Solution Approach
Solution Overview
There are 2 main approaches to solve this problem:
- Dynamic Programming (3D) - Complex approach
- Dynamic Programming (Iterative) - Complex approach
Approach 1: Dynamic Programming (3D)
The key insight for solving this problem is to use dynamic programming to count the number of paths that lead the ball out of the grid boundary.
Let's define a 3D DP array dp[i][j][k] where dp[i][j][k] represents the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
The base case is when the ball is already out of the grid boundary, which contributes 1 to the count.
For each state (i, j, k), we consider all four possible directions to move the ball: up, down, left, and right. For each direction, we check if the new position is out of the grid boundary. If it is, we add 1 to the count. Otherwise, we recursively calculate the number of ways to move the ball out of the grid boundary from the new position with k-1 moves remaining.
To avoid redundant calculations, we use memoization to store the results of subproblems.
Since the answer can be very large, we need to take the modulo of 10^9 + 7 at each step to avoid integer overflow.
Algorithm:
- Define a 3D DP array dp[i][j][k] where dp[i][j][k] represents the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
- Initialize all values in the DP array to -1 (indicating that the subproblem has not been solved yet).
- Define a recursive function findPaths(i, j, k) that returns the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
- Base case: If the ball is already out of the grid boundary (i < 0 || i >= m || j < 0 || j >= n), return 1.
- Base case: If there are no moves remaining (k == 0), return 0.
- If dp[i][j][k] != -1, return dp[i][j][k] (memoization).
- For each of the four directions (up, down, left, right), recursively calculate the number of ways to move the ball out of the grid boundary from the new position with k-1 moves remaining.
- Sum up the results from all four directions, take the modulo of 10^9 + 7, and store it in dp[i][j][k].
- Return dp[i][j][k].
Time Complexity:
Where m is the number of rows, n is the number of columns, and maxMove is the maximum number of moves allowed. We need to compute the DP value for each possible state, and there are m * n * maxMove possible states.
Space Complexity:
We use a 3D DP array of size m * n * maxMove to store the results of subproblems.
Approach 2: Dynamic Programming (Iterative)
We can also solve this problem using an iterative approach instead of recursion.
Let's define a 3D DP array dp[k][i][j] where dp[k][i][j] represents the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
We initialize dp[0][startRow][startColumn] = 1 (there is 1 way to start from the initial position).
For each move k from 1 to maxMove, we iterate through all positions (i, j) in the grid and update dp[k][i][j] based on the values of dp[k-1] for the four adjacent positions.
For each position (i, j), we check if any of the four adjacent positions are out of the grid boundary. If an adjacent position is out of the grid boundary, we add dp[k-1][i][j] to the result.
The final answer is the sum of dp[k][i][j] for all positions (i, j) that are out of the grid boundary, for all moves k from 1 to maxMove.
Algorithm:
- Define a 3D DP array dp[k][i][j] where dp[k][i][j] represents the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
- Initialize dp[0][startRow][startColumn] = 1 (there is 1 way to start from the initial position).
- For each move k from 1 to maxMove:
- a. For each position (i, j) in the grid:
- i. For each of the four directions (up, down, left, right):
- 1. Calculate the new position (ni, nj).
- 2. If the new position is out of the grid boundary, add dp[k-1][i][j] to the result.
- 3. Otherwise, add dp[k-1][i][j] to dp[k][ni][nj].
- Return the result.
Time Complexity:
Where m is the number of rows, n is the number of columns, and maxMove is the maximum number of moves allowed. We need to compute the DP value for each possible state, and there are m * n * maxMove possible states.
Space Complexity:
We use a 3D DP array of size m * n * maxMove to store the results of subproblems.
Pseudocode
12345678910111213141516171819202122232425262728293031323334353637function findPaths(m, n, maxMove, startRow, startColumn): // Define the modulo value MOD = 10^9 + 7 // Initialize the DP array dp = new 3D array of size m x n x (maxMove + 1), initialized with -1 // Define the recursive function function dfs(i, j, k): // Base case: If the ball is out of the grid boundary, return 1 if i < 0 || i >= m || j < 0 || j >= n: return 1 // Base case: If there are no moves remaining, return 0 if k == 0: return 0 // If the subproblem has already been solved, return the result if dp[i][j][k] != -1: return dp[i][j][k] // Initialize the result result = 0 // Consider all four directions result = (result + dfs(i - 1, j, k - 1)) % MOD // Up result = (result + dfs(i + 1, j, k - 1)) % MOD // Down result = (result + dfs(i, j - 1, k - 1)) % MOD // Left result = (result + dfs(i, j + 1, k - 1)) % MOD // Right // Store the result in the DP array dp[i][j][k] = result return result // Call the recursive function with the initial position and maximum moves return dfs(startRow, startColumn, maxMove)String Problems
Learning Objective
Understand different approaches to solve the out of boundary paths problem and analyze their efficiency.
Solution Approaches
Approach 1: Dynamic Programming (3D)
The key insight for solving this problem is to use dynamic programming to count the number of paths that lead the ball out of the grid boundary.
Let's define a 3D DP array dp[i][j][k] where dp[i][j][k] represents the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
The base case is when the ball is already out of the grid boundary, which contributes 1 to the count.
For each state (i, j, k), we consider all four possible directions to move the ball: up, down, left, and right. For each direction, we check if the new position is out of the grid boundary. If it is, we add 1 to the count. Otherwise, we recursively calculate the number of ways to move the ball out of the grid boundary from the new position with k-1 moves remaining.
