2679. Sum in a Matrix

Problem Description

In this problem, we are provided with a 0-indexed 2D integer array nums, which represents a matrix. The goal is to calculate a score based on specific operations performed on this matrix. We start with a score of 0, and we repeatedly perform the following two-part operation until the matrix is empty:

1. From each row in the matrix, select and remove the largest number. If any row has multiple largest numbers (i.e., a tie), any of them can be selected.
2. After removing these numbers from each row, identify the largest number among them and add that number to the score.

The required output is the final score once there are no more numbers left in the matrix.

It's important to keep track of the largest numbers being selected during each operation and ensuring that the correct value (the maximum from these selections) is added appropriately to the score.

Intuition

The intuition behind the solution comes from the way the operations on the matrix are defined. Since we always need to select the largest number from each row and find the overall maximum to add to the score, sorting the rows can simplify the problem. By sorting each row, we guarantee that the largest number in each row will be at the end.

This operation effectively "flips" the matrix so that each row becomes a column. Then, the algorithm iterates through the new "rows" (original columns) and computes the maximum value. This value is the one that should be added to the score since it represents the largest value that would be selected from the original rows during an iteration.

By continuing to sum these maxima for all "new rows," we accumulate the total score. The benefit of this approach is that the whole process takes place with a complexity close to that of the sorting operation itself, which is efficient compared to a naive approach that might involve multiple iterations for each step.

The use of Python's built-in functions like sort(), max(), and map() allows for a concise and efficient implementation of this intuition.

Learn more about Sorting and Heap (Priority Queue) patterns.

Solution Approach

The implementation of the solution follows the steps that correspond to its intuitive approach described previously. Let's walk through each part of the implementation:

1. Sort each row of the nums matrix. In Python, this can be achieved using the sort() method, which sorts the elements of a list in ascending order. By sorting each row, we ensure that the largest number in each row will be at the end.

   • This is done with the following line of code:

     1for row in nums:
     2    row.sort()

2. The next step is to find the largest number that was removed from each row and add that to the score. Since the rows are sorted, we can use the fact that in Python, the max() function can be applied to a list of lists by using the zip(*iterables) function. The zip(*nums) effectively transposes the matrix, so the last elements of each original row (which are the largest due to the sort) become the elements of the new "rows."

   • This "transposition" and finding the maximum occurs in the following line of code:

     1return sum(map(max, zip(*nums)))

   • The map() function applies max to each new row (actually a column of the original matrix) and finds the maximum element, which corresponds to the maximum number we removed from the original rows.

   • Finally, the sum() function takes the iterable produced by map() (which consists of the largest numbers from each operation) and sums them up, thus calculating the final score.

The combination of sort(), max(), zip(), and map() provides an elegant and efficient solution. It uses sorting to rearrange the elements, a matrix transpose operation to work across rows as if they were columns, the max() function to grab the largest values in a straightforward manner, and sum() to accumulate these values into the final score.

Example Walkthrough

Consider the following small matrix for our example:

1nums = [
2    [4, 5, 6],
3    [1, 2, 9],
4    [7, 8, 3]
5]

Let's walk through the steps of the solution approach using this matrix:

1. Sort each row of the nums matrix in ascending order:

   Before sorting:

   1[
   2    [4, 5, 6],
   3    [1, 2, 9],
   4    [7, 8, 3]
   5]

   After sorting each row:

   1[
   2    [4, 5, 6],
   3    [1, 2, 9],
   4    [3, 7, 8]
   5]

   As per the first step, rows are sorted, and now the largest number in each row is at the end.

2. Using zip(*nums), we transpose the matrix so we can easily access the largest numbers (the last elements of each row):

   Transposed and zipped matrix (effectively "flip" so that each row is now a column):

   1[
   2    [4, 1, 3],
   3    [5, 2, 7],
   4    [6, 9, 8]
   5]

3. For each new row (originally a column), we select the maximum value (which, in this case, would be the originally last item of each row due to sorting).

   Largest numbers, which are the maxima of the transposed rows:

   • The max of the first transposed row is 4
   • The max of the second transposed row is 7
   • The max of the third transposed row is 9

4. Finally, we sum up these maxima to get the final score, which is 4 + 7 + 9 = 20.

The sum of these maxima gives us the solution to the problem, which is the final score. In this example, the score is 20.

Solution Implementation

Time and Space Complexity

Time Complexity:

The time complexity of the matrixSum method involves two steps: sorting the rows of the matrix and finding the maximum element of each column after transposition.

1. Sorting: Each row is sorted individually using row.sort(), which typically uses Tim Sort, an algorithm with a worst-case time complexity of O(n log n) for sorting a list of n elements. If m represents the number of rows and n represents the number of columns in nums, then the sorting step has a time complexity of O(m * n log n).

2. Finding max and summing: The zip(*nums) function is used to transpose the matrix, and map(max, ...) is used to find the maximum element in each column. Since there are n columns, and finding the max takes O(m) time for each column, this step has a time complexity of O(m * n). The final summing of these values is done in O(n).

Combining the two steps, the overall time complexity is O(m * n log n) + O(m * n) + O(n), which simplifies to O(m * n log n) because O(m * n log n) is the dominating term.

Space Complexity:

1. Sorting: Sorting is done in-place for each row, so no additional space is proportional to the size of the input matrix is used. Therefore, it does not increase the asymptotic space complexity.

2. Transposing and finding max: The zip function returns an iterator of tuples, which, when combined with map, doesn't create a list of the entire transposed matrix but rather creates one tuple at a time. Thus, the space required for this operation is O(n) for storing the maximums of each column.

Thus, the space complexity of the matrixSum method is O(n).

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