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Minimum Operations to Make Columns Strictly Increasing

Number: 3691

Difficulty: Easy

Paid? No

Companies: N/A

Problem Description

Given an m x n matrix grid containing non-negative integers, you can increment any cell by 1 in one operation. The goal is to determine the minimum number of operations needed so that every column of grid becomes strictly increasing from the top row to the bottom row.

Key Insights

  • The columns can be processed independently, as operations in one column do not affect another.
  • For each column, starting from the first row, ensure each subsequent cell is at least one unit greater than its predecessor.
  • If grid[i][j] is less than or equal to the previous cell's value, you must increase it to exactly (previous cell's value + 1) to minimize additional operations.
  • Sum the adjustments across all columns to get the total minimum operations.

Space and Time Complexity

Time Complexity: O(m * n), where m is the number of rows and n is the number of columns.
Space Complexity: O(1) additional space, using only constant extra storage.

Solution

The solution applies a greedy approach by processing each column separately. For each column:

  1. Start from the top value.
  2. For each subsequent cell, check if it is greater than the current required value.
  3. If the cell's value is less than or equal to the required value, increase it to exactly (current required value + 1) and add the difference to the operation count.
  4. Update the current required value for the column accordingly.
  5. Finally, add the operations for each column together for the overall answer.

Code Solutions

# Python solution with line-by-line comments

def min_operations(grid):
    # Determine the number of rows and columns
    m = len(grid)
    n = len(grid[0])
    total_operations = 0
    # Process each column independently
    for col in range(n):
        # Set the initial required value as the first element of the column
        current_value = grid[0][col]
        # Iterate through the remaining rows of the column
        for row in range(1, m):
            # If the current cell is not strictly greater than the previous value
            if grid[row][col] <= current_value:
                # Calculate the number of operations needed to make it strictly larger
                operations_needed = (current_value + 1) - grid[row][col]
                total_operations += operations_needed
                # Update current_value to the new adjusted value
                current_value += 1
            else:
                # Update current_value with the cell's value if no increment is needed
                current_value = grid[row][col]
    return total_operations

# Example usage:
grid1 = [[3,2],[1,3],[3,4],[0,1]]
print(min_operations(grid1))  # Expected output: 15

grid2 = [[3,2,1],[2,1,0],[1,2,3]]
print(min_operations(grid2))  # Expected output: 12
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