Problem Description

You have a matrix grid with m rows and n columns, where each cell contains a non-negative integer.

You can perform operations on this matrix. In each operation, you can choose any cell grid[i][j] and increase its value by 1.

Your goal is to make each column of the matrix strictly increasing. This means that for every column, each element must be strictly greater than the element above it (i.e., grid[i][j] < grid[i+1][j] for all valid i).

Return the minimum total number of operations needed to achieve this.

For example, if you have a column with values [3, 2, 5], you would need to increase the second element from 2 to at least 4 (making it [3, 4, 5]), which requires 2 operations. The column would then be strictly increasing since 3 < 4 < 5.

The solution works by processing each column independently. For each column, it tracks the previous element's value and ensures each subsequent element is at least one greater than the previous. If an element is already greater than needed, no operations are required for that element. Otherwise, the element is increased to previous + 1, and the number of operations is counted.

Intuition

The key insight is that we can process each column independently since operations on one column don't affect other columns. This simplifies the problem significantly.

For a column to be strictly increasing, each element must be greater than the one above it. When we encounter an element that violates this rule (it's not greater than the previous element), we need to increase it. But by how much?

The minimum increase would be to make it exactly one more than the previous element. Why? Because:

1. Making it equal to the previous element wouldn't satisfy the "strictly increasing" requirement
2. Making it more than previous + 1 would use extra operations unnecessarily

Consider a column [5, 3, 7]. When we reach 3, it's not greater than 5. The minimum value it needs to be is 6 (which is 5 + 1). This requires 6 - 3 = 3 operations.

Now, when we move to the next element 7, we compare it with our updated previous value 6 (not the original 5). Since 7 > 6, no operations are needed.

This greedy approach works because:

• We only increase values when absolutely necessary
• We increase by the minimum amount possible
• Once we fix an element to maintain the strictly increasing property, we never need to revisit it

The algorithm maintains a pre variable that tracks what the previous element's value should be after any necessary operations. This way, we can make decisions for each element based on the corrected values, not the original ones.

Learn more about Greedy patterns.

Solution Approach

The solution uses a column-wise traversal approach with a greedy strategy to minimize operations.

Algorithm Steps:

1. Transpose the Matrix: Use zip(*grid) to iterate through columns instead of rows. This Python idiom effectively transposes the matrix, allowing us to process each column as a sequence.

2. Process Each Column Independently: For each column, maintain a variable pre initialized to -1 (since all values are non-negative, -1 ensures the first element always satisfies pre < cur).

3. Traverse Elements in Current Column: For each element cur in the column:

   • Case 1: If pre < cur, the strictly increasing property is already satisfied. Simply update pre = cur to track the current element's value.
   • Case 2: If pre >= cur, we need to increment cur to maintain strict increasing order:
     • Set pre = pre + 1 (the minimum value needed for strict increasing)
     • Add pre - cur to the answer (number of operations needed)

4. Accumulate Operations: The variable ans accumulates the total operations across all columns.

Example Walkthrough:

Consider a column [3, 5, 2, 8]:

• Start with pre = -1
• Element 3: Since -1 < 3, set pre = 3, operations = 0
• Element 5: Since 3 < 5, set pre = 5, operations = 0
• Element 2: Since 5 >= 2, need to increase 2 to 6 (which is 5 + 1)
  • Set pre = 6
  • Add 6 - 2 = 4 operations
• Element 8: Since 6 < 8, set pre = 8, operations = 0
• Total operations for this column: 4

Time Complexity: O(m × n) where m is the number of rows and n is the number of columns, as we visit each element once.

Space Complexity: O(1) excluding the input, as we only use a few variables for tracking.

Example Walkthrough

Let's walk through a complete example with a 3×3 matrix to illustrate the solution approach.

Given matrix:

grid = [[3, 1, 6],
        [2, 5, 4],
        [4, 3, 8]]

We need to make each column strictly increasing from top to bottom.

