LeetCode #3276 — HARD

Select Cells in Grid With Maximum Score

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given a 2D matrix grid consisting of positive integers.

You have to select one or more cells from the matrix such that the following conditions are satisfied:

No two selected cells are in the same row of the matrix.
The values in the set of selected cells are unique.

Your score will be the sum of the values of the selected cells.

Return the maximum score you can achieve.

Example 1:

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

Output: 8

Explanation:

We can select the cells with values 1, 3, and 4 that are colored above.

Example 2:

Input: grid = [[8,7,6],[8,3,2]]

Output: 15

Explanation:

We can select the cells with values 7 and 8 that are colored above.

Constraints:

1 <= grid.length, grid[i].length <= 10
1 <= grid[i][j] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given a 2D matrix grid consisting of positive integers. You have to select one or more cells from the matrix such that the following conditions are satisfied: No two selected cells are in the same row of the matrix. The values in the set of selected cells are unique. Your score will be the sum of the values of the selected cells. Return the maximum 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 · Bit Manipulation

Example 1

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

Example 2

[[8,7,6],[8,3,2]]

Step 02

Core Insight

What unlocks the optimal approach

Sort all the cells in the grid by their values and keep track of their original positions.
Try dynamic programming with the following states: the current cell that we might select and a bitmask representing all the rows from which we have already selected a cell so far.

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

Largest constraint values

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 #3276: Select Cells in Grid With Maximum Score
class Solution {
    public int maxScore(List<List<Integer>> grid) {
        int m = grid.size();
        int mx = 0;
        boolean[][] g = new boolean[101][m + 1];
        for (int i = 0; i < m; ++i) {
            for (int x : grid.get(i)) {
                g[x][i] = true;
                mx = Math.max(mx, x);
            }
        }
        int[][] f = new int[mx + 1][1 << m];
        for (int i = 1; i <= mx; ++i) {
            for (int j = 0; j < 1 << m; ++j) {
                f[i][j] = f[i - 1][j];
                for (int k = 0; k < m; ++k) {
                    if (g[i][k] && (j >> k & 1) == 1) {
                        f[i][j] = Math.max(f[i][j], f[i - 1][j ^ 1 << k] + i);
                    }
                }
            }
        }
        return f[mx][(1 << m) - 1];
    }
}

// Accepted solution for LeetCode #3276: Select Cells in Grid With Maximum Score
func maxScore(grid [][]int) int {
	m := len(grid)
	mx := 0
	g := [101][11]bool{}
	for i, row := range grid {
		for _, x := range row {
			g[x][i] = true
			mx = max(mx, x)
		}
	}
	f := make([][]int, mx+1)
	for i := range f {
		f[i] = make([]int, 1<<m)
	}
	for i := 1; i <= mx; i++ {
		for j := 0; j < 1<<m; j++ {
			f[i][j] = f[i-1][j]
			for k := 0; k < m; k++ {
				if g[i][k] && (j>>k&1) == 1 {
					f[i][j] = max(f[i][j], f[i-1][j^1<<k]+i)
				}
			}
		}
	}
	return f[mx][1<<m-1]
}

# Accepted solution for LeetCode #3276: Select Cells in Grid With Maximum Score
class Solution:
    def maxScore(self, grid: List[List[int]]) -> int:
        g = defaultdict(set)
        mx = 0
        for i, row in enumerate(grid):
            for x in row:
                g[x].add(i)
                mx = max(mx, x)
        m = len(grid)
        f = [[0] * (1 << m) for _ in range(mx + 1)]
        for i in range(1, mx + 1):
            for j in range(1 << m):
                f[i][j] = f[i - 1][j]
                for k in g[i]:
                    if j >> k & 1:
                        f[i][j] = max(f[i][j], f[i - 1][j ^ 1 << k] + i)
        return f[-1][-1]

// Accepted solution for LeetCode #3276: Select Cells in Grid With Maximum Score
// 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 #3276: Select Cells in Grid With Maximum Score
// class Solution {
//     public int maxScore(List<List<Integer>> grid) {
//         int m = grid.size();
//         int mx = 0;
//         boolean[][] g = new boolean[101][m + 1];
//         for (int i = 0; i < m; ++i) {
//             for (int x : grid.get(i)) {
//                 g[x][i] = true;
//                 mx = Math.max(mx, x);
//             }
//         }
//         int[][] f = new int[mx + 1][1 << m];
//         for (int i = 1; i <= mx; ++i) {
//             for (int j = 0; j < 1 << m; ++j) {
//                 f[i][j] = f[i - 1][j];
//                 for (int k = 0; k < m; ++k) {
//                     if (g[i][k] && (j >> k & 1) == 1) {
//                         f[i][j] = Math.max(f[i][j], f[i - 1][j ^ 1 << k] + i);
//                     }
//                 }
//             }
//         }
//         return f[mx][(1 << m) - 1];
//     }
// }

// Accepted solution for LeetCode #3276: Select Cells in Grid With Maximum Score
function maxScore(grid: number[][]): number {
    const m = grid.length;
    let mx = 0;
    const g: boolean[][] = Array.from({ length: 101 }, () => Array(m + 1).fill(false));
    for (let i = 0; i < m; ++i) {
        for (const x of grid[i]) {
            g[x][i] = true;
            mx = Math.max(mx, x);
        }
    }
    const f: number[][] = Array.from({ length: mx + 1 }, () => Array(1 << m).fill(0));
    for (let i = 1; i <= mx; ++i) {
        for (let j = 0; j < 1 << m; ++j) {
            f[i][j] = f[i - 1][j];
            for (let k = 0; k < m; ++k) {
                if (g[i][k] && ((j >> k) & 1) === 1) {
                    f[i][j] = Math.max(f[i][j], f[i - 1][j ^ (1 << k)] + i);
                }
            }
        }
    }
    return f[mx][(1 << m) - 1];
}

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 × 2^m × \textitmx)

Space

O(\textitmx × 2^m)

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.