LeetCode #3148 — MEDIUM

Maximum Difference Score in a Grid

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an m x n matrix grid consisting of positive integers. You can move from a cell in the matrix to any other cell that is either to the bottom or to the right (not necessarily adjacent). The score of a move from a cell with the value c1 to a cell with the value c2 is c2 - c1.

You can start at any cell, and you have to make at least one move.

Return the maximum total score you can achieve.

Example 1:

Input: grid = [[9,5,7,3],[8,9,6,1],[6,7,14,3],[2,5,3,1]]

Output: 9

Explanation: We start at the cell (0, 1), and we perform the following moves:
- Move from the cell (0, 1) to (2, 1) with a score of 7 - 5 = 2.
- Move from the cell (2, 1) to (2, 2) with a score of 14 - 7 = 7.
The total score is 2 + 7 = 9.

Example 2:

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

Output: -1

Explanation: We start at the cell (0, 0), and we perform one move: (0, 0) to (0, 1). The score is 3 - 4 = -1.

Constraints:

m == grid.length
n == grid[i].length
2 <= m, n <= 1000
4 <= m * n <= 10⁵
1 <= grid[i][j] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an m x n matrix grid consisting of positive integers. You can move from a cell in the matrix to any other cell that is either to the bottom or to the right (not necessarily adjacent). The score of a move from a cell with the value c1 to a cell with the value c2 is c2 - c1. You can start at any cell, and you have to make at least one move. Return the maximum total score you can achieve.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[[9,5,7,3],[8,9,6,1],[6,7,14,3],[2,5,3,1]]

Example 2

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

Core Insight

What unlocks the optimal approach

Any path from a cell <code>(x1, y1)</code> to another cell <code>(x2, y2)</code> will always have a score of <code>grid[x2][y2] - grid[x1][y1]</code>.
Let’s say we fix the starting cell <code>(x1, y1)</code>, how to the find a cell <code>(x2, y2)</code> such that the value <code>grid[x2][y2] - grid[x1][y1]</code> is the maximum possible?

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 #3148: Maximum Difference Score in a Grid
class Solution {
    public int maxScore(List<List<Integer>> grid) {
        int m = grid.size(), n = grid.get(0).size();
        final int inf = 1 << 30;
        int ans = -inf;
        int[][] f = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int mi = inf;
                if (i > 0) {
                    mi = Math.min(mi, f[i - 1][j]);
                }
                if (j > 0) {
                    mi = Math.min(mi, f[i][j - 1]);
                }
                ans = Math.max(ans, grid.get(i).get(j) - mi);
                f[i][j] = Math.min(grid.get(i).get(j), mi);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3148: Maximum Difference Score in a Grid
func maxScore(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, n)
	}
	const inf int = 1 << 30
	ans := -inf
	for i, row := range grid {
		for j, x := range row {
			mi := inf
			if i > 0 {
				mi = min(mi, f[i-1][j])
			}
			if j > 0 {
				mi = min(mi, f[i][j-1])
			}
			ans = max(ans, x-mi)
			f[i][j] = min(x, mi)
		}
	}
	return ans
}

# Accepted solution for LeetCode #3148: Maximum Difference Score in a Grid
class Solution:
    def maxScore(self, grid: List[List[int]]) -> int:
        f = [[0] * len(grid[0]) for _ in range(len(grid))]
        ans = -inf
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                mi = inf
                if i:
                    mi = min(mi, f[i - 1][j])
                if j:
                    mi = min(mi, f[i][j - 1])
                ans = max(ans, x - mi)
                f[i][j] = min(x, mi)
        return ans

// Accepted solution for LeetCode #3148: Maximum Difference Score in a Grid
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3148: Maximum Difference Score in a Grid
// class Solution {
//     public int maxScore(List<List<Integer>> grid) {
//         int m = grid.size(), n = grid.get(0).size();
//         final int inf = 1 << 30;
//         int ans = -inf;
//         int[][] f = new int[m][n];
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 int mi = inf;
//                 if (i > 0) {
//                     mi = Math.min(mi, f[i - 1][j]);
//                 }
//                 if (j > 0) {
//                     mi = Math.min(mi, f[i][j - 1]);
//                 }
//                 ans = Math.max(ans, grid.get(i).get(j) - mi);
//                 f[i][j] = Math.min(grid.get(i).get(j), mi);
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3148: Maximum Difference Score in a Grid
function maxScore(grid: number[][]): number {
    const [m, n] = [grid.length, grid[0].length];
    const f: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
    let ans = -Infinity;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            let mi = Infinity;
            if (i) {
                mi = Math.min(mi, f[i - 1][j]);
            }
            if (j) {
                mi = Math.min(mi, f[i][j - 1]);
            }
            ans = Math.max(ans, grid[i][j] - mi);
            f[i][j] = Math.min(mi, grid[i][j]);
        }
    }
    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(m × n)

Space

O(m × n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.