Processing Column 0: [3, 2, 4]

• Initialize pre = -1
• Element 3: Since -1 < 3, no operations needed. Set pre = 3
• Element 2: Since 3 >= 2, we need to increase 2 to at least 4 (which is 3 + 1)
  • Operations needed: 4 - 2 = 2
  • Set pre = 4
• Element 4: Since 4 >= 4, we need to increase 4 to at least 5 (which is 4 + 1)
  • Operations needed: 5 - 4 = 1
  • Set pre = 5
• Column 0 total operations: 2 + 1 = 3
• Final column 0: [3, 4, 5]

Processing Column 1: [1, 5, 3]

• Initialize pre = -1
• Element 1: Since -1 < 1, no operations needed. Set pre = 1
• Element 5: Since 1 < 5, no operations needed. Set pre = 5
• Element 3: Since 5 >= 3, we need to increase 3 to at least 6 (which is 5 + 1)
  • Operations needed: 6 - 3 = 3
  • Set pre = 6
• Column 1 total operations: 3
• Final column 1: [1, 5, 6]

Processing Column 2: [6, 4, 8]

• Initialize pre = -1
• Element 6: Since -1 < 6, no operations needed. Set pre = 6
• Element 4: Since 6 >= 4, we need to increase 4 to at least 7 (which is 6 + 1)
  • Operations needed: 7 - 4 = 3
  • Set pre = 7
• Element 8: Since 7 < 8, no operations needed. Set pre = 8
• Column 2 total operations: 3
• Final column 2: [6, 7, 8]

Final Result:

• Total operations: 3 + 3 + 3 = 9
• Final matrix after operations:

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

Each column is now strictly increasing: 3 < 4 < 5, 1 < 5 < 6, and 6 < 7 < 8.

Solution Implementation

Time and Space Complexity

The time complexity is O(m × n), where m is the number of rows and n is the number of columns in the matrix grid. This is because the code iterates through each element of the grid exactly once. The zip(*grid) operation transposes the grid to iterate column by column, which takes O(m × n) time, and then the nested loops process each element once.

The space complexity is O(m) for the transposed columns created by zip(*grid). When zip(*grid) is called, it creates iterator objects that hold references to the columns. Each column contains m elements (where m is the number of rows), and we process one column at a time. However, if we consider only the additional space used beyond the input, and noting that zip creates iterators rather than materializing all columns at once, the auxiliary space complexity can be considered O(1) as we only use a constant amount of extra variables (ans, pre, cur) regardless of input size.

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

Common Pitfalls

1. Modifying the Original Grid

A common mistake is trying to modify the grid in-place while calculating operations. This can lead to incorrect results when processing subsequent columns.

Incorrect Approach:

def minimumOperations(self, grid: List[List[int]]) -> int:
    total_operations = 0
    m, n = len(grid), len(grid[0])
  
    for j in range(n):
        for i in range(1, m):
            if grid[i][j] <= grid[i-1][j]:
                old_value = grid[i][j]
                grid[i][j] = grid[i-1][j] + 1  # WRONG: Modifying grid
                total_operations += grid[i][j] - old_value
  
    return total_operations

Why it's wrong: Modifying the grid affects the calculation for other columns if you're not careful about the order of processing.

Solution: Track the "effective" value without modifying the grid, as shown in the correct solution using the previous_value variable.

2. Not Handling the Strictly Increasing Requirement

Some might mistakenly allow equal values, implementing non-decreasing instead of strictly increasing.

Incorrect Check:

if previous_value <= current_value:  # WRONG: Allows equal values
    previous_value = current_value

Correct Check:

if previous_value < current_value:  # Correct: Ensures strict inequality
    previous_value = current_value

3. Integer Overflow in Other Languages

While Python handles large integers automatically, implementing this in languages like C++ or Java requires careful consideration of integer overflow when accumulating operations.

Potential Issue in C++:

int total_operations = 0;  // May overflow for large grids

Solution for C++:

long long total_operations = 0;  // Use long long to prevent overflow

4. Incorrect Initial Value for previous_value

Using 0 as the initial value instead of -1 can cause issues when the first element of a column is 0.

Incorrect:

previous_value = 0  # WRONG: Won't work if first element is 0

Correct:

previous_value = -1  # Correct: Works for all non-negative integers

5. Processing Rows Instead of Columns

Misreading the problem and making rows strictly increasing instead of columns.

Incorrect Direction:

for row in grid:  # WRONG: Processing rows
    previous_value = -1
    for current_value in row:
        # ... process row elements

Correct Direction:

for column in zip(*grid):  # Correct: Processing columns
    previous_value = -1
    for current_value in column:
        # ... process column elements