What is the time complexity of Maximum Score From Grid Operations?

The optimal solution has a time complexity of O(n^3).

What data structures are used to solve Maximum Score From Grid Operations?

This problem uses Array, Dynamic Programming, Matrix, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Score From Grid Operations?

Insight 1: The problem can be solved using Dynamic Programming (DP) to explore different combinations of operations and keep track of the maximum achievable score. Insight 2: The state of the DP can be defined based on which column is being considered and the height up to which the previous columns were colored black.

Which companies ask LeetCode 3225. Maximum Score From Grid Operations in interviews?

This problem is frequently asked at Hudson River Trading.

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Maximum Score From Grid Operations - LeetCode 3225 Solution

Q: What is the space complexity of Maximum Score From Grid Operations?

The space complexity is O(n^2).

Maximum Score From Grid Operations - Complete Solution Guide

Maximum Score From Grid Operations is LeetCode problem 3225, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 2D matrix grid of size n x n . Initially, all cells of the grid are colored white. In one operation, you can select any cell of indices (i, j) , and color black all the cells of the j th column starting from the top row down to the i th row. The grid score is the sum of all grid[i][j] such that cell (i, j) is white and it has a horizontally adjacent black cell. Return the maximum score that can be achieved after some number of operations. Example 1: Input: grid = [[0,0,0,0,0],[0,

Detailed Explanation

The problem presents a 2D grid of size n x n, initially all white. The goal is to maximize the 'score' of the grid after performing some number of operations. Each operation involves selecting a cell (i, j) and coloring black all cells in the j-th column from the top row down to the i-th row. The score is defined as the sum of values of white cells that have at least one horizontally adjacent black cell. The challenge is to find the optimal sequence of operations that yields the highest possible score.

Solution Approach

The solution utilizes a dynamic programming approach to determine the optimal coloring strategy for each column. A 2D DP table `dp[i][j]` is used to store the maximum score achievable up to the current column `c` given that the previous column `c-1` was colored black up to row `i` and the current column `c` is colored black up to row `j`. The algorithm iterates through each column, computing the `dp` table based on the maximum scores achievable in the previous columns. Prefix sums are employed to calculate the column sums efficiently. The prefix max of the dp table, prefix max of the dp table with the current column values added, and suffix max of the dp table are used to determine the best heights for the current column given the previous column height.

Step-by-Step Algorithm

Step 1: Calculate the prefix sum for each column. `P[c][r]` stores the sum of `grid[i][c]` for `i = 0` to `r-1`.
Step 2: Initialize a DP table `dp[i][j]` where `dp[i][j]` represents the maximum score achievable considering the columns up to the current one, given that the previous column was colored black up to row `i` and current up to row `j`.
Step 3: Base case: Calculate the initial DP values for the first column (c=0). The `dp[h0][h1]` value represents the added score for the first column alone given h0 for the before column and h1 for current column. If the row upto which coloring happened in the first column is greater than the row upto which hypothetically coloring happened before, then that value is added to the max score.
Step 4: Iterate through the remaining columns (c = 1 to n-2) to populate the `dp` table. For each column `c` and height `hc`, calculate the maximum score achievable considering two cases:
Step 5: Case 1: `hc_minus_1 <= hc_plus_1`. The score can be obtained using `prefix_max_A` array.
Step 6: Case 2: `hc_minus_1 > hc_plus_1`. The score can be obtained using `suffix_max_B` and `suffix_max_A_upto_hc` arrays.
Step 7: Use the accumulated DP values to determine the maximum possible total score considering the last column n-1.
Step 8: Return the maximum score found.

Key Insights

Insight 1: The problem can be solved using Dynamic Programming (DP) to explore different combinations of operations and keep track of the maximum achievable score.
Insight 2: The state of the DP can be defined based on which column is being considered and the height up to which the previous columns were colored black.
Insight 3: Prefix sums are used to efficiently calculate the sum of values in a column up to a certain row, which avoids redundant calculations within the DP.
Insight 4: The key is to efficiently compute the maximum possible score for each column considering all possible heights up to which the previous columns might be blacked out.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming, Matrix, Prefix Sum.

