0] to [R-1, C-1].

Example 1:

Input: A = [[5,4,5],[1,2,6],[7,4,6]]
Output: 4
Explanation: 
The path with the maximum minimum value is highlighted in yellow.

Example 2:

Input: A = [[2,2,1,2,2],[1,2,2,2,1],[1,1,1,2,2]]
Output: 2

Example 3:

Input: A = [[3,4,6,3,4],[0,2,1,1,7],[8,8,3,2,7],[4,9,6,1,1],[5,6,5,2,6]]
Output: 3

Constraints:

1 <= R, C <= 50
0 <= A[i][j] <= 10^9

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 matrix, and what is the range of values within the matrix? Can matrix values be negative, zero, or floating-point numbers?
By 'path', do you mean a sequence of adjacent cells (up, down, left, right) from the top-left to the bottom-right, or are diagonal moves allowed?
If there are multiple paths with the same maximum minimum value, should I return any one of them, or is there a specific criterion to choose one?
If no path exists from the top-left to the bottom-right, what value should I return? For example, should I return negative infinity, zero, or throw an exception?
Is the input matrix guaranteed to be non-empty and rectangular (i.e., all rows have the same number of columns)? What should I do if the input matrix is null or empty?

Brute Force Solution

Approach

The brute force strategy for finding the path with the maximum minimum value involves exploring every possible path through the grid. We determine the smallest value along each of these paths, and then, select the path where that smallest value is the largest.

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

Imagine you're exploring a maze, and you can only move right or down.
Start at the top-left corner and try going down, then right.
For each path you take, remember the smallest (or weakest) value you encountered along the way.
Continue trying all possible routes to the bottom-right corner.
Once you've explored all possible paths from the start to the end, compare the weakest value of each path.
The path with the highest weakest value is the one we are looking for.

Code Implementation

def find_max_min_path_brute_force(grid):
    rows = len(grid)
    cols = len(grid[0])

    def explore_paths(row_index, col_index, current_path, min_value_so_far):
        if row_index >= rows or col_index >= cols:
            return -1  # Path is invalid

        current_value = grid[row_index][col_index]
        new_min_value = min(min_value_so_far, current_value) 

        if row_index == rows - 1 and col_index == cols - 1:
            return new_min_value # Reached the end

        # Explore going down
        down_value = explore_paths(row_index + 1, col_index, current_path + [(row_index + 1, col_index)], new_min_value)

        # Explore going right
        right_value = explore_paths(row_index, col_index + 1, current_path + [(row_index, col_index + 1)], new_min_value)

        # Return the larger of the two paths.
        return max(down_value, right_value)

    # The starting value must be considered.
    initial_min = grid[0][0]
    max_min_value = explore_paths(0, 0, [(0, 0)], initial_min) 

    return max_min_value

Big(O) Analysis

Time Complexity

O(2^(m+n)) – The brute force approach explores all possible paths from the top-left to the bottom-right of the grid, where 'm' is the number of rows and 'n' is the number of columns. At each cell, we can move either right or down, resulting in two choices. Since we must make a total of (m-1) downward moves and (n-1) rightward moves to reach the destination, the total number of paths is equivalent to the number of ways to interleave these moves, which grows exponentially. Therefore, the time complexity is O(2^(m+n)).

Space Complexity

O(2^(M+N)) – The brute force approach explores every possible path from the top-left to the bottom-right corner, where M and N are the dimensions of the grid. In the worst-case, we generate all possible paths. Each path effectively becomes a stored sequence of moves (right or down). The number of such paths can grow exponentially. Therefore, the space required to hold each path during exploration and comparison contributes to an auxiliary space complexity of O(2^(M+N)).

Optimal Solution

Approach

The best path through a grid prioritizes high values while guaranteeing connectivity. Instead of exploring all possible paths, we start with high minimum-value thresholds, and then progressively lower these threshold values until a valid path from start to finish is discovered using a connected component search. This ensures efficiency by immediately identifying the best possible minimum value that enables a complete path.

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

Start by guessing the highest possible minimum score, which is the lowest value between the starting and ending cell.
Treat all values in the grid below the current minimum score guess as obstacles, effectively blocking those cells.
Check if there is a path from the start to the end, avoiding obstacles, using a connected component search like a flood fill or similar approach.
If a path exists at that minimum score, it might be possible to do better, so remember the current score, and try increasing the guess and repeating the connectivity check.
If a path does not exist, reduce the minimum score, since the current one blocks too many cells, and then repeat the connectivity check until a path is found.
The highest minimum score that permits a valid path from the start to the end is the result.

