algoadvance

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

Perform the following operation until the grid becomes empty:

1. Find the greatest value in each row.
2. Delete these greatest values in each row.
3. Sum up the greatest values found in step 1.

Return the sum of the greatest values encountered during the operation.

Example:

Input: grid = [[10, 20, 30], [5, 15, 25]]
Output: 55
Explanation: 
- First greatest values: [30, 25] => sum = 55
  New grid: [[10, 20], [5, 15]]
- Second greatest values: [20, 15] => sum = 35
  New grid: [[10], [5]]
- Third greatest values: [10, 5] => sum = 15
  New grid: [[], []]
Total sum = 55 + 35 + 15 = 105

Clarifying Questions

1. Can the elements be repeated within the grid?
   • Yes, since we are not given any constraints about unique values.
2. Is the matrix guaranteed to have at least one row and one column?
   • Yes, since it is described as an m x n matrix with both m and n positive.
3. What should we do if the grid is already empty initially?
   • It is mentioned that the matrix is m x n with positive integers, so this scenario does not apply.

Strategy

1. While the grid is not empty:
   • Traverse each row to find the maximum value.
   • Remove this maximum value from the row.
   • Add this maximum value to a running sum.
2. Continue the above operation until all rows are empty.
3. Finally, return the total sum.

Code

def delete_greatest_values(grid):
    total_sum = 0
    while any(grid):
        max_values = []
        for row in grid:
            if row:  # Ensure row is not empty
                max_val = max(row)
                max_values.append(max_val)
                row.remove(max_val)  # Remove the greatest value in the row
        total_sum += sum(max_values)
    return total_sum

# Example usage
grid = [[10, 20, 30], [5, 15, 25]]
print(delete_greatest_values(grid))  # Output should be 105

Time Complexity

• Finding the maximum value in each row: O(n) for each row.
• Removing the maximum value from each row: This operation is generally O(n), as list removal operations can be costly due to shifts.
• Total iterations: Each operation reduces the column count, so at most n operations.

Thus, the overall time complexity is O(m * n^2) where m is the number of rows and n is the number of columns, because each removal operation can potentially take O(n) time.

Space Complexity

• Auxiliary space used for max_values list: O(m)
• No additional space scales with the size of the input grid.

So the space complexity is O(m).

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