#3402

Easy

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Operations to Make Columns Strictly Increasing

Calculate the minimum increments to make each column of a matrix strictly increasing using a greedy invariant approach.

Problem Statement

You are given an m x n matrix grid containing non-negative integers. You can increase any cell by 1 per operation.

Return the minimum number of operations needed so that each column is strictly increasing from top to bottom, maintaining grid[i + 1][j] > grid[i][j] for every column.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Example details omitted.

Constraints

m == grid.length
n == grid[i].length
1 <= m, n <= 50
0 <= grid[i][j] < 2500

Solution Approach

Iterate Column by Column

Process each column independently. Start from the top row and move downward, comparing each cell with the one above to enforce strict increase.

Apply Greedy Increments

If the current cell is not greater than the previous cell, increment it minimally to satisfy the invariant. Accumulate the total increments for the final answer.

Return Total Operations

After all columns are processed, sum the increments from each column. This sum represents the minimum number of operations needed to make all columns strictly increasing.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(m * n) since each cell is visited once per column. Space complexity is O(1) beyond input storage, as only running totals are maintained.

What Interviewers Usually Probe

The candidate identifies a per-column approach and avoids unnecessary row comparisons across columns.
They correctly implement minimal increments instead of blindly adding arbitrary values.
They justify why processing top-down ensures the greedy invariant holds for all subsequent rows.

Common Pitfalls or Variants

Common pitfalls

Incrementing cells without checking the previous row can violate the strictly increasing requirement.
Attempting to modify rows simultaneously may overcount operations.
Neglecting edge cases where multiple rows have equal values leads to incorrect totals.

Follow-up variants

Allowing decrements as well as increments to achieve strictly increasing columns.
Finding minimum operations for strictly increasing rows instead of columns.
Handling larger matrices with m, n up to 10^3, requiring optimized traversal.

FAQ

What is the main pattern in Minimum Operations to Make Columns Strictly Increasing?

The problem uses a greedy choice plus invariant validation pattern, ensuring each cell exceeds the one above it with minimal increments.

Can I process rows instead of columns?

No, the problem specifically requires strictly increasing columns; row processing would violate the intended pattern.

How do I handle equal consecutive values in a column?

Increment the lower or equal value minimally so it becomes strictly greater than the one above, counting the increments toward the total.

What is the optimal time complexity for this problem?

O(m * n) since each cell must be checked once per column to maintain the strictly increasing invariant.

Does this approach work for large matrices?

Yes, it scales linearly with m * n, but extremely large matrices may require careful iteration to avoid timeouts.

terminal

Solution

Solution 1: Column-wise Calculation

We can traverse the matrix column by column. For each column, we calculate the minimum number of operations required to make it strictly increasing. Specifically, for each column, we maintain a variable $\textit{pre}$ to represent the value of the previous element in the current column. Then, we traverse the current column from top to bottom. For the current element $\textit{cur}$, if $\textit{pre} < \textit{cur}$, it means the current element is already greater than the previous element, so we only need to update $\textit{pre} = \textit{cur}$. Otherwise, we need to increase the current element to $\textit{pre} + 1$ and add the number of increases to the answer.

1

2

3

4

5

6

7

8

9

10

11

12

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

Continue Practicing

#3462 maximum sum with at most k elements #2931 maximum spending after buying items #2573 find the string with lcp