LeetCode 3402 - Minimum Operations to Make Columns Strictly Increasing

The problem gives us a matrix grid of size m x n containing non-negative integers. Each column must become strictly increasing, meaning for each column j, the condition grid[i][j] < grid[i+1][j] must hold for all 0 <= i < m-1.

leetcodeeasyarraygreedymatrix

LeetCode Problem 3402

Difficulty: 🟢 Easy
Topics: Array, Greedy, Matrix

Solution

Problem Understanding

The problem gives us a matrix grid of size m x n containing non-negative integers. Each column must become strictly increasing, meaning for each column j, the condition grid[i][j] < grid[i+1][j] must hold for all 0 <= i < m-1. We are allowed to perform operations that increment any element by 1, and the goal is to calculate the minimum total number of operations needed.

The input grid represents the initial state of the matrix, and the output is a single integer representing the sum of all increments required to satisfy the strictly increasing condition in all columns.

Constraints indicate that both m and n are relatively small (<=50) and element values are below 2500. This means a solution that iterates over all elements and performs simple arithmetic operations will be efficient enough, but nested loops with unnecessary complexity might be wasteful. Key edge cases include columns that are already strictly increasing, columns with repeated values, or single-row matrices where no operations are needed.

Approaches

Brute Force

A brute-force approach would be to repeatedly scan each column from top to bottom and, whenever a row violates the strictly increasing condition, increment it until it satisfies grid[i][j] > grid[i-1][j]. While correct, this approach involves potentially many repeated increments, leading to redundant operations if not carefully implemented. Given the constraints, this would be inefficient but feasible due to small m and n.

Optimal

The optimal approach is greedy. For each column, we iterate from the second row to the last. We check if the current element grid[i][j] is less than or equal to the previous element grid[i-1][j]. If it is, we compute the difference needed to make it strictly greater and add that to the total operations. Then, we update the current element to reflect this increment. This ensures each column becomes strictly increasing with the minimum number of operations, and since each column is handled independently, the total operations are simply the sum over all columns.

Approach Time Complexity Space Complexity Notes
Brute Force O(m * n * max_value) O(1) Increment elements repeatedly until column is strictly increasing
Optimal O(m * n) O(1) Greedy update each element if it violates strictly increasing condition

Algorithm Walkthrough

  1. Initialize a variable total_ops to 0. This will accumulate the total number of operations.

  2. Iterate over each column j from 0 to n-1.

  3. For each column, iterate over each row i from 1 to m-1:

  4. Compare grid[i][j] with the previous element grid[i-1][j].

  5. If grid[i][j] is less than or equal to grid[i-1][j], compute the difference needed to make it strictly greater: needed_ops = grid[i-1][j] - grid[i][j] + 1.

  6. Add needed_ops to total_ops.

  7. Update grid[i][j] by adding needed_ops to it, so subsequent rows are compared correctly.

  8. Return total_ops as the final answer.

Why it works: Each column is updated greedily from top to bottom. By ensuring that each element is at least one greater than the previous, the column becomes strictly increasing. Since each update is minimal, the sum of all updates represents the minimum total operations.

Python Solution

from typing import List

class Solution:
    def minimumOperations(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        total_ops = 0
        
        for j in range(n):
            for i in range(1, m):
                if grid[i][j] <= grid[i-1][j]:
                    needed_ops = grid[i-1][j] - grid[i][j] + 1
                    total_ops += needed_ops
                    grid[i][j] += needed_ops
        
        return total_ops

The Python solution follows the algorithm exactly. total_ops accumulates the total increments, and each column is processed independently. The update grid[i][j] += needed_ops ensures that subsequent rows are compared against the correct updated value.

Go Solution

func minimumOperations(grid [][]int) int {
    m, n := len(grid), len(grid[0])
    totalOps := 0
    
    for j := 0; j < n; j++ {
        for i := 1; i < m; i++ {
            if grid[i][j] <= grid[i-1][j] {
                neededOps := grid[i-1][j] - grid[i][j] + 1
                totalOps += neededOps
                grid[i][j] += neededOps
            }
        }
    }
    
    return totalOps
}

The Go solution mirrors the Python version closely. Go requires explicit variable declarations and uses int types. Slice indexing and update logic is identical, and no additional data structures are needed.

Worked Examples

Example 1: grid = [[3,2],[1,3],[3,4],[0,1]]

i Column 0 updates Column 1 updates
1 1 <= 3 → +3 → 4 3 > 2 → 0
2 3 <= 4 → +2 → 5 4 > 3 → 0
3 0 <= 5 → +6 → 6 1 <= 4 → +4 → 5

Total operations = 3 + 2 + 6 + 4 = 15

Example 2: grid = [[3,2,1],[2,1,0],[1,2,3]]

i Column 0 updates Column 1 updates Column 2 updates
1 2 <= 3 → +2 → 4 1 <= 2 → +2 → 3 0 <= 1 → +2 → 2
2 1 <= 4 → +4 → 5 2 <= 3 → +2 → 4 3 > 2 → 0

Total operations = 2+4+2+2+2=12

Complexity Analysis

Measure Complexity Explanation
Time O(m * n) Each element is visited once in a nested loop over rows and columns
Space O(1) Updates are done in-place, no additional data structures needed

The time complexity is efficient due to the small bounds of m and n (up to 50), and the in-place updates ensure minimal space usage.

Test Cases

# Provided examples
assert Solution().minimumOperations([[3,2],[1,3],[3,4],[0,1]]) == 15  # Example 1
assert Solution().minimumOperations([[3,2,1],[2,1,0],[1,2,3]]) == 12  # Example 2

# Edge cases
assert Solution().minimumOperations([[1]]) == 0  # Single element
assert Solution().minimumOperations([[1,2,3]]) == 0  # Single row, already increasing
assert Solution().minimumOperations([[3],[2],[1]]) == 3  # Single column, decreasing
assert Solution().minimumOperations([[0,0],[0,0],[0,0]]) == 9  # All zeros
Test Why
Single element No operations needed, smallest input
Single row Already strictly increasing horizontally doesn't matter; columns fine
Single column decreasing Requires minimal increments to enforce strictly increasing
All zeros Tests multiple columns and cumulative increments

Edge Cases

First, a single-row matrix does not require any operations because there are no consecutive rows to violate the strictly increasing property. Our implementation correctly handles this because the inner loop starts from row 1.

Second, a single-column matrix requires incrementing each row below the first if it is not strictly larger than the previous. This tests that our column-wise approach works even when n=1.

Third, a matrix with all identical values requires progressively larger increments for lower rows to achieve strict increase. Our algorithm handles this by updating each element in-place to satisfy the condition with minimal increments, ensuring correctness even in uniform grids.

Problem Understanding the current value is greater than the one above, resulting in zero operations. This confirms that it does not overcount operations unnecessarily.