Code Implementation

def maximum_minimum_path(grid):
    rows = len(grid)
    cols = len(grid[0])
    left, right = -1, -1
    
    start_value = grid[0][0]
    end_value = grid[rows - 1][cols - 1]
    
    left = min(start_value, end_value)
    right = 1000000000
    
    while left <= right:
        mid = (left + right) // 2
        
        visited = [[False] * cols for _ in range(rows)]

        def depth_first_search(row, col):
            if row < 0 or row >= rows or col < 0 or col >= cols or visited[row][col] or grid[row][col] < mid:
                return False
            
            visited[row][col] = True
            
            if row == rows - 1 and col == cols - 1:
                return True
            
            # Explore all 4 directions
            return (depth_first_search(row + 1, col) or
                    depth_first_search(row - 1, col) or
                    depth_first_search(row, col + 1) or
                    depth_first_search(row, col - 1))
        
        # Check if a path exists with the current minimum value.
        if depth_first_search(0, 0):
            left = mid + 1 # Try a higher min value.
        else:
            right = mid - 1 # Lower min value because current is too high.

    # The right pointer holds the maximum minimum value found.
    return right

Big(O) Analysis

Time Complexity

O(m * n * log(V)) – The algorithm performs a binary search on the possible minimum values, where V is the range of values in the grid, leading to a log(V) factor. For each minimum value guess, we perform a connected component search (e.g., DFS or BFS) on the m * n grid. In the worst case, the connected component search visits every cell in the grid, giving a complexity of O(m * n). Therefore, the overall time complexity is O(m * n * log(V)), where m and n are the dimensions of the grid and V is the range of possible minimum values (from the smallest to largest grid cell value).

Space Complexity

O(N) – The algorithm uses a connected component search, such as flood fill, to check for a path. Flood fill typically employs a queue or recursion to keep track of cells to visit, and in the worst case, it might visit all cells in the grid. If the grid has dimensions R rows and C columns, then N = R * C represents the total number of cells. Therefore, the auxiliary space used by the queue or the recursion stack can grow linearly with the number of cells in the grid, up to a size of N. Consequently, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty grid	Return -1 immediately as there is no path.
Grid with a single cell	Return the value of that single cell if it's the only cell.
Grid with start or end value of negative infinity	The min-value will always be negative infinity so return negative infinity.
All cells in the grid have the same value	The maximum minimum value will be that single repeating value.
No path exists from start to end	Return -1 as there is no valid path to consider.
Integer overflow when calculating minimum values	Ensure the minimum value calculation uses a data type that can handle the range of potential minimum values without overflow.
Large grid exceeding memory limits	Consider optimizing memory usage with techniques like iterative deepening A* or streaming graph algorithms if the grid is too large to hold in memory.
Grid with negative values	The algorithm should correctly handle and propagate negative values when determining the minimum path value.

Empty grid

How to Handle:

Return -1 immediately as there is no path.

Grid with a single cell

How to Handle:

Return the value of that single cell if it's the only cell.

Grid with start or end value of negative infinity

How to Handle:

The min-value will always be negative infinity so return negative infinity.

All cells in the grid have the same value

How to Handle:

The maximum minimum value will be that single repeating value.

No path exists from start to end

How to Handle:

Return -1 as there is no valid path to consider.

Integer overflow when calculating minimum values

How to Handle:

Ensure the minimum value calculation uses a data type that can handle the range of potential minimum values without overflow.

Large grid exceeding memory limits

How to Handle:

Consider optimizing memory usage with techniques like iterative deepening A* or streaming graph algorithms if the grid is too large to hold in memory.

Grid with negative values

How to Handle:

The algorithm should correctly handle and propagate negative values when determining the minimum path value.

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 matrix, and what is the range of values within the matrix? Can matrix values be negative, zero, or floating-point numbers?
By 'path', do you mean a sequence of adjacent cells (up, down, left, right) from the top-left to the bottom-right, or are diagonal moves allowed?
If there are multiple paths with the same maximum minimum value, should I return any one of them, or is there a specific criterion to choose one?
If no path exists from the top-left to the bottom-right, what value should I return? For example, should I return negative infinity, zero, or throw an exception?
Is the input matrix guaranteed to be non-empty and rectangular (i.e., all rows have the same number of columns)? What should I do if the input matrix is null or empty?

Brute Force Solution

Approach

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

Imagine you're exploring a maze, and you can only move right or down.
Start at the top-left corner and try going down, then right.
For each path you take, remember the smallest (or weakest) value you encountered along the way.
Continue trying all possible routes to the bottom-right corner.
Once you've explored all possible paths from the start to the end, compare the weakest value of each path.
The path with the highest weakest value is the one we are looking for.

Code Implementation

def find_max_min_path_brute_force(grid):
    rows = len(grid)
    cols = len(grid[0])

    def explore_paths(row_index, col_index, current_path, min_value_so_far):
        if row_index >= rows or col_index >= cols:
            return -1  # Path is invalid

        current_value = grid[row_index][col_index]
        new_min_value = min(min_value_so_far, current_value) 

        if row_index == rows - 1 and col_index == cols - 1:
            return new_min_value # Reached the end

        # Explore going down
        down_value = explore_paths(row_index + 1, col_index, current_path + [(row_index + 1, col_index)], new_min_value)

        # Explore going right
        right_value = explore_paths(row_index, col_index + 1, current_path + [(row_index, col_index + 1)], new_min_value)

        # Return the larger of the two paths.
        return max(down_value, right_value)

    # The starting value must be considered.
    initial_min = grid[0][0]
    max_min_value = explore_paths(0, 0, [(0, 0)], initial_min) 

    return max_min_value

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by guessing the highest possible minimum score, which is the lowest value between the starting and ending cell.
Treat all values in the grid below the current minimum score guess as obstacles, effectively blocking those cells.
Check if there is a path from the start to the end, avoiding obstacles, using a connected component search like a flood fill or similar approach.
If a path exists at that minimum score, it might be possible to do better, so remember the current score, and try increasing the guess and repeating the connectivity check.
If a path does not exist, reduce the minimum score, since the current one blocks too many cells, and then repeat the connectivity check until a path is found.
The highest minimum score that permits a valid path from the start to the end is the result.

Code Implementation

def maximum_minimum_path(grid):
    rows = len(grid)
    cols = len(grid[0])
    left, right = -1, -1
    
    start_value = grid[0][0]
    end_value = grid[rows - 1][cols - 1]
    
    left = min(start_value, end_value)
    right = 1000000000
    
    while left <= right:
        mid = (left + right) // 2
        
        visited = [[False] * cols for _ in range(rows)]

        def depth_first_search(row, col):
            if row < 0 or row >= rows or col < 0 or col >= cols or visited[row][col] or grid[row][col] < mid:
                return False
            
            visited[row][col] = True
            
            if row == rows - 1 and col == cols - 1:
                return True
            
            # Explore all 4 directions
            return (depth_first_search(row + 1, col) or
                    depth_first_search(row - 1, col) or
                    depth_first_search(row, col + 1) or
                    depth_first_search(row, col - 1))
        
        # Check if a path exists with the current minimum value.
        if depth_first_search(0, 0):
            left = mid + 1 # Try a higher min value.
        else:
            right = mid - 1 # Lower min value because current is too high.

    # The right pointer holds the maximum minimum value found.
    return right

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty grid	Return -1 immediately as there is no path.
Grid with a single cell	Return the value of that single cell if it's the only cell.
Grid with start or end value of negative infinity	The min-value will always be negative infinity so return negative infinity.
All cells in the grid have the same value	The maximum minimum value will be that single repeating value.
No path exists from start to end	Return -1 as there is no valid path to consider.
Integer overflow when calculating minimum values	Ensure the minimum value calculation uses a data type that can handle the range of potential minimum values without overflow.
Large grid exceeding memory limits	Consider optimizing memory usage with techniques like iterative deepening A* or streaming graph algorithms if the grid is too large to hold in memory.
Grid with negative values	The algorithm should correctly handle and propagate negative values when determining the minimum path value.

Empty grid

How to Handle:

Return -1 immediately as there is no path.

Grid with a single cell

How to Handle:

Return the value of that single cell if it's the only cell.

Grid with start or end value of negative infinity

How to Handle:

The min-value will always be negative infinity so return negative infinity.

All cells in the grid have the same value

How to Handle:

The maximum minimum value will be that single repeating value.

No path exists from start to end

How to Handle:

Return -1 as there is no valid path to consider.

Integer overflow when calculating minimum values

How to Handle:

Ensure the minimum value calculation uses a data type that can handle the range of potential minimum values without overflow.

Large grid exceeding memory limits

How to Handle:

Consider optimizing memory usage with techniques like iterative deepening A* or streaming graph algorithms if the grid is too large to hold in memory.

Grid with negative values

How to Handle:

The algorithm should correctly handle and propagate negative values when determining the minimum path value.