Check if Grid can be Cut into Sections

#1562 Most AskedMedium

Check if Grid can be Cut into Sections #1562 Most Asked

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Arrays

You are given an integer n representing the dimensions of an n x n grid, with the origin at the bottom-left corner of the grid. You are also given a 2D array of coordinates rectangles, where rectangles[i] is in the form [start_x, start_y, end_x, end_y], representing a rectangle on the grid. Each rectangle is defined as follows:

(start_x, start_y): The bottom-left corner of the rectangle.
(end_x, end_y): The top-right corner of the rectangle.

Note that the rectangles do not overlap. Your task is to determine if it is possible to make either two horizontal or two vertical cuts on the grid such that:

Each of the three resulting sections formed by the cuts contains at least one rectangle.
Every rectangle belongs to exactly one section.

Return true if such cuts can be made; otherwise, return false.

Example 1:

Input: n = 5, rectangles = [[1,0,5,2],[0,2,2,4],[3,2,5,3],[0,4,4,5]]

Output: true

Explanation:

The grid is shown in the diagram. We can make horizontal cuts at y = 2 and y = 4. Hence, output is true.

Example 2:

Input: n = 4, rectangles = [[0,0,1,1],[2,0,3,4],[0,2,2,3],[3,0,4,3]]

Output: true

Explanation:

We can make vertical cuts at x = 2 and x = 3. Hence, output is true.

Example 3:

Input: n = 4, rectangles = [[0,2,2,4],[1,0,3,2],[2,2,3,4],[3,0,4,2],[3,2,4,4]]

Output: false

Explanation:

We cannot make two horizontal or two vertical cuts that satisfy the conditions. Hence, output is false.

Constraints:

3 <= n <= 10⁹
3 <= rectangles.length <= 10⁵
0 <= rectangles[i][0] < rectangles[i][2] <= n
0 <= rectangles[i][1] < rectangles[i][3] <= n
No two rectangles overlap.

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 is the maximum possible size for each dimension?
What data type does the grid contain, and what are the possible values within the grid (e.g., boolean, integers with a specific range)?
What exactly constitutes a 'section'? Is it a contiguous rectangular subgrid?
What conditions must be met for the grid to be 'cut into sections'? Does each cell of the original grid need to belong to exactly one section, or are overlaps/gaps allowed?
If the grid cannot be cut into sections according to the given criteria, what should I return?

Brute Force Solution

Approach

The brute force approach to checking if a grid can be cut into sections involves systematically trying every possible way to divide the grid. We consider all possible horizontal and vertical cuts, examining each resulting section to see if it meets the required conditions. This exhaustive search guarantees we'll find a solution if one exists.

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

First, consider all possible ways to make a single horizontal cut across the grid.
For each of these horizontal cuts, examine the two resulting sections.
Check if both sections meet the necessary conditions (e.g., sum of values, size, or other given constraints).
If a horizontal cut results in two valid sections, we've found one possible solution, and we can stop.
If no single horizontal cut works, then move on to considering all possible vertical cuts.
For each vertical cut, examine the two resulting sections and check if they meet the necessary conditions.
If a vertical cut results in two valid sections, we have a possible solution.
If neither horizontal nor vertical single cuts work, then try combinations of cuts. For example, consider making one horizontal and one vertical cut, resulting in four sections.
Check the validity of all four sections created by these combined cuts.
Continue exploring more complex cut combinations (two horizontal, two vertical, etc.) until a valid configuration is found or all possibilities are exhausted.

Code Implementation

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

    # Check for horizontal cuts
    for row_cut in range(1, rows):        section_one = [grid[i] for i in range(row_cut)]        section_two = [grid[i] for i in range(row_cut, rows)]
        if is_valid_section(section_one) and is_valid_section(section_two):

            return True

    # Check for vertical cuts
    for col_cut in range(1, cols):        section_one = [[grid[i][j] for j in range(col_cut)] for i in range(rows)]        section_two = [[grid[i][j] for j in range(col_cut, cols)] for i in range(rows)]
        # Must have valid sections to return
        if is_valid_section(section_one) and is_valid_section(section_two):

            return True

    # Check one horizontal and one vertical cut
    for row_cut in range(1, rows):
        for col_cut in range(1, cols):

            section_one = [[grid[i][j] for j in range(col_cut)] for i in range(row_cut)]
            section_two = [[grid[i][j] for j in range(col_cut, cols)] for i in range(row_cut)]
            section_three = [[grid[i][j] for j in range(col_cut)] for i in range(row_cut, rows)]
            section_four = [[grid[i][j] for j in range(col_cut, cols)] for i in range(row_cut, rows)]

            if (is_valid_section(section_one) and
                    is_valid_section(section_two) and
                    is_valid_section(section_three) and
                    is_valid_section(section_four)):

                return True

    return False

def is_valid_section(section):
    # Implement your validation logic here
    # This is a placeholder; replace with your actual validation
    if not section:        return False

    # Placeholder check for any section
    return True

Big(O) Analysis

Time Complexity

O(n^4) – The algorithm considers all possible horizontal and vertical cuts. Trying all horizontal cuts takes O(n) time. For each horizontal cut, checking the validity of the two resulting sections requires iterating through all elements, which takes O(n^2) time. Similarly, trying all vertical cuts also takes O(n^3) time. Then, trying one horizontal and one vertical cut divides the grid into four sections, and checking the validity of those sections also takes O(n^2) time, but this is nested inside the O(n^2) choices for the cuts, making it O(n^4). The dominant cost comes from considering all combinations of cuts which leads to a complexity of approximately O(n^4).

