Select Cells in Grid With Maximum Score - Practice Coding Problems

Select Cells in Grid With Maximum Score - Problem

You're given a 2D grid filled with positive integers, and your goal is to become a strategic treasure hunter! 🏆

Your mission: Select cells from the grid to maximize your score (sum of selected values), but there are two crucial rules:

No two cells from the same row - each row can contribute at most one treasure
All selected values must be unique - no duplicates allowed in your collection

Think of it as choosing the best representatives from each row, where each representative must have a unique value that hasn't been chosen before.

Example: In grid [[1,2,3],[4,3,2],[1,1,1]], you could select 3 from row 0 and 4 from row 1 for a score of 7, but you can't select another 3 since values must be unique!

Input & Output

example_1.py — Basic Grid

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

› Output: 8

💡 Note: Select 3 from row 0 (index [0,2]), 4 from row 1 (index [1,0]), and skip row 2 (all values are 1, but 1 hasn't been used yet, so we could pick 1 for a total of 3+4+1=8). Actually, the optimal is to pick 2 from row 0, 4 from row 1, and 1 from row 2 for total 2+4+1=7, or pick 3 from row 0 and 4 from row 1 for 3+4=7. Wait, let me recalculate: pick 3 from row 0, 4 from row 1, 1 from row 2 = 8 total.

example_2.py — Duplicate Values

$ Input: grid = [[7,9,8,6,2],[6,1,1,2,8]]

› Output: 24

💡 Note: From row 0, select 9 (highest unique value). From row 1, select 1 (since 6, 2, 8 are already available in row 0, pick a value not in row 0). Actually, select 9 from row 0 and 1 from row 1 gives 9+1=10. Better: select 7 from row 0 and 8 from row 1 gives 15. Even better: select 9 from row 0 and 6 from row 1 gives 15. Optimal: select 8 from row 0 and 6 from row 1, but wait - we want to maximize, so select 9 from row 0 and 8 from row 1 but 8 appears in both rows. Select 9 from row 0 and 6 from row 1 = 15.

example_3.py — Single Row

$ Input: grid = [[8,7,6]]

› Output: 8

💡 Note: With only one row, select the maximum value which is 8.

Visualization

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Understanding the Visualization

Catalog Artifacts

List all unique artifact values and create a tracking system

Hall by Hall

Process each exhibition hall (row) one by one

Smart Selection

For each hall, try selecting each artifact while tracking which values are already taken

Maximize Value

Keep track of the best possible exhibition value for each combination of selected artifacts

Key Takeaway

🎯 Key Insight: The bitmask DP approach efficiently explores all valid combinations by encoding 'which values have been used' as bits, avoiding redundant subproblem calculations and guaranteeing optimal results.

Time & Space Complexity

Time Complexity

⏱️

O((max_cols)^rows)

For each row, we try all possible column choices, leading to exponential combinations

✓ Linear Growth

Space Complexity

O(rows)

Recursion stack depth and space to track current selection

✓ Linear Space

Constraints

1 ≤ grid.length, grid[i].length ≤ 10
1 ≤ grid[i][j] ≤ 100
All values in the grid are positive integers
You must select at least one cell to get a non-zero score

Asked in

G Google 45 f Meta 32 a Amazon 28 ⊞ Microsoft 22

The optimal solution uses Dynamic Programming with Bitmask to efficiently track which values have been selected while processing each row. This approach has time complexity O(rows × 2^unique_values × cols) and guarantees the maximum score by exploring all valid combinations without redundant calculations.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O((max_cols)^rows)	O(rows)	Generate all possible selections and find the maximum valid score
Dynamic Programming with Bitmask	O(rows × 2^unique_values × cols)	O(rows × 2^unique_values)	Use DP with bitmask to track used values and optimize overlapping subproblems

Brute Force (Try All Combinations) — Algorithm Steps

Generate all possible combinations of selecting 0 or 1 cell from each row
For each combination, check if all selected values are unique
If valid, calculate the sum and track the maximum
Return the maximum score found

Visualization

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Step-by-Step Walkthrough

Start at Row 0

Consider all cells in first row plus option to skip

Branch for Each Choice

For each valid choice, create a new branch

Check Uniqueness

Verify selected values don't conflict

Calculate Score

Sum up values when reaching leaf nodes

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

int backtrack(int row, int** grid, int* gridColSize, int gridSize, bool* used, int maxVal, int score) {
    if (row == gridSize) {
        return score;
    }
    
    // Skip this row
    int maxScore = backtrack(row + 1, grid, gridColSize, gridSize, used, maxVal, score);
    
    // Pick a cell from this row
    for (int col = 0; col < gridColSize[row]; col++) {
        int val = grid[row][col];
        if (val <= maxVal && !used[val]) {
            used[val] = true;
            int newScore = backtrack(row + 1, grid, gridColSize, gridSize, used, maxVal, score + val);
            if (newScore > maxScore) {
                maxScore = newScore;
            }
            used[val] = false;
        }
    }
    
    return maxScore;
}

int main() {
    // Input parsing and function call
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O((max_cols)^rows)

For each row, we try all possible column choices, leading to exponential combinations

✓ Linear Growth

Space Complexity

O(rows)

Recursion stack depth and space to track current selection

✓ Linear Space

Constraints

1 ≤ grid.length, grid[i].length ≤ 10
1 ≤ grid[i][j] ≤ 100
All values in the grid are positive integers
You must select at least one cell to get a non-zero score

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Smart Actions

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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