To avoid redundant calculations, we use memoization to store the results of subproblems.
Since the answer can be very large, we need to take the modulo of 10^9 + 7 at each step to avoid integer overflow.
Algorithm:
- Define a 3D DP array dp[i][j][k] where dp[i][j][k] represents the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
- Initialize all values in the DP array to -1 (indicating that the subproblem has not been solved yet).
- Define a recursive function findPaths(i, j, k) that returns the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
- Base case: If the ball is already out of the grid boundary (i < 0 || i >= m || j < 0 || j >= n), return 1.
- Base case: If there are no moves remaining (k == 0), return 0.
- If dp[i][j][k] != -1, return dp[i][j][k] (memoization).
- For each of the four directions (up, down, left, right), recursively calculate the number of ways to move the ball out of the grid boundary from the new position with k-1 moves remaining.
- Sum up the results from all four directions, take the modulo of 10^9 + 7, and store it in dp[i][j][k].
- Return dp[i][j][k].
Approach 2: Dynamic Programming (Iterative)
We can also solve this problem using an iterative approach instead of recursion.
Let's define a 3D DP array dp[k][i][j] where dp[k][i][j] represents the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
We initialize dp[0][startRow][startColumn] = 1 (there is 1 way to start from the initial position).
For each move k from 1 to maxMove, we iterate through all positions (i, j) in the grid and update dp[k][i][j] based on the values of dp[k-1] for the four adjacent positions.
For each position (i, j), we check if any of the four adjacent positions are out of the grid boundary. If an adjacent position is out of the grid boundary, we add dp[k-1][i][j] to the result.
The final answer is the sum of dp[k][i][j] for all positions (i, j) that are out of the grid boundary, for all moves k from 1 to maxMove.
Algorithm:
- Define a 3D DP array dp[k][i][j] where dp[k][i][j] represents the number of ways to move the ball out of the grid boundary when the ball is at position (i, j) with k moves remaining.
- Initialize dp[0][startRow][startColumn] = 1 (there is 1 way to start from the initial position).
- For each move k from 1 to maxMove:
- a. For each position (i, j) in the grid:
- i. For each of the four directions (up, down, left, right):
- 1. Calculate the new position (ni, nj).
- 2. If the new position is out of the grid boundary, add dp[k-1][i][j] to the result.
- 3. Otherwise, add dp[k-1][i][j] to dp[k][ni][nj].
- Return the result.
Complexity Analysis
Dynamic Programming (3D) Approach
Time Complexity
Where m is the number of rows, n is the number of columns, and maxMove is the maximum number of moves allowed. We need to compute the DP value for each possible state, and there are m * n * maxMove possible states.
Space Complexity
We use a 3D DP array of size m * n * maxMove to store the results of subproblems.
Dynamic Programming (Iterative) Approach
Time Complexity
Where m is the number of rows, n is the number of columns, and maxMove is the maximum number of moves allowed. We need to compute the DP value for each possible state, and there are m * n * maxMove possible states.
Space Complexity
We use a 3D DP array of size m * n * maxMove to store the results of subproblems.
Pseudocode
Dynamic Programming (3D) Approach
12345678910111213141516171819202122232425262728293031323334353637function findPaths(m, n, maxMove, startRow, startColumn): // Define the modulo value MOD = 10^9 + 7 // Initialize the DP array dp = new 3D array of size m x n x (maxMove + 1), initialized with -1 // Define the recursive function function dfs(i, j, k): // Base case: If the ball is out of the grid boundary, return 1 if i < 0 || i >= m || j < 0 || j >= n: return 1 // Base case: If there are no moves remaining, return 0 if k == 0: return 0 // If the subproblem has already been solved, return the result if dp[i][j][k] != -1: return dp[i][j][k] // Initialize the result result = 0 // Consider all four directions result = (result + dfs(i - 1, j, k - 1)) % MOD // Up result = (result + dfs(i + 1, j, k - 1)) % MOD // Down result = (result + dfs(i, j - 1, k - 1)) % MOD // Left result = (result + dfs(i, j + 1, k - 1)) % MOD // Right // Store the result in the DP array dp[i][j][k] = result return result // Call the recursive function with the initial position and maximum moves return dfs(startRow, startColumn, maxMove)Dynamic Programming (Iterative) Approach
1234567891011121314151617181920212223242526272829303132333435363738function findPaths(m, n, maxMove, startRow, startColumn): // Define the modulo value MOD = 10^9 + 7 // Initialize the DP array dp = new 3D array of size (maxMove + 1) x m x n, initialized with 0 // Initialize the starting position dp[0][startRow][startColumn] = 1 // Initialize the result result = 0 // Define the four directions: up, down, left, right directions = [(-1, 0), (1, 0), (0, -1), (0, 1)] // Iterate through all moves for k from 1 to maxMove: // Iterate through all positions in the grid for i from 0 to m - 1: for j from 0 to n - 1: // Skip if there are no ways to reach this position with k-1 moves if dp[k-1][i][j] == 0: continue // Consider all four directions for (di, dj) in directions: ni = i + di nj = j + dj // If the new position is out of the grid boundary, add to the result if ni < 0 || ni >= m || nj < 0 || nj >= n: result = (result + dp[k-1][i][j]) % MOD else: // Otherwise, add to the DP array dp[k][ni][nj] = (dp[k][ni][nj] + dp[k-1][i][j]) % MOD return result