Companies

Asked at: Hudson River Trading.

Maximum Score From Grid Operations | LeetCode 3225

Hard

Array Dynamic Programming Matrix Prefix Sum

Time: O(n^3)Space: O(n^2)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long maximumScore(int[][] grid) {
        int n = grid.length;
        if (n == 1) {
            return 0;
        }

        int[][] P = new int[n][n + 1];
        for (int c = 0; c < n; c++) {
            for (int r = 0; r < n; r++) {
                P[c][r + 1] = P[c][r] + grid[r][c];
            }
        }

        long[][] dp = new long[n + 1][n + 1];

        // Base case: c = 0
        for (int h0 = 0; h0 <= n; h0++) {
            for (int h1 = 0; h1 <= n; h1++) {
                if (h1 > h0) {
                    dp[h0][h1] = P[0][h1] - P[0][h0];
                }
            }
        }

        // DP for columns c = 1 to n - 2
        for (int c = 1; c < n - 1; c++) {
            long[][] next_dp = new long[n + 1][n + 1];
            for (int hc = 0; hc <= n; hc++) {
                long[] A = new long[n + 1];
                for (int i = 0; i <= n; i++) {
                    A[i] = dp[i][hc];
                }

                long[] prefix_max_A = new long[n + 1];
                Arrays.fill(prefix_max_A, -1);
                prefix_max_A[0] = A[0];
                for (int i = 1; i <= n; i++) {
                    prefix_max_A[i] = Math.max(prefix_max_A[i - 1], A[i]);
                }

                long[] B = new long[n + 1];
                for (int i = 0; i <= n; i++) {
                    B[i] = A[i] + P[c][i];
                }

                long[] suffix_max_B = new long[n + 2];
                Arrays.fill(suffix_max_B, -1);
                for (int i = n; i >= 0; i--) {
                    suffix_max_B[i] = Math.max(suffix_max_B[i + 1], B[i]);
                }

                long[] suffix_max_A_upto_hc = new long[hc + 2];
                Arrays.fill(suffix_max_A_upto_hc, -1);
                if (hc >= 0) {
                    suffix_max_A_upto_hc[hc] = A[hc];
                    for (int i = hc - 1; i >= 0; i--) {
                        suffix_max_A_upto_hc[i] = Math.max(suffix_max_A_upto_hc[i + 1], A[i]);
                    }
                }

                for (int hc_plus_1 = 0; hc_plus_1 <= n; hc_plus_1++) {
                    // Case 1: hc_minus_1 <= hc_plus_1
                    long score_c_1 = (hc_plus_1 > hc) ? (long)P[c][hc_plus_1] - P[c][hc] : 0;
                    long max_dp_1 = prefix_max_A[hc_plus_1];
                    long res1 = score_c_1 + max_dp_1;

                    // Case 2: hc_minus_1 > hc_plus_1
                    long res2 = -1;
                    if (hc_plus_1 < n) {
                        if (hc_plus_1 >= hc) {
                            res2 = (long)-P[c][hc] + suffix_max_B[hc_plus_1 + 1];
                        } else { // hc_plus_1 < hc
                            long term_k_gt_hc = (long)-P[c][hc] + suffix_max_B[hc + 1];
                            long term_k_le_hc = -1;
                            if (hc_plus_1 + 1 <= hc + 1) {
                                term_k_le_hc = suffix_max_A_upto_hc[hc_plus_1 + 1];
                            }
                            res2 = Math.max(term_k_gt_hc, term_k_le_hc);
                        }
                    }

                    next_dp[hc][hc_plus_1] = Math.max(res1, res2);
                }
            }
            dp = next_dp;
        }

        long max_score = 0;
        for (int hn_minus_2 = 0; hn_minus_2 <= n; hn_minus_2++) {
            for (int hn_minus_1 = 0; hn_minus_1 <= n; hn_minus_1++) {
                long final_score_contrib = (hn_minus_2 > hn_minus_1) ? (long)P[n - 1][hn_minus_2] - P[n - 1][hn_minus_1] : 0;
                long total_score = dp[hn_minus_2][hn_minus_1] + final_score_contrib;
                max_score = Math.max(max_score, total_score);
            }
        }

        return max_score;
    }
}

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Problem Info

Description

Problem 3225: Maximum Score From Grid Operations - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix Prefix Sum

