Jobs

Interviews

Discussions

Minimize Maximum Value in a Grid

Hard

Asked by:

31 views

Topics:

Binary SearchArrays

You are given an m x n integer matrix grid. You are allowed to perform the following operation any number of times:

Choose any row and rotate it by one or more cells in either direction (clockwise or counter-clockwise).

The goal is to minimize the maximum value in the grid. After performing the operations, what is the minimum possible value for the maximum value in the grid?

Return the minimum possible value for the maximum value in the grid.

For example, if the grid is [[1, 2, 3], [5, 6, 1]], one possible sequence of operations is:

Rotate the first row clockwise by one cell: [[3, 1, 2], [5, 6, 1]].
Rotate the second row counter-clockwise by two cells: [[3, 1, 2], [1, 5, 6]].

After performing these operations, the maximum value in the grid is 6, which is the minimum possible value for the maximum value in the grid.

Example 1:

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

Example 2:

Input: grid = [[9,1,6,1,7],[3,9,3,9,7],[3,4,9,3,2]]
Output: 7

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 100
1 <= grid[i][j] <= 10⁵

Try coding it in LeetCode

Minimize Maximum Value in a Grid

Hard

Asked by:

31 views

Topics:

Binary SearchArrays

You are given an m x n integer matrix grid. You are allowed to perform the following operation any number of times:

Choose any row and rotate it by one or more cells in either direction (clockwise or counter-clockwise).

The goal is to minimize the maximum value in the grid. After performing the operations, what is the minimum possible value for the maximum value in the grid?

Return the minimum possible value for the maximum value in the grid.

For example, if the grid is [[1, 2, 3], [5, 6, 1]], one possible sequence of operations is:

Rotate the first row clockwise by one cell: [[3, 1, 2], [5, 6, 1]].
Rotate the second row counter-clockwise by two cells: [[3, 1, 2], [1, 5, 6]].

After performing these operations, the maximum value in the grid is 6, which is the minimum possible value for the maximum value in the grid.

Example 1:

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

Example 2:

Input: grid = [[9,1,6,1,7],[3,9,3,9,7],[3,4,9,3,2]]
Output: 7

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 100
1 <= grid[i][j] <= 10⁵

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 (number of rows and columns), and what are the upper bounds for these dimensions?
What is the data type and range of values within the grid? Can the grid contain negative values, zero, or only positive integers?
Can you provide an example of how the 'minimize maximum value' is defined in this context? Specifically, what operation or constraint are we applying to minimize the largest value encountered?
Is the grid guaranteed to be rectangular (i.e., all rows have the same number of columns)? What should I return if the grid is empty?
Are there any specific constraints on how we can traverse the grid (e.g., only moving right or down, or allowing movement in all four directions)? If there are specific movement rules, what is the starting and ending location for the traversal?

Brute Force Solution

Approach

The brute force method is all about trying absolutely everything to find the best answer. For this grid problem, it means we explore every possible way to assign values to the empty spaces, and then pick the assignment that results in the smallest possible maximum value in any row or column.

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

Start by considering the very first empty space in the grid.
Try assigning it the smallest possible value.
Then, move to the next empty space and try assigning it the smallest possible value, and so on.
Keep doing this until you've filled every empty space with a value.
Once the grid is full, check the maximum value in each row and each column.
Remember that maximum value.
Now, go back and try a different value for the very first empty space.
Repeat this entire process, trying every possible value for every empty space.
Each time you fill the grid, compare the maximum value you found to the smallest maximum value you've seen so far.
In the end, the assignment of values that resulted in the smallest maximum row or column value is your solution.

Code Implementation

def minimize_maximum_value_brute_force(grid):    rows = len(grid)
    cols = len(grid[0])
    empty_cells = []
    for row_index in range(rows):
        for col_index in range(cols):
            if grid[row_index][col_index] == -1:
                empty_cells.append((row_index, col_index))

    min_max_value = float('inf')

    def solve(index):
        nonlocal min_max_value

        if index == len(empty_cells):
            # Grid is filled, calculate maximum row/col sum
            max_value = 0
            for row_index in range(rows):
                max_value = max(max_value, max(grid[row_index]))
            for col_index in range(cols):
                column_max = 0
                for row_index in range(rows):
                    column_max = max(column_max, grid[row_index][col_index])
                max_value = max(max_value, column_max)

