1301. Number of Paths with Max Score LeetCode Solution

In this guide, you will get 1301. Number of Paths with Max Score LeetCode Solution with the best time and space complexity. The solution to Number of Paths with Max Score problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Number of Paths with Max Score solution in C++
4. Number of Paths with Max Score solution in Java
5. Number of Paths with Max Score solution in Python
6. Additional Resources

Problem Statement of Number of Paths with Max Score

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 can collect, and the second is the number of such paths that you can take to get that maximum sum, taken modulo 10^9 + 7.
In case there is no path, return [0, 0].

Example 1:
Input: board = [“E23″,”2X2″,”12S”]
Output: [7,1]
Example 2:
Input: board = [“E12″,”1X1″,”21S”]
Output: [4,2]
Example 3:
Input: board = [“E11″,”XXX”,”11S”]
Output: [0,0]

Constraints:

2 <= board.length == board[i].length <= 100

Complexity Analysis

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

1301. Number of Paths with Max Score LeetCode Solution in C++

class Solution {
 public:
  vector<int> pathsWithMaxScore(vector<string>& board) {
    constexpr int kMod = 1'000'000'007;
    constexpr int dirs[3][2] = {{0, 1}, {1, 0}, {1, 1}};
    const int n = board.size();
    // dp[i][j] := the maximum sum from (n - 1, n - 1) to (i, j)
    vector<vector<int>> dp(n + 1, vector<int>(n + 1, -1));
    // count[i][j] := the number of paths to get dp[i][j] from (n - 1, n - 1) to
    // (i, j)
    vector<vector<int>> count(n + 1, vector<int>(n + 1));

    dp[0][0] = 0;
    dp[n - 1][n - 1] = 0;
    count[n - 1][n - 1] = 1;

    for (int i = n - 1; i >= 0; --i)
      for (int j = n - 1; j >= 0; --j) {
        if (board[i][j] == 'S' || board[i][j] == 'X')
          continue;
        for (const auto& [dx, dy] : dirs) {
          const int x = i + dx;
          const int y = j + dy;
          if (dp[i][j] < dp[x][y]) {
            dp[i][j] = dp[x][y];
            count[i][j] = count[x][y];
          } else if (dp[i][j] == dp[x][y]) {
            count[i][j] += count[x][y];
            count[i][j] %= kMod;
          }
        }
        // If there's path(s) from 'S' to (i, j) and the cell is not 'E'.
        if (dp[i][j] != -1 && board[i][j] != 'E') {
          dp[i][j] += board[i][j] - '0';
          dp[i][j] %= kMod;
        }
      }

    return {dp[0][0], count[0][0]};
  }
};
/* code provided by PROGIEZ */

1301. Number of Paths with Max Score LeetCode Solution in Java

class Solution {
  public int[] pathsWithMaxScore(List<String> board) {
    final int kMod = 1_000_000_007;
    final int[][] dirs = {{0, 1}, {1, 0}, {1, 1}};
    final int n = board.size();
    // dp[i][j] := the maximum sum from (n - 1, n - 1) to (i, j)
    int[][] dp = new int[n + 1][n + 1];
    Arrays.stream(dp).forEach(A -> Arrays.fill(A, -1));
    // count[i][j] := the number of paths to get dp[i][j] from (n - 1, n - 1) to (i, j)
    int[][] count = new int[n + 1][n + 1];

    dp[0][0] = 0;
    dp[n - 1][n - 1] = 0;
    count[n - 1][n - 1] = 1;

    for (int i = n - 1; i >= 0; --i)
      for (int j = n - 1; j >= 0; --j) {
        if (board.get(i).charAt(j) == 'S' || board.get(i).charAt(j) == 'X')
          continue;
        for (int[] dir : dirs) {
          final int x = i + dir[0];
          final int y = j + dir[1];
          if (dp[i][j] < dp[x][y]) {
            dp[i][j] = dp[x][y];
            count[i][j] = count[x][y];
          } else if (dp[i][j] == dp[x][y]) {
            count[i][j] += count[x][y];
            count[i][j] %= kMod;
          }
        }
        // If there's path(s) from 'S' to (i, j) and the cell is not 'E'.
        if (dp[i][j] != -1 && board.get(i).charAt(j) != 'E') {
          dp[i][j] += board.get(i).charAt(j) - '0';
          dp[i][j] %= kMod;
        }
      }

    return new int[] {dp[0][0], count[0][0]};
  }
}
// code provided by PROGIEZ

1301. Number of Paths with Max Score LeetCode Solution in Python

class Solution:
  def pathsWithMaxScore(self, board: list[str]) -> list[int]:
    kMod = 1_000_000_007
    dirs = ((0, 1), (1, 0), (1, 1))
    n = len(board)
    # dp[i][j] := the maximum sum from (n - 1, n - 1) to (i, j)
    dp = [[-1] * (n + 1) for _ in range(n + 1)]
    # count[i][j] := the number of paths to get dp[i][j] from (n - 1, n - 1) to
    # (i, j)
    count = [[0] * (n + 1) for _ in range(n + 1)]

    dp[0][0] = 0
    dp[n - 1][n - 1] = 0
    count[n - 1][n - 1] = 1

    for i in reversed(range(n)):
      for j in reversed(range(n)):
        if board[i][j] == 'S' or board[i][j] == 'X':
          continue
        for dx, dy in dirs:
          x = i + dx
          y = j + dy
          if dp[i][j] < dp[x][y]:
            dp[i][j] = dp[x][y]
            count[i][j] = count[x][y]
          elif dp[i][j] == dp[x][y]:
            count[i][j] += count[x][y]
            count[i][j] %= kMod

        # If there's path(s) from 'S' to (i, j) and the cell is not 'E'.
        if dp[i][j] != -1 and board[i][j] != 'E':
          dp[i][j] += int(board[i][j])
          dp[i][j] %= kMod

    return [dp[0][0], count[0][0]]
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

Happy Coding! Keep following PROGIEZ for more updates and solutions.