2371. Minimize Maximum Value in a Grid

Problem Description

The problem presents a challenge to minimize the maximum value in a given m x n matrix, grid, while maintaining the relative ordering of numbers in their respective rows and columns. In this context, "relative order" means that if a number is greater than another number in the same row or column in the original matrix, it must remain greater in the transformed matrix. The objective is to do so by replacing each element in the matrix with a new positive integer and ensuring that the new maximum number in the matrix is as small as possible.

A fundamental requirement is that the original matrix contains distinct positive integers. This constraint simplifies the problem, as we do not need to consider the case where two elements are equal. The challenge is to determine the smallest possible replacements that satisfy the aforementioned relative order conditions.

Intuition

To maintain the relative order within rows and columns, we look at the individual elements' ranks when compared to other elements within the same rows and columns. The key observation is that the smallest number can be replaced by 1, the second smallest by 2, and so on, without changing their relative positions. However, simply increasing each subsequent element by 1 does not always work, especially when an element is not the smallest in its row or column.

The approach for finding the solution is twofold: Firstly, sort all of the elements in the grid based on their value while keeping track of their original positions. This allows us to process the elements in increasing order, ensuring that each replacement maintains the overall relative ordering structure of the grid.

Secondly, maintain two arrays row_max and col_max, where row_max[i] indicates the largest number assigned so far in row i, and col_max[j] the largest number in column j. As we process each element in sorted order, we determine its new value by taking the maximum of row_max[i] and col_max[j] for its position (i, j) and then add 1 to it. This ensures that the new number will be larger than any previously assigned number in the same row or column, maintaining the relative order. The two arrays are then updated with this new value to reflect the new maximum values for that row and column.

By using this method, we can iterate through all the elements in the grid only once, after sorting, and consistently assign the minimum required number that maintains both the ordering and the constraint of minimizing the maximum number in the resultant matrix.

Learn more about Union Find, Graph, Topological Sort and Sorting patterns.

Solution Approach

The solution approach takes advantage of a few powerful algorithms and data structures to organize and process data efficiently. Here's how the solution implements the approach step by step:

1. Sorting: The original matrix elements are expanded into a list of tuples nums, where each tuple contains an element value along with its row and column indices. This list is then sorted primarily by the value, which means the tuples will be arranged in ascending order according to the elements of the matrix they represent.

   1nums = [(v, i, j) for i, row in enumerate(grid) for j, v in enumerate(row)]
   2nums.sort()

   This organizing strategy is crucial because it allows for the iteration over the matrix elements in order of their size. Since each element is associated with its original position, we can apply the exact replacements required, maintaining the relative order.

2. Tracking Row and Column Maximums: Two arrays row_max and col_max are used to keep track of the highest value that has been assigned to each row and column, respectively. Initiated with zeros, these arrays will update with each assignment made in the following steps:

   1row_max = [0] * m
   2col_max = [0] * n

3. Constructing the Result Matrix: A new matrix ans is initialized with zero values, where the solution will be built incrementally.

   1ans = [[0] * n for _ in range(m)]

4. Iterating and Assigning New Values: The algorithm then iterates over this sorted list of tuples. For each tuple, which corresponds to an element from the original grid, we determine the replacement number. This number is one greater than the maximum of the current values of row_max[i] and col_max[j]. We then update both row_max and col_max at index i and j to reflect this new maximum.

   1for _, i, j in nums:
   2    ans[i][j] = max(row_max[i], col_max[j]) + 1
   3    row_max[i] = col_max[j] = ans[i][j]

In this implementation, Python list comprehensions, tuple unpacking, and sorting mechanisms are powerful tools used to streamline the solution process. This approach ensures that each element's relative value to its neighbors is preserved while the overall maximum value in the matrix is minimized. The end result is a matrix ans that meets all the stipulated requirements and can be returned as the solution to the problem.

Thus, the solution makes effective use of familiar data structures (such as lists) and algorithmic patterns (like sorting and iteration) to solve a relatively complex problem in an efficient and straightforward manner.

Example Walkthrough

Let's take a 2x3 matrix as an example to illustrate the solution approach.

Imagine we have the following matrix grid:

18 4 6
23 5 9

The challenge is to minimize the maximum number after replacing numbers while maintaining their relative ordering.

1. Sorting:

   • Expand the elements of grid into a list of tuples containing the value with its row and column index:

   1nums = [(8, 0, 0), (4, 0, 1), (6, 0, 2), (3, 1, 0), (5, 1, 1), (9, 1, 2)]

   • After sorting nums by value, we get:

   1nums = [(3, 1, 0), (4, 0, 1), (5, 1, 1), (6, 0, 2), (8, 0, 0), (9, 1, 2)]

2. Tracking Row and Column Maximums:

   • Initialize row_max and col_max to keep track of the max values so far:

   1row_max = [0, 0]
   2col_max = [0, 0, 0]

3. Constructing the Result Matrix:

   • Initialize the ans matrix with zeroes to build the solution:

   1ans = [[0, 0, 0], [0, 0, 0]]

4. Iterating and Assigning New Values:

   • Iterate over sorted nums and replace values in ans while updating row_max and col_max:

     • For (3, 1, 0) -> ans becomes:

       1ans = [[0, 0, 0], [1, 0, 0]]

       And row_max and col_max update to:

       1row_max = [0, 1]
       2col_max = [1, 0, 0]

     • For (4, 0, 1) -> ans becomes:

       1ans = [[0, 2, 0], [1, 0, 0]]

       And row_max and col_max update to:

       1row_max = [2, 1]
       2col_max = [1, 2, 0]

     • Repeat this process for each element in nums.

The final ans matrix after completing the iterations will look like this:

12 2 3
21 2 4

In this new ans matrix, the relative ordering is maintained, and the maximum value, which is 4, is minimized compared to the original grid. The ans matrix can be returned as the solution to the problem.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code has several distinct operations, each contributing to the overall time complexity.

1. First, the list comprehension iterates over each element in the m x n grid to create a list of tuples. This operation has a time complexity of O(m * n).

2. Next, the sort operation on this list of tuples has a time complexity of O(m * n log(m * n)), since it sorts the list with respect to the values in the grid.

3. Then, the code iterates over the sorted list nums with m * n elements, performing constant-time operations inside the loop. This contributes O(m * n) to the time complexity.

4. The updates to row_max and col_max are also constant-time operations, happening m * n times.

Combining all the above steps, the overall time complexity of the code is O(m * n log(m * n)) + O(m * n) + O(m * n), which simplifies to O(m * n log(m * n)) since the log term will dominate at scale.

Space Complexity

The space complexity can be analyzed based on the data structures used:

1. The list nums which stores the tuples has a space complexity of O(m * n) because it stores each element of the grid.

2. Two arrays, row_max and col_max, each with a length of m and n respectively, resulting in O(m) and O(n) space used.

3. The ans list is a 2D list of the same size as the input grid, yielding a space complexity of O(m * n).

Therefore, the total space complexity is O(m * n) + O(m) + O(n), which simplifies to O(m * n) when m and n are of the same order.

Learn more about how to find time and space complexity quickly using problem constraints.