Delete Greatest Value in Each Row

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ArraysGreedy Algorithms

You are given an m x n matrix grid consisting of positive integers.

Perform the following operation until grid becomes empty:

Delete the element with the greatest value from each row. If multiple such elements exist, delete any of them.
Add the maximum of deleted elements to the answer.

Note that the number of columns decreases by one after each operation.

Return the answer after performing the operations described above.

Example 1:

Input: grid = [[1,2,4],[3,3,1]]
Output: 8
Explanation: The diagram above shows the removed values in each step.
- In the first operation, we remove 4 from the first row and 3 from the second row (notice that, there are two cells with value 3 and we can remove any of them). We add 4 to the answer.
- In the second operation, we remove 2 from the first row and 3 from the second row. We add 3 to the answer.
- In the third operation, we remove 1 from the first row and 1 from the second row. We add 1 to the answer.
The final answer = 4 + 3 + 1 = 8.

Example 2:

Input: grid = [[10]]
Output: 10
Explanation: The diagram above shows the removed values in each step.
- In the first operation, we remove 10 from the first row. We add 10 to the answer.
The final answer = 10.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 50
1 <= grid[i][j] <= 100

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 input `grid` (i.e., maximum number of rows and columns)?
What is the range of integer values within the `grid`? Can they be negative, zero, or very large?
If the `grid` is empty or has zero rows or zero columns, what should I return?
If multiple elements in a row have the same greatest value, which one should be deleted (e.g., the first one encountered)?
Do I need to modify the input `grid` in-place, or can I create a new `grid` or return the result in some other format?

Brute Force Solution

Approach

The goal is to find the largest number in each row of a table and then sum up these largest numbers. The brute force approach looks at every single number in the table to determine the largest in each row, without trying to be clever or efficient.

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

Look at the first row of the table.
Go through each number in that row, one by one.
Keep track of which number is the biggest so far.
Once you've looked at all the numbers in the first row, you'll know the largest number in that row.
Do the same thing for the second row, the third row, and so on, until you find the largest number in every single row.
Finally, add up all the largest numbers you found from each row to get the final answer.

Code Implementation

def delete_greatest_value_in_each_row(grid):
    total_greatest_sum = 0

    # Iterate through each row in the grid
    for row in grid:
        largest_value_in_row = row[0]

        # Find the largest value in the current row
        for value in row:
            if value > largest_value_in_row:
                largest_value_in_row = value

        # Add the largest value to the total sum
        total_greatest_sum += largest_value_in_row

    return total_greatest_sum

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm iterates through each row of the table, where 'm' is the number of rows. For each row, it iterates through all 'n' elements to find the maximum value in that row. Since it performs this row-wise maximum finding for every row, the total number of operations is proportional to m*n. Therefore, the overall time complexity is O(m*n).

Space Complexity

O(1) – The provided plain English explanation outlines a brute force approach that iterates through each row and element to find the maximum value in each row. It mentions keeping track of the 'biggest so far' in each row. This implies using a single variable to store the current maximum value found while iterating through a row, and another variable to accumulate the sum of the largest values. These are constant space operations. Thus, the algorithm uses constant auxiliary space, independent of the input size N, where N represents the total number of elements in the input table.

Optimal Solution

Approach

The key is to find the largest number in each row first. Then, we can just keep track of the maximum of these largest values as we go, deleting one at a time.

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

For each row of numbers, identify the biggest number and remember it.
Find the biggest number out of all the biggest numbers you just found for each row. This is the number you will 'delete' for this round, as it's the greatest.
Add this greatest value to your total.
Remove or mark as deleted the greatest value that was selected.
Repeat the process of finding the biggest number in each row, identifying the greatest across rows, and adding it to the total, until all the numbers are gone.
The final total is the answer.