Companies

Hudson River Trading

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3225

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Calculate the prefix sum for each column. `P[c][r]` stores the sum of `grid[i][c]` for `i = 0` to `r-1`.
2
Step 2: Initialize a DP table `dp[i][j]` where `dp[i][j]` represents the maximum score achievable considering the columns up to the current one, given that the previous column was colored black up to row `i` and current up to row `j`.
3
Step 3: Base case: Calculate the initial DP values for the first column (c=0). The `dp[h0][h1]` value represents the added score for the first column alone given h0 for the before column and h1 for current column. If the row upto which coloring happened in the first column is greater than the row upto which hypothetically coloring happened before, then that value is added to the max score.
4
Step 4: Iterate through the remaining columns (c = 1 to n-2) to populate the `dp` table. For each column `c` and height `hc`, calculate the maximum score achievable considering two cases:
5
Step 5: Case 1: `hc_minus_1 <= hc_plus_1`. The score can be obtained using `prefix_max_A` array.
6
Step 6: Case 2: `hc_minus_1 > hc_plus_1`. The score can be obtained using `suffix_max_B` and `suffix_max_A_upto_hc` arrays.
7
Step 7: Use the accumulated DP values to determine the maximum possible total score considering the last column n-1.
8
Step 8: Return the maximum score found.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^3)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is O(n^3) because of the nested loops in the dynamic programming step where we iterate over each column (O(n)), and for each column, we iterate over all possible heights for the current column (O(n)) and the previous column (O(n)), in addition to building the prefix/suffix maxes. The space complexity is O(n^2) because we store the DP table `dp` of size (n+1) x (n+1), and also create temporary arrays such as the prefix/suffix maxes, with size O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive solution with memoization can also be used. This approach involves defining a recursive function that explores different coloring options and stores the results in a memoization table to avoid redundant computations.

Time: O(n^3)Space: O(n^2)

Trade-offs: Recursion with memoization can be more intuitive for some, but might have slightly higher overhead due to function call stack. The DP approach might be preferred for iterative implementation.

✨

Key Insights

1Insight 1: The problem can be solved using Dynamic Programming (DP) to explore different combinations of operations and keep track of the maximum achievable score.
2Insight 2: The state of the DP can be defined based on which column is being considered and the height up to which the previous columns were colored black.
3Insight 3: Prefix sums are used to efficiently calculate the sum of values in a column up to a certain row, which avoids redundant calculations within the DP.
4Insight 4: The key is to efficiently compute the maximum possible score for each column considering all possible heights up to which the previous columns might be blacked out.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: When the grid size is 1 (n=1). In this case, no operations are possible, so the score is 0.
Edge case 2: All grid values are 0. The score will always be 0 as there are no values to contribute to the score. This is handled implicitly by the algorithm.
Edge case 3: Large grid values to avoid overflow. Using long long or similar data types helps prevent integer overflow issues when summing the values.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all possible heights for the colored regions. The algorithm needs to explore all possible combinations of coloring to find the optimal one.

✗Common mistake 2: Incorrectly calculating the score due to forgetting the horizontally adjacent black cell requirement.

✗Common mistake 3: Integer overflow. The grid values can be large, so using `long long` for accumulations is important.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Optimizing resource allocation in a grid-like environment where certain actions affect neighboring resources.

🏦Real-world scenario 2: Game development. Finding the optimal strategy in a game where actions influence the state of neighboring cells, and the score is based on adjacent relationships.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Score From Grid Operations"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?