Problem Description

You are given an m x n matrix grid consisting of non-negative integers.

In one operation, you can increment the value of any grid[i][j] by 1.

Return the minimum number of operations needed to make all columns of grid strictly increasing.

Intuition

To solve this problem, we focus on each column independently. Our goal is to adjust the values in such a way that each column becomes strictly increasing. The crucial step is to determine how to decide the required changes for each element in a column:

1. Column-wise Calculation: Since each column needs to be strictly increasing, we go through the grid column by column. This allows us to make localized decisions without affecting other columns.

2. Maintaining Order: For each column, we maintain a variable pre to keep track of the previous element as we iterate through the column elements. This helps us determine whether each current element is already greater than the preceding one.

3. Increment Operations:

   • If the current element is already larger than the previous (pre < cur), we simply update pre to the current value.
   • If the current element is not greater, we increment it to pre + 1 to make it greater than the previous element. The number of incrementation steps needed is then added to our answer.

This approach guarantees that each column becomes strictly increasing with the minimal number of operations needed.

Learn more about Greedy patterns.

Solution Approach

Solution 1: Column-wise Calculation

The provided solution employs a clear approach to solve the problem by iterating through each column of the grid. The algorithm is structured as follows:

1. Initialize a Counter: Start with a counter ans = 0 to accumulate the total number of operations needed across all columns.

2. Column Iteration: For each column, treat it as a separate list of integers. This is efficiently done by using Python's zip(*grid), which allows iteration over columns directly.

3. Track Previous Element: Within each column, keep track of the previous value with a variable pre. Initially, this is set to -1 to ensure that any non-negative number in the column will be greater at the start.

4. Check and Update:

   • Traverse the column element by element.
   • If the current element (cur) is greater than pre, update pre to the current element since this meets the strictly increasing condition.
   • If not, increment the current element to pre + 1 and add the increment to ans. This ensures the element becomes larger than pre, thereby maintaining the strictly increasing requirement.

5. Complete Iteration: After iterating through each column and making necessary adjustments, return the accumulated count ans as the total minimum operations needed.

The algorithm efficiently ensures that each column is transformed into a strictly increasing sequence with minimal operations, leveraging a straightforward linear pass through the matrix.

class Solution:
    def minimumOperations(self, grid: List[List[int]]) -> int:
        ans = 0
        for col in zip(*grid):
            pre = -1
            for cur in col:
                if pre < cur:
                    pre = cur
                else:
                    pre += 1
                    ans += pre - cur
        return ans

This solution efficiently addresses the problem using the zip functionality for easy column access and a simple tracking mechanism to count necessary increments.

Example Walkthrough

Let's consider a small example with a 3x3 grid:

grid = [
    [1, 3, 1],
    [2, 0, 5],
    [2, 4, 3]
]

The goal is to make each column strictly increasing with the fewest operations.

1. Initialize a Counter:

   • Set ans = 0 to track total operations.

2. Column Iteration:

   • Use zip(*grid) to iterate through columns.

3. First Column [1, 2, 2]:

   • Initialize pre = -1.
   • For 1: 1 > -1, update pre = 1.
   • For 2: 2 > 1, update pre = 2.
   • For 2: 2 not > 2, increment pre to 3 and ans = 1.

   After handling the first column, the modified first column is [1, 2, 3] and ans = 1.

4. Second Column [3, 0, 4]:

   • Initialize pre = -1.
   • For 3: 3 > -1, update pre = 3.
   • For 0: 0 not > 3, increment pre to 4 and ans = 5.
   • For 4: 4 > 4, wouldn't happen, increment pre to 5 and ans = 6.

   After handling the second column, the modified second column is [3, 4, 4] and ans = 6.

5. Third Column [1, 5, 3]:

   • Initialize pre = -1.
   • For 1: 1 > -1, update pre = 1.
   • For 5: 5 > 1, update pre = 5.
   • For 3: 3 not > 5, increment pre to 6 and ans = 9.

   After handling the third column, the modified third column is [1, 5, 6] and ans = 9.

By following these steps, we've transformed the grid into:

[
    [1, 3, 1],
    [2, 4, 5],
    [3, 5, 6]
]

Each column is now strictly increasing, and the total number of operations needed is 9.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(m \times 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 over each column (zip(*grid) results in iterating over columns) and then iterates each element within that column, leading to a nested iteration covering all elements in the grid.

The space complexity is O(1) as the solution uses a constant amount of extra space for variables like ans and pre, regardless of the size of the input grid.

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