grid47 Exploring patterns and algorithms
Sep 10, 2024
7 min read
Solution to LeetCode 576: Out of Boundary Paths Problem
You are given an m x n grid and a ball placed at position [startRow, startColumn]. The ball can move to one of four adjacent cells or out of the grid boundary. You are allowed at most maxMove moves. Return the number of ways the ball can move out of the grid boundary modulo 10^9 + 7.
Problem
Approach
Steps
Complexity
Input: The input consists of five integers representing the grid dimensions, maximum allowed moves, and the ball's starting position.
Example: Input: m = 3, n = 3, maxMove = 2, startRow = 1, startColumn = 1
Constraints:
• 1 <= m, n <= 50
• 0 <= maxMove <= 50
• 0 <= startRow < m
• 0 <= startColumn < n
Output: Return the number of ways to move the ball out of the grid boundary within the allowed moves, modulo 10^9 + 7.
Example: Output: 12
Constraints:
• The result must be an integer modulo 10^9 + 7.
Goal: Determine the number of paths to move the ball out of the grid boundary within maxMove moves.
Steps:
• Use a recursive approach to track the number of moves left and the ball's position.
• If the ball moves out of the grid boundary, count it as one valid path.
• Use memoization to avoid redundant calculations.
Goal: The input constraints ensure that the grid size, starting position, and allowed moves are within valid ranges.
Steps:
• 1 <= m, n <= 50
• 0 <= maxMove <= 50
• 0 <= startRow < m
• 0 <= startColumn < n
Assumptions:
• The grid dimensions and ball position are valid.
• The ball can move in any direction, even if it immediately crosses the boundary.
• Input: Input: m = 3, n = 3, maxMove = 2, startRow = 1, startColumn = 1
• Explanation: There are 12 possible ways for the ball to move out of the grid boundary within 2 moves.
• Input: Input: m = 2, n = 3, maxMove = 3, startRow = 0, startColumn = 2
• Explanation: The ball can move out of the grid boundary in 18 ways from the starting position.
Approach: The approach involves using Depth-First Search (DFS) with memoization to count the paths while avoiding redundant calculations.
Observations:
• The ball can move in four directions, and paths may overlap if revisiting positions.
• The number of moves is limited, so a recursive solution with memoization will be efficient.
• A grid-based recursive approach with boundary checks should solve this problem efficiently.
Steps:
• Define a recursive function that tracks the current position and remaining moves.
• If the ball moves out of bounds, increment the path count.
• If moves are exhausted without crossing the boundary, return 0.
• Use a 3D memoization array to store results for states defined by moves left and position.
• Iterate through all four possible moves (up, down, left, right) recursively.
Empty Inputs:
• There are no empty inputs due to constraints; the grid always has dimensions.
Large Inputs:
• For large grids (m, n = 50) and maxMove = 50, the algorithm must use memoization to remain efficient.
Special Values:
• The ball starts at the grid boundary; it might move out immediately.
Constraints:
• The solution must handle all inputs efficiently within the given constraints.
classSolution {
int mod = (int) 1e9+7;
public:int findPaths(int m, int n, int mv, int i, int j) {
vector<vector<vector<int>>> mm(mv +1, vector<vector<int>>(m, vector<int>(n, -1)));
returndfs(m, n, mv, 0, i, j, mm);
}
intdfs(int m, int n, int mx, int mv, int x, int y, vector<vector<vector<int>>>& mm) {
if (x <0|| y <0|| x >= m || y >= n) return1;
if (mv == mx) return0;
if (mm[mv][x][y] >-1) return mm[mv][x][y];
int res =0;
res = ( res + dfs(m, n, mx, mv +1, x +1, y, mm) ) % mod;
res = ( res + dfs(m, n, mx, mv +1, x -1, y, mm) ) % mod;
res = ( res + dfs(m, n, mx, mv +1, x, y +1, mm) ) % mod;
res = ( res + dfs(m, n, mx, mv +1, x, y -1, mm) ) % mod;
return mm[mv][x][y] = (res % mod);
}
1 : Class Definition
classSolution {
Defines the main `Solution` class which contains the function `findPaths` to solve the problem.
2 : Modulus Initialization
int mod = (int) 1e9+7;
Initializes the modulus `mod` to ( 10^9 + 7 ) for ensuring the result stays within integer limits.
3 : Access Modifier
public:
The public section of the class where the functions are accessible outside the class.
4 : Function Declaration - findPaths
intfindPaths(int m, int n, int mv, int i, int j) {
The `findPaths` function initializes the dynamic programming table and calls the `dfs` function to calculate the number of possible paths.
Creates a 3D memoization table `mm` to store results for subproblems. Each element is initialized to -1 indicating that it hasn't been computed yet.
6 : Recursive Call to DFS
returndfs(m, n, mv, 0, i, j, mm);
Calls the `dfs` function to start the recursive traversal and path calculation from the given starting position `(i, j)` with `mv` remaining moves.
7 : Function Declaration - dfs
intdfs(int m, int n, int mx, int mv, int x, int y, vector<vector<vector<int>>>& mm) {
The `dfs` function performs depth-first search to explore all possible paths from the current position `(x, y)` within `mv` moves.
8 : Boundary Check
if (x <0|| y <0|| x >= m || y >= n) return1;
Checks if the current position `(x, y)` is out of bounds of the grid. If so, returns 1 as the base case of the recursive call, indicating one valid path.
9 : Max Moves Check
if (mv == mx) return0;
Checks if the current number of moves has reached the maximum allowed moves `mx`. If so, it returns 0 since no more moves are left.
10 : Memoization Check
if (mm[mv][x][y] >-1) return mm[mv][x][y];
Checks if the result for the current position `(x, y)` with `mv` moves has been previously computed. If so, it returns the stored result to avoid redundant calculations.
11 : Path Calculation
int res =0;
Initializes the variable `res` to store the total number of valid paths from the current position.
12 : Recursive Calls - Down
res = ( res + dfs(m, n, mx, mv +1, x +1, y, mm) ) % mod;
Recursively calls `dfs` for moving down in the grid, incrementing the move count by 1.
13 : Recursive Calls - Up
res = ( res + dfs(m, n, mx, mv +1, x -1, y, mm) ) % mod;
Recursively calls `dfs` for moving up in the grid, incrementing the move count by 1.
14 : Recursive Calls - Right
res = ( res + dfs(m, n, mx, mv +1, x, y +1, mm) ) % mod;
Recursively calls `dfs` for moving right in the grid, incrementing the move count by 1.
15 : Recursive Calls - Left
res = ( res + dfs(m, n, mx, mv +1, x, y -1, mm) ) % mod;
Recursively calls `dfs` for moving left in the grid, incrementing the move count by 1.
16 : Memoization Result Storage
return mm[mv][x][y] = (res % mod);
Stores the computed result for the current position `(x, y)` with `mv` moves in the memoization table.
Best Case: O(m * n * maxMove)
Average Case: O(m * n * maxMove)
Worst Case: O(m * n * maxMove)
Description: Each grid position and number of remaining moves is calculated once, thanks to memoization.
Best Case: O(m * n * maxMove)
Worst Case: O(m * n * maxMove)
Description: The space complexity is driven by the memoization array, which has dimensions m x n x maxMove.
