Below is the implementation of the out of boundary paths:
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132/** * Find the number of paths to move the ball out of the grid boundary. * Dynamic Programming (3D) approach with recursion and memoization. * * @param {number} m - Number of rows in the grid * @param {number} n - Number of columns in the grid * @param {number} maxMove - Maximum number of moves allowed * @param {number} startRow - Starting row position * @param {number} startColumn - Starting column position * @return {number} - Number of paths to move the ball out of the grid boundary */function findPaths(m, n, maxMove, startRow, startColumn) { // Define the modulo value const MOD = 1000000007; // Initialize the DP array const dp = Array(m).fill().map(() => Array(n).fill().map(() => Array(maxMove + 1).fill(-1) ) ); /** * Recursive function to find the number of paths. * * @param {number} i - Current row position * @param {number} j - Current column position * @param {number} k - Number of moves remaining * @return {number} - Number of paths to move the ball out of the grid boundary */ 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 let 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);} /** * Find the number of paths to move the ball out of the grid boundary. * Dynamic Programming (Iterative) approach. * * @param {number} m - Number of rows in the grid * @param {number} n - Number of columns in the grid * @param {number} maxMove - Maximum number of moves allowed * @param {number} startRow - Starting row position * @param {number} startColumn - Starting column position * @return {number} - Number of paths to move the ball out of the grid boundary */function findPathsIterative(m, n, maxMove, startRow, startColumn) { // Define the modulo value const MOD = 1000000007; // Initialize the DP array const dp = Array(maxMove + 1).fill().map(() => Array(m).fill().map(() => Array(n).fill(0) ) ); // Initialize the starting position dp[0][startRow][startColumn] = 1; // Initialize the result let result = 0; // Define the four directions: up, down, left, right const directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]; // Iterate through all moves for (let k = 1; k <= maxMove; k++) { // Iterate through all positions in the grid for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { // 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 (const [di, dj] of directions) { const ni = i + di; const 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;} // Test casesconsole.log(findPaths(2, 2, 2, 0, 0)); // 6console.log(findPaths(1, 3, 3, 0, 1)); // 12 console.log(findPathsIterative(2, 2, 2, 0, 0)); // 6console.log(findPathsIterative(1, 3, 3, 0, 1)); // 12
Let's break down the implementation:
Implement the out of boundary paths solution in different programming languages.
Below is the implementation of the out of boundary paths in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132/** * Find the number of paths to move the ball out of the grid boundary. * Dynamic Programming (3D) approach with recursion and memoization. * * @param {number} m - Number of rows in the grid * @param {number} n - Number of columns in the grid * @param {number} maxMove - Maximum number of moves allowed * @param {number} startRow - Starting row position * @param {number} startColumn - Starting column position * @return {number} - Number of paths to move the ball out of the grid boundary */function findPaths(m, n, maxMove, startRow, startColumn) { // Define the modulo value const MOD = 1000000007; // Initialize the DP array const dp = Array(m).fill().map(() => Array(n).fill().map(() => Array(maxMove + 1).fill(-1) ) ); /** * Recursive function to find the number of paths. * * @param {number} i - Current row position * @param {number} j - Current column position * @param {number} k - Number of moves remaining * @return {number} - Number of paths to move the ball out of the grid boundary */ 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 let 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);} /** * Find the number of paths to move the ball out of the grid boundary. * Dynamic Programming (Iterative) approach. * * @param {number} m - Number of rows in the grid * @param {number} n - Number of columns in the grid * @param {number} maxMove - Maximum number of moves allowed * @param {number} startRow - Starting row position * @param {number} startColumn - Starting column position * @return {number} - Number of paths to move the ball out of the grid boundary */function findPathsIterative(m, n, maxMove, startRow, startColumn) { // Define the modulo value const MOD = 1000000007; // Initialize the DP array const dp = Array(maxMove + 1).fill().map(() => Array(m).fill().map(() => Array(n).fill(0) ) ); // Initialize the starting position dp[0][startRow][startColumn] = 1; // Initialize the result let result = 0; // Define the four directions: up, down, left, right const directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]; // Iterate through all moves for (let k = 1; k <= maxMove; k++) { // Iterate through all positions in the grid for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { // 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 (const [di, dj] of directions) { const ni = i + di; const 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;} // Test casesconsole.log(findPaths(2, 2, 2, 0, 0)); // 6console.log(findPaths(1, 3, 3, 0, 1)); // 12 console.log(findPathsIterative(2, 2, 2, 0, 0)); // 6console.log(findPathsIterative(1, 3, 3, 0, 1)); // 12
Understand that we need to find the number of paths to move the ball out of the grid boundary within a maximum number of moves.
Initialize a 3D DP array to store the number of paths for each position and remaining moves.
Define a recursive function that calculates the number of paths for each state (position and remaining moves).
If the ball is out of the grid boundary, return 1. If there are no moves remaining, return 0.
For each state, consider all four possible directions to move the ball and sum up the results.
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the out of boundary paths problem using a different approach than shown above.
Handle the case where maxMove is 0 (return 0, as the ball cannot move out of the grid boundary).
Handle the case where the ball starts at a position adjacent to the grid boundary (it can move out in 1 step).
Handle the case where the grid is large and the maximum number of moves is small (the ball may not be able to reach the boundary).
Below is the implementation of the out of boundary paths:
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132/** * Find the number of paths to move the ball out of the grid boundary. * Dynamic Programming (3D) approach with recursion and memoization. * * @param {number} m - Number of rows in the grid * @param {number} n - Number of columns in the grid * @param {number} maxMove - Maximum number of moves allowed * @param {number} startRow - Starting row position * @param {number} startColumn - Starting column position * @return {number} - Number of paths to move the ball out of the grid boundary */function findPaths(m, n, maxMove, startRow, startColumn) { // Define the modulo value const MOD = 1000000007; // Initialize the DP array const dp = Array(m).fill().map(() => Array(n).fill().map(() => Array(maxMove + 1).fill(-1) ) ); /** * Recursive function to find the number of paths. * * @param {number} i - Current row position * @param {number} j - Current column position * @param {number} k - Number of moves remaining * @return {number} - Number of paths to move the ball out of the grid boundary */ 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 let 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);} /** * Find the number of paths to move the ball out of the grid boundary. * Dynamic Programming (Iterative) approach. * * @param {number} m - Number of rows in the grid * @param {number} n - Number of columns in the grid * @param {number} maxMove - Maximum number of moves allowed * @param {number} startRow - Starting row position * @param {number} startColumn - Starting column position * @return {number} - Number of paths to move the ball out of the grid boundary */function findPathsIterative(m, n, maxMove, startRow, startColumn) { // Define the modulo value const MOD = 1000000007; // Initialize the DP array const dp = Array(maxMove + 1).fill().map