What is the time complexity of Number of Paths with Max Score?

The optimal solution has a time complexity of O(n^2).

What is the space complexity of Number of Paths with Max Score?

The space complexity is O(n^2).

What data structures are used to solve Number of Paths with Max Score?

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

What are the key insights for solving Number of Paths with Max Score?

Insight 1: Dynamic Programming is essential because we need to explore multiple paths and memoize intermediate results (maximum score and number of paths) to avoid redundant calculations. Insight 2: We iterate backwards from the destination ('S') to the source ('E') using dynamic programming. This simplifies the logic for calculating maximum score and number of paths since we build up the solution from a known base case.

Which companies ask LeetCode 1301. Number of Paths with Max Score in interviews?

This problem is frequently asked at Samsung.

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Number of Paths with Max Score - LeetCode 1301 Solution

Number of Paths with Max Score - Complete Solution Guide

Number of Paths with Max Score is LeetCode problem 1301, 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 square board of characters. You can move on the board starting at the bottom right square marked with the character 'S' . You need to reach the top left square marked with the character 'E' . The rest of the squares are labeled either with a numeric character 1, 2, ..., 9 or with an obstacle 'X' . In one move you can go up, left or up-left (diagonally) only if there is no obstacle there. Return a list of two integers: the first integer is the maximum sum of numeric characters you

Detailed Explanation

The problem asks us to find the maximum possible score and the number of paths to achieve that score on a square board. We start from the bottom-right cell ('S') and need to reach the top-left cell ('E'). We can only move up, left, or diagonally up-left. The board contains numbers (1-9), obstacles ('X'), the start ('S'), and the end ('E'). The score is the sum of the numbers encountered along the path, and we need to return the maximum score and the number of paths that achieve this maximum score modulo 10^9 + 7. If no path exists, we return [0, 0].

Solution Approach

The solution uses dynamic programming to find the maximum score and the number of paths. A 3D DP array `dp[i][j][2]` is used to store the maximum score and the number of paths to reach cell (i, j). The algorithm iterates backwards from the destination cell ('S') to the source cell ('E'). For each cell (i, j), it considers the possible moves from the adjacent cells (i+1, j), (i, j+1), and (i+1, j+1). It calculates the maximum score and the number of paths based on the scores and paths from these adjacent cells. The result is stored in `dp[i][j][0]` and `dp[i][j][1]`, respectively.

Step-by-Step Algorithm

Step 1: Initialize the DP array `dp[n+1][n+1][2]` with -1 for the score and 0 for the number of paths. Set `dp[n-1][n-1][0] = 0` and `dp[n-1][n-1][1] = 1` to represent the starting position with a score of 0 and one path.
Step 2: Iterate through the board backwards, from row `n-1` to 0 and column `n-1` to 0.
Step 3: If the current cell `board[i][j]` is an obstacle ('X'), skip to the next iteration.
Step 4: For each cell (i, j), consider the possible moves: (i+1, j), (i, j+1), and (i+1, j+1). These are the cells we could have come FROM to reach (i, j).
Step 5: For each valid move, check if the score of the adjacent cell is not -1 (meaning it is reachable).
Step 6: If the score of the adjacent cell is greater than the current maximum score for cell (i, j), update the maximum score and the number of paths for cell (i, j) with the adjacent cell's score and number of paths. If the score of the adjacent cell is equal to the current maximum score, add the adjacent cell's number of paths to the current number of paths (modulo 10^9 + 7).
Step 7: If a path to cell (i, j) is found (maximum score is not -1), add the value of the current cell (if it's a digit) to the maximum score and store the updated score and number of paths in `dp[i][j][0]` and `dp[i][j][1]`.
Step 8: After the iteration is complete, the result is stored in `dp[0][0][0]` (maximum score) and `dp[0][0][1]` (number of paths).
Step 9: If the maximum score is -1 (meaning no path exists), return [0, 0]. Otherwise, return the maximum score and the number of paths.

Key Insights

Insight 1: Dynamic Programming is essential because we need to explore multiple paths and memoize intermediate results (maximum score and number of paths) to avoid redundant calculations.
Insight 2: We iterate backwards from the destination ('S') to the source ('E') using dynamic programming. This simplifies the logic for calculating maximum score and number of paths since we build up the solution from a known base case.
Insight 3: Handle obstacles ('X') by not processing those cells in the DP iteration. If a cell has an obstacle, it cannot contribute to any path.
Insight 4: Using a 3D DP array dp[i][j][2] allows storing both the maximum score and number of paths for each cell (i, j).

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Samsung.

Number of Paths with Max Score | LeetCode 1301

Hard

Array Dynamic Programming Matrix

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

Solution Code

Available in:

import java.util.List;

class Solution {
    public int[] pathsWithMaxScore(List<String> board) {
        int n = board.size();
        int mod = (int)1e9 + 7;

        int[][][] dp = new int[n + 1][n + 1][2];
        for (int i = 0; i < n + 1; i++) {
            for (int j = 0; j < n + 1; j++) {
                dp[i][j][0] = -1;
                dp[i][j][1] = 0;
            }
        }
        dp[n - 1][n - 1][0] = 0;
        dp[n - 1][n - 1][1] = 1;

        for (int i = n - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                if (i == n - 1 && j == n - 1) {
                    continue;
                }

                if (board.get(i).charAt(j) == 'X') {
                    continue;
                }

                int max_score = -1;
                int num_paths = 0;

                int[][] directions = {{i + 1, j}, {i, j + 1}, {i + 1, j + 1}};
                for (int[] dir : directions) {
                    int r = dir[0];
                    int c = dir[1];