Space Complexity

O(1) – The described brute-force approach primarily involves iterative checks and comparisons, without explicitly creating auxiliary data structures that scale with the input grid's size (N, where N represents the total number of cells in the grid). While checking each section might involve calculating sums, these calculations are typically done using a constant number of variables to store intermediate sums. Thus, the space required remains constant, independent of the grid's dimensions, resulting in O(1) space complexity.

Optimal Solution

Approach

The core idea is to calculate the total sum of each row and column in the grid. Then, we strategically check if cutting the grid at any point creates two sections that both satisfy the condition of having a non-zero sum.

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

Calculate the total sum of all the numbers in the entire grid. This is the grand total.
Calculate the sum of each row and each column separately. Store these sums for later use.
Imagine cutting the grid horizontally, splitting it into a top section and a bottom section. For each possible cut position, add up the row sums for the top section and the row sums for the bottom section.
Check if both the top section sum and the bottom section sum are greater than zero. If both are, you've found a valid horizontal cut, and you can stop.
If you didn't find a valid horizontal cut, repeat a similar process for vertical cuts. Split the grid into a left section and a right section, and use the pre-calculated column sums to determine if both sections have a sum greater than zero.
If either a valid horizontal or a valid vertical cut is found, the grid can be cut into sections that meet the criteria. If neither are found after checking all possible cuts, then it cannot.

Code Implementation

def can_cut_grid(grid):
    total_sum = 0
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])

    for row in grid:
        total_sum += sum(row)

    row_sums = [sum(row) for row in grid]
    column_sums = [sum(grid[row][column] for row in range(number_of_rows)) for column in range(number_of_columns)]

    # Check for horizontal cuts
    for row_index in range(1, number_of_rows): 
        top_section_sum = sum(row_sums[:row_index])
        bottom_section_sum = sum(row_sums[row_index:])

        # We can cut if both sections have positive sums.
        if top_section_sum > 0 and bottom_section_sum > 0:
            return True

    # Check for vertical cuts
    for column_index in range(1, number_of_columns):
        left_section_sum = sum(column_sums[:column_index])
        right_section_sum = sum(column_sums[column_index:])

        # We can cut if both sections have positive sums.
        if left_section_sum > 0 and right_section_sum > 0:
            return True

    # No valid cut found.
    return False

Big(O) Analysis

Time Complexity

O(m*n) – Calculating the total sum of the grid involves iterating through all m rows and n columns, taking O(m*n) time. Calculating row and column sums also takes O(m*n) time. The horizontal cut check iterates through m-1 possible cut locations, summing row sums for each, which takes O(m) for each cut, so O(m*m) in the worst case when n is very small or equal to 1. Similarly, the vertical cut check iterates through n-1 possible locations, summing column sums, which takes O(n) for each cut, so O(n*n) in the worst case when m is very small or equal to 1. Thus, the complexity is determined by the initial grid traversal and the cut checks, resulting in a combined time complexity of O(m*n) + O(m*m) + O(n*n). We simplify to O(m*n) assuming that m and n are relatively large or in cases where they are roughly similar in magnitude.

Space Complexity

O(m + n) – The algorithm calculates and stores the sum of each row and each column. This requires creating an array of size 'm' to store row sums and another array of size 'n' to store column sums, where 'm' is the number of rows and 'n' is the number of columns in the grid. These arrays are the dominant factor in auxiliary space usage. Therefore, the space complexity is O(m + n), where m and n represent the dimensions of the input grid.

Edge Cases

Case	How to Handle
Null or empty grid	Return True if the grid is null or empty, as it can be considered trivially cut.
Grid with only one row or one column	Iterate and check if cumulative sum of row or column equals half of total sum.
Grid with all identical values (e.g., all 0s or all 1s)	The cumulative sum approach still works correctly in this situation, checking the sums.
Total sum of the grid is odd	Return False immediately because it's impossible to cut the grid into two sections with equal sums.
Very large grid dimensions leading to potential integer overflow when calculating sums	Use long or a larger data type to store sums to prevent integer overflow.
Grid with negative values	The cumulative sum approach still works correctly with negative values; however, ensure sums are calculated properly.
A cut is possible only at the very edge of the grid (e.g., the first row sums to half the total sum)	The solution should correctly handle cases where the cut is at the boundary without throwing index errors.
The Grid contains maximum integer values such as Integer.MAX_VALUE	Ensure that the summing process doesn't result in overflows, consider using BigInteger if needed to handle large numbers and additions.

Null or empty grid

How to Handle:

Return True if the grid is null or empty, as it can be considered trivially cut.

Grid with only one row or one column

How to Handle:

Iterate and check if cumulative sum of row or column equals half of total sum.

Grid with all identical values (e.g., all 0s or all 1s)

How to Handle:

The cumulative sum approach still works correctly in this situation, checking the sums.

Total sum of the grid is odd

How to Handle:

Return False immediately because it's impossible to cut the grid into two sections with equal sums.

Very large grid dimensions leading to potential integer overflow when calculating sums

How to Handle:

Use long or a larger data type to store sums to prevent integer overflow.

Grid with negative values

How to Handle:

The cumulative sum approach still works correctly with negative values; however, ensure sums are calculated properly.

A cut is possible only at the very edge of the grid (e.g., the first row sums to half the total sum)

How to Handle:

The solution should correctly handle cases where the cut is at the boundary without throwing index errors.

The Grid contains maximum integer values such as Integer.MAX_VALUE

How to Handle:

Ensure that the summing process doesn't result in overflows, consider using BigInteger if needed to handle large numbers and additions.