            # Update minimum maximum value seen so far
            min_max_value = min(min_max_value, max_value)
            return

        row_index, col_index = empty_cells[index]
        for value in range(1, 101):
            # Try assigning every possible value
            grid[row_index][col_index] = value
            solve(index + 1)
            # Backtrack: Reset cell to -1 for next try
            grid[row_index][col_index] = -1

    solve(0)
    return min_max_value

Big(O) Analysis

Time Complexity

O(m^k) – Let 'm' be the maximum possible value that can be assigned to an empty cell, and let 'k' be the number of empty cells in the grid. The brute force approach explores every possible assignment of values to the empty cells. Since each empty cell can take 'm' possible values, and there are 'k' empty cells, the total number of possible assignments is m^k. For each assignment, we need to check the maximum value in each row and column, which takes O(n) time where n is the size of the grid (assuming a square grid). However, the dominant factor is the exploration of all m^k possible assignments, making the overall time complexity O(m^k).

Space Complexity

O(N) – The brute force approach explores every possible assignment using recursion. The depth of the recursion is determined by the number of empty spaces in the grid, which in the worst case can be all N cells, where N represents the total number of cells in the grid. Each level of recursion requires stack space to store the state of the variables, leading to O(N) space for the call stack. No other significant auxiliary data structures are used.

Optimal Solution

Approach

We're trying to find the smallest possible largest value we can get in the grid. The key idea is to use Binary Search to efficiently narrow down the range of possible maximum values, and then use Depth-First Search (DFS) to validate if a maximum value is valid.

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

First, figure out the smallest and largest possible values any cell could have in the grid.
Use Binary Search to guess a maximum value within that range. This means we'll repeatedly pick a middle value and see if it works.
For each guess, we'll treat it as the maximum value allowed for any cell in the grid.
Using Depth-First Search (DFS), try to find a path from the top-left to the bottom-right cell of the grid where every cell's value is no more than our guessed maximum.
If we find such a path, it means our guessed maximum is too big (we can find a solution with a smaller maximum), so we try a smaller guess in the Binary Search.
If we can't find a path, it means our guessed maximum is too small (we need a larger maximum to allow a path), so we try a larger guess in the Binary Search.
Keep repeating steps 2-6, narrowing down the range, until we find the smallest possible maximum value that allows a path from top-left to bottom-right.

Code Implementation

def minimize_maximum_value_in_grid(grid):
    rows = len(grid)
    cols = len(grid[0])
    
    def can_reach_destination(maximum_allowed_value):
        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] > maximum_allowed_value:
                return False

            if row == rows - 1 and col == cols - 1:
                return True

            visited[row][col] = True

            # Explore adjacent cells
            if 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):
                return True

            return False

        # Start DFS from the top-left cell
        return depth_first_search(0, 0)

    minimum_value = min(min(row) for row in grid)
    maximum_value = max(max(row) for row in grid)

    # Binary search to find the smallest maximum value
    while minimum_value < maximum_value:
        middle_value = (minimum_value + maximum_value) // 2
        
        # Check if we can reach the destination with the current middle value
        if can_reach_destination(middle_value):
            # If we can, try a smaller maximum value
            maximum_value = middle_value
        else:
            # If we can't, we need a larger maximum value
            minimum_value = middle_value + 1
    
    # 'minimum_value' is the smallest possible maximum value
    return minimum_value

Big(O) Analysis

Time Complexity

O(mn log(max_value - min_value)) – The algorithm uses binary search to find the minimized maximum value within a range defined by the minimum and maximum values in the grid. The binary search contributes a factor of log(max_value - min_value). For each guess in the binary search, a Depth-First Search (DFS) is performed on the grid. DFS, in the worst case, visits all cells in the m x n grid, taking O(mn) time. Therefore, the overall time complexity is O(mn log(max_value - min_value)), where m and n are the dimensions of the grid, and max_value and min_value are the maximum and minimum values within the grid, respectively.

Space Complexity

O(N) – The dominant space usage comes from the Depth-First Search (DFS). In the worst-case scenario, the DFS might visit all cells in the grid, leading to a recursion depth proportional to the number of cells. This results in a call stack that can grow up to N, where N is the total number of cells in the grid. The DFS also implicitly uses a visited set which could potentially store every cell in the grid. Therefore, the auxiliary space required is O(N).