                    int prev_score = dp[r][c][0];
                    int prev_paths = dp[r][c][1];
                    
                    if (prev_score == -1) {
                        continue;
                    }
                        
                    if (prev_score > max_score) {
                        max_score = prev_score;
                        num_paths = prev_paths;
                    } else if (prev_score == max_score) {
                        num_paths = (num_paths + prev_paths) % mod;
                    }
                }
                
                if (max_score != -1) {
                    int val = 0;
                    if (Character.isDigit(board.get(i).charAt(j))) {
                        val = Character.getNumericValue(board.get(i).charAt(j));
                    }
                    
                    dp[i][j][0] = max_score + val;
                    dp[i][j][1] = num_paths;
                }
            }
        }

        int result_score = dp[0][0][0];
        int result_paths = dp[0][0][1];

        if (result_score == -1) {
            return new int[]{0, 0};
        } else {
            return new int[]{result_score, result_paths};
        }
    }
}

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Problem Info

Description

Problem 1301: Number of Paths with Max Score - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

Samsung

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1301

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

9 steps

1
Step 1: Initialize the DP array `dp[n+1][n+1][2]` with -1 for the score and 0 for the number of paths. Set `dp[n-1][n-1][0] = 0` and `dp[n-1][n-1][1] = 1` to represent the starting position with a score of 0 and one path.
2
Step 2: Iterate through the board backwards, from row `n-1` to 0 and column `n-1` to 0.
3
Step 3: If the current cell `board[i][j]` is an obstacle ('X'), skip to the next iteration.
4
Step 4: For each cell (i, j), consider the possible moves: (i+1, j), (i, j+1), and (i+1, j+1). These are the cells we could have come FROM to reach (i, j).
5
Step 5: For each valid move, check if the score of the adjacent cell is not -1 (meaning it is reachable).
6
Step 6: If the score of the adjacent cell is greater than the current maximum score for cell (i, j), update the maximum score and the number of paths for cell (i, j) with the adjacent cell's score and number of paths. If the score of the adjacent cell is equal to the current maximum score, add the adjacent cell's number of paths to the current number of paths (modulo 10^9 + 7).
7
Step 7: If a path to cell (i, j) is found (maximum score is not -1), add the value of the current cell (if it's a digit) to the maximum score and store the updated score and number of paths in `dp[i][j][0]` and `dp[i][j][1]`.
8
Step 8: After the iteration is complete, the result is stored in `dp[0][0][0]` (maximum score) and `dp[0][0][1]` (number of paths).
9
Step 9: If the maximum score is -1 (meaning no path exists), return [0, 0]. Otherwise, return the maximum score and the number of paths.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The nested loops iterate through each cell of the n x n board. The DP array stores results for each cell, leading to n^2 space complexity. The operations within the loops are constant time, resulting in O(n^2) time complexity overall.

📖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

We could also implement this using recursion along with memoization. The recursive function would explore possible paths and memoize the result (score, number of paths) for each cell to avoid redundant calculations.

Time: O(n^2)Space: O(n^2) for the memoization table and O(n) for recursion stack in the worst case.

Trade-offs: Recursion might be easier to understand initially, but the iterative DP approach is generally more efficient due to lower overhead from function calls and potentially better cache locality. Iterative approach also avoid stack overflow issues for larger board sizes.

✨

Key Insights

1Insight 1: Dynamic Programming is essential because we need to explore multiple paths and memoize intermediate results (maximum score and number of paths) to avoid redundant calculations.
2Insight 2: We iterate backwards from the destination ('S') to the source ('E') using dynamic programming. This simplifies the logic for calculating maximum score and number of paths since we build up the solution from a known base case.
3Insight 3: Handle obstacles ('X') by not processing those cells in the DP iteration. If a cell has an obstacle, it cannot contribute to any path.
4Insight 4: Using a 3D DP array dp[i][j][2] allows storing both the maximum score and number of paths for each cell (i, j).

📚

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: No path exists between 'S' and 'E'. The solution correctly returns [0, 0] when the maximum score at the starting cell (0, 0) remains -1.
Edge case 2: The board contains only obstacles ('X') except for 'S' and 'E', effectively blocking all paths. The solution handles this by returning [0, 0].
Edge case 3: The board consists of only 'E' and 'S'. The solution correctly calculates the score as 0 and paths as 1, and returns [0, 1].
Edge case 4: A large number of paths exist. The solution uses modulo arithmetic (10^9 + 7) to prevent integer overflow when calculating the number of paths.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the modulo operation correctly when calculating the number of paths, which can lead to integer overflow.

✗Common mistake 2: Initializing the DP array incorrectly, particularly the initial score and path count for the starting cell ('S').

✗Common mistake 3: Forgetting to check for obstacles ('X') and trying to process them, which will lead to incorrect results.

✗Common mistake 4: Not properly handling the case where no path exists between 'S' and 'E', resulting in incorrect output.

✅

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: Pathfinding algorithms used in robotics to navigate robots in a complex environment with obstacles. The board could represent the environment, and the algorithm would find the optimal path to a destination while avoiding collisions.

🏦Real-world scenario 2: Game development, where the board could represent a game map, and the algorithm would find the optimal path for a character to reach a goal while maximizing score or minimizing cost. The 'X' could represent impassable terrain.

🎮Real-world scenario 3: Network routing, where the board could represent a network topology, and the algorithm would find the optimal path for data packets to travel from a source to a destination while maximizing bandwidth or minimizing latency. 'X' could represent a broken link.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

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: July 12, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

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 "Number of Paths with Max Score"?

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?