LeetCode #3402 — EASY

Minimum Operations to Make Columns Strictly Increasing

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

You are given a 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.

Example 1:

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

Output: 15

Explanation:

To make the 0^th column strictly increasing, we can apply 3 operations on grid[1][0], 2 operations on grid[2][0], and 6 operations on grid[3][0].
To make the 1^st column strictly increasing, we can apply 4 operations on grid[3][1].

Example 2:

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

Output: 12

Explanation:

To make the 0^th column strictly increasing, we can apply 2 operations on grid[1][0], and 4 operations on grid[2][0].
To make the 1^st column strictly increasing, we can apply 2 operations on grid[1][1], and 2 operations on grid[2][1].
To make the 2^nd column strictly increasing, we can apply 2 operations on grid[1][2].

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given a 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[[3,2],[1,3],[3,4],[0,1]]

Example 2

[[3,2,1],[2,1,0],[1,2,3]]

Core Insight

What unlocks the optimal approach

<code>grid[i + 1][j]</code> must be at least equal to <code>grid[i][j] + 1<code>.
Iterate on <code>i</code> in increasing order, and set <code>grid[i + 1][j] = max(grid[i][j]+1, grid[i + 1][j])<code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3402: Minimum Operations to Make Columns Strictly Increasing
class Solution {
    public int minimumOperations(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int ans = 0;
        for (int j = 0; j < n; ++j) {
            int pre = -1;
            for (int i = 0; i < m; ++i) {
                int cur = grid[i][j];
                if (pre < cur) {
                    pre = cur;
                } else {
                    ++pre;
                    ans += pre - cur;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3402: Minimum Operations to Make Columns Strictly Increasing
func minimumOperations(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	for j := 0; j < n; j++ {
		pre := -1
		for i := 0; i < m; i++ {
			cur := grid[i][j]
			if pre < cur {
				pre = cur
			} else {
				pre++
				ans += pre - cur
			}
		}
	}
	return
}

# Accepted solution for LeetCode #3402: Minimum Operations to Make Columns Strictly Increasing
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

// Accepted solution for LeetCode #3402: Minimum Operations to Make Columns Strictly Increasing
fn minimum_operations(grid: Vec<Vec<i32>>) -> i32 {
    let mut ret = 0;
    let (rows, cols) = (grid.len(), grid[0].len());

    for j in 0..cols {
        let mut prev = grid[0][j];
        for i in 1..rows {
            if prev >= grid[i][j] {
                ret += prev - grid[i][j] + 1;
                prev += 1;
            } else {
                prev = grid[i][j];
            }
        }
    }

    ret
}

fn main() {
    let grid = vec![vec![3, 2], vec![1, 3], vec![3, 4], vec![0, 1]];
    let ret = minimum_operations(grid);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let grid = vec![vec![3, 2], vec![1, 3], vec![3, 4], vec![0, 1]];
        let ret = minimum_operations(grid);
        assert_eq!(ret, 15);
    }
    {
        let grid = vec![vec![3, 2, 1], vec![2, 1, 0], vec![1, 2, 3]];
        let ret = minimum_operations(grid);
        assert_eq!(ret, 12);
    }
}

// Accepted solution for LeetCode #3402: Minimum Operations to Make Columns Strictly Increasing
function minimumOperations(grid: number[][]): number {
    const [m, n] = [grid.length, grid[0].length];
    let ans: number = 0;
    for (let j = 0; j < n; ++j) {
        let pre: number = -1;
        for (let i = 0; i < m; ++i) {
            const cur = grid[i][j];
            if (pre < cur) {
                pre = cur;
            } else {
                ++pre;
                ans += pre - cur;
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.