Edge Cases

Case	How to Handle
Empty grid (rows or columns are zero)	Return 0 or an appropriate default value (e.g., positive infinity if minimizing), since there is nothing to minimize.
Grid with a single element	Return the value of that single element as it is both the minimum and maximum possible result.
Grid with all identical values	The minimized maximum value will be this identical value; no special handling needed.
Grid with very large values that could lead to integer overflow	Use long long data type for intermediate calculations if possible, or check for overflow before returning result.
Grid with extremely skewed values (one very large value)	The large value may dominate the binary search or other optimization strategy; ensure calculations and conditions are correct.
Negative values in the grid	The algorithm should handle negative values correctly without causing incorrect minimum or maximum calculation.
Very large grid sizes (approaching memory limits)	Ensure the algorithm is memory-efficient and avoid unnecessary copies or large data structures that could lead to out-of-memory errors.
The target value cannot be achieved, the constraints are impossible to meet.	If no valid assignment exists, the algorithm should be able to return a valid answer indicating an impossibility, such as returning -1 or a flag.

Empty grid (rows or columns are zero)

How to Handle:

Return 0 or an appropriate default value (e.g., positive infinity if minimizing), since there is nothing to minimize.

Grid with a single element

How to Handle:

Return the value of that single element as it is both the minimum and maximum possible result.

Grid with all identical values

How to Handle:

The minimized maximum value will be this identical value; no special handling needed.

Grid with very large values that could lead to integer overflow

How to Handle:

Use long long data type for intermediate calculations if possible, or check for overflow before returning result.

Grid with extremely skewed values (one very large value)

How to Handle:

The large value may dominate the binary search or other optimization strategy; ensure calculations and conditions are correct.

Negative values in the grid

How to Handle:

The algorithm should handle negative values correctly without causing incorrect minimum or maximum calculation.

Very large grid sizes (approaching memory limits)

How to Handle:

Ensure the algorithm is memory-efficient and avoid unnecessary copies or large data structures that could lead to out-of-memory errors.

The target value cannot be achieved, the constraints are impossible to meet.

How to Handle:

If no valid assignment exists, the algorithm should be able to return a valid answer indicating an impossibility, such as returning -1 or a flag.

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 (number of rows and columns), and what are the upper bounds for these dimensions?
What is the data type and range of values within the grid? Can the grid contain negative values, zero, or only positive integers?
Can you provide an example of how the 'minimize maximum value' is defined in this context? Specifically, what operation or constraint are we applying to minimize the largest value encountered?
Is the grid guaranteed to be rectangular (i.e., all rows have the same number of columns)? What should I return if the grid is empty?
Are there any specific constraints on how we can traverse the grid (e.g., only moving right or down, or allowing movement in all four directions)? If there are specific movement rules, what is the starting and ending location for the traversal?

Brute Force Solution

Approach

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

Start by considering the very first empty space in the grid.
Try assigning it the smallest possible value.
Then, move to the next empty space and try assigning it the smallest possible value, and so on.
Keep doing this until you've filled every empty space with a value.
Once the grid is full, check the maximum value in each row and each column.
Remember that maximum value.
Now, go back and try a different value for the very first empty space.
Repeat this entire process, trying every possible value for every empty space.
Each time you fill the grid, compare the maximum value you found to the smallest maximum value you've seen so far.
In the end, the assignment of values that resulted in the smallest maximum row or column value is your solution.

Code Implementation

def minimize_maximum_value_brute_force(grid):    rows = len(grid)
    cols = len(grid[0])
    empty_cells = []
    for row_index in range(rows):
        for col_index in range(cols):
            if grid[row_index][col_index] == -1:
                empty_cells.append((row_index, col_index))

    min_max_value = float('inf')

    def solve(index):
        nonlocal min_max_value

        if index == len(empty_cells):
            # Grid is filled, calculate maximum row/col sum
            max_value = 0
            for row_index in range(rows):
                max_value = max(max_value, max(grid[row_index]))
            for col_index in range(cols):
                column_max = 0
                for row_index in range(rows):
                    column_max = max(column_max, grid[row_index][col_index])
                max_value = max(max_value, column_max)