Code Implementation

def delete_greatest_value(grid):
    total_sum = 0
    while grid and grid[0]:
        maximums_per_row = []
        
        # Find the maximum value in each row.
        for row in grid:
            maximums_per_row.append(max(row))

        greatest_maximum = max(maximums_per_row)
        total_sum += greatest_maximum
        
        #Remove the greatest value from each row
        for row in grid:
            if greatest_maximum in row:
                row.remove(greatest_maximum)

        # Clean up any empty rows after deletion
        grid = [row for row in grid if row]

    return total_sum

Big(O) Analysis

Time Complexity

O(m*n*log(n)) – Given an input matrix of size m x n, where m is the number of rows and n is the number of columns, for each iteration we need to find the largest element in each row, which takes O(n*log(n)) since sorting each row could be done in n*log(n) time. After finding the largest element in each row, determining the overall maximum among those m largest values takes O(m). Deleting the maximum identified elements takes constant time. This process is repeated until the matrix is empty, which requires 'n' iterations in the worst case. Therefore, the time complexity can be expressed as O(n * (n*log(n) + m)). Assuming that in the worst case n is far larger than m, this simplifies to O(m*n*log(n)).

Space Complexity

O(m) – The described solution involves identifying the largest number in each row and keeping track of these largest numbers. This implies maintaining a collection (implicitly an array or list) of size 'm', where 'm' is the number of rows in the input matrix. The algorithm also keeps track of the greatest of these largest numbers; this requires a constant amount of space. Thus, the dominant space usage is storing the largest numbers from each row, requiring space proportional to the number of rows. Therefore, the auxiliary space complexity is O(m).

Edge Cases

Case	How to Handle
Null or empty input matrix	Return 0 if the input matrix is null or empty, as there's no greatest value to sum.
Matrix with only one row	Return the largest element of the single row as the sum in this case.
Matrix with rows of varying lengths	Handle the jagged array gracefully, iterating only within valid bounds of each row and padding with negative infinity or similar for comparisons to determine the greatest element.
Matrix with all rows containing only identical values	The greatest value will be the same for each row, so the sum will be that value multiplied by the number of rows.
Matrix with negative numbers	The algorithm should correctly identify the largest value even if it's negative or if all values are negative; it should not assume non-negativity.
Matrix with very large numbers leading to potential integer overflow during summing	Use a data type that can accommodate large sums (e.g., long) or return an error if the sum exceeds the maximum representable value for the chosen data type.
Extremely large matrix dimensions causing memory issues	Consider an iterative approach to minimize memory usage, processing rows one at a time instead of loading the entire matrix into memory, or throwing an exception if the size is beyond manageable limits.
A row contains only Integer.MIN_VALUE.	Handle `Integer.MIN_VALUE` correctly by checking if the row's max value is `Integer.MIN_VALUE` and proceeding normally; `Integer.MIN_VALUE` is still a valid value.

Null or empty input matrix

How to Handle:

Return 0 if the input matrix is null or empty, as there's no greatest value to sum.

Matrix with only one row

How to Handle:

Return the largest element of the single row as the sum in this case.

Matrix with rows of varying lengths

How to Handle:

Handle the jagged array gracefully, iterating only within valid bounds of each row and padding with negative infinity or similar for comparisons to determine the greatest element.

Matrix with all rows containing only identical values

How to Handle:

The greatest value will be the same for each row, so the sum will be that value multiplied by the number of rows.

Matrix with negative numbers

How to Handle:

The algorithm should correctly identify the largest value even if it's negative or if all values are negative; it should not assume non-negativity.

Matrix with very large numbers leading to potential integer overflow during summing

How to Handle:

Use a data type that can accommodate large sums (e.g., long) or return an error if the sum exceeds the maximum representable value for the chosen data type.

Extremely large matrix dimensions causing memory issues

How to Handle:

Consider an iterative approach to minimize memory usage, processing rows one at a time instead of loading the entire matrix into memory, or throwing an exception if the size is beyond manageable limits.

A row contains only Integer.MIN_VALUE.

How to Handle:

Handle `Integer.MIN_VALUE` correctly by checking if the row's max value is `Integer.MIN_VALUE` and proceeding normally; `Integer.MIN_VALUE` is still a valid value.