            # Update minimum maximum value seen so far
            min_max_value = min(min_max_value, max_value)
            return

        row_index, col_index = empty_cells[index]
        for value in range(1, 101):
            # Try assigning every possible value
            grid[row_index][col_index] = value
            solve(index + 1)
            # Backtrack: Reset cell to -1 for next try
            grid[row_index][col_index] = -1

    solve(0)
    return min_max_value

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, figure out the smallest and largest possible values any cell could have in the grid.
Use Binary Search to guess a maximum value within that range. This means we'll repeatedly pick a middle value and see if it works.
For each guess, we'll treat it as the maximum value allowed for any cell in the grid.
Using Depth-First Search (DFS), try to find a path from the top-left to the bottom-right cell of the grid where every cell's value is no more than our guessed maximum.
If we find such a path, it means our guessed maximum is too big (we can find a solution with a smaller maximum), so we try a smaller guess in the Binary Search.
If we can't find a path, it means our guessed maximum is too small (we need a larger maximum to allow a path), so we try a larger guess in the Binary Search.
Keep repeating steps 2-6, narrowing down the range, until we find the smallest possible maximum value that allows a path from top-left to bottom-right.

Code Implementation

def minimize_maximum_value_in_grid(grid):
    rows = len(grid)
    cols = len(grid[0])
    
    def can_reach_destination(maximum_allowed_value):
        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] > maximum_allowed_value:
                return False

            if row == rows - 1 and col == cols - 1:
                return True

            visited[row][col] = True

            # Explore adjacent cells
            if 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):
                return True

            return False

        # Start DFS from the top-left cell
        return depth_first_search(0, 0)

    minimum_value = min(min(row) for row in grid)
    maximum_value = max(max(row) for row in grid)

    # Binary search to find the smallest maximum value
    while minimum_value < maximum_value:
        middle_value = (minimum_value + maximum_value) // 2
        
        # Check if we can reach the destination with the current middle value
        if can_reach_destination(middle_value):
            # If we can, try a smaller maximum value
            maximum_value = middle_value
        else:
            # If we can't, we need a larger maximum value
            minimum_value = middle_value + 1
    
    # 'minimum_value' is the smallest possible maximum value
    return minimum_value

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty grid (rows or columns are zero)	Return 0 or an appropriate default value (e.g., positive infinity if minimizing), since there is nothing to minimize.
Grid with a single element	Return the value of that single element as it is both the minimum and maximum possible result.
Grid with all identical values	The minimized maximum value will be this identical value; no special handling needed.
Grid with very large values that could lead to integer overflow	Use long long data type for intermediate calculations if possible, or check for overflow before returning result.
Grid with extremely skewed values (one very large value)	The large value may dominate the binary search or other optimization strategy; ensure calculations and conditions are correct.
Negative values in the grid	The algorithm should handle negative values correctly without causing incorrect minimum or maximum calculation.
Very large grid sizes (approaching memory limits)	Ensure the algorithm is memory-efficient and avoid unnecessary copies or large data structures that could lead to out-of-memory errors.
The target value cannot be achieved, the constraints are impossible to meet.	If no valid assignment exists, the algorithm should be able to return a valid answer indicating an impossibility, such as returning -1 or a flag.

Empty grid (rows or columns are zero)

How to Handle:

Return 0 or an appropriate default value (e.g., positive infinity if minimizing), since there is nothing to minimize.

Grid with a single element

How to Handle:

Return the value of that single element as it is both the minimum and maximum possible result.

Grid with all identical values

How to Handle:

The minimized maximum value will be this identical value; no special handling needed.

Grid with very large values that could lead to integer overflow

How to Handle:

Use long long data type for intermediate calculations if possible, or check for overflow before returning result.

Grid with extremely skewed values (one very large value)

How to Handle:

The large value may dominate the binary search or other optimization strategy; ensure calculations and conditions are correct.

Negative values in the grid

How to Handle:

The algorithm should handle negative values correctly without causing incorrect minimum or maximum calculation.

Very large grid sizes (approaching memory limits)

How to Handle:

Ensure the algorithm is memory-efficient and avoid unnecessary copies or large data structures that could lead to out-of-memory errors.

The target value cannot be achieved, the constraints are impossible to meet.

How to Handle:

If no valid assignment exists, the algorithm should be able to return a valid answer indicating an impossibility, such as returning -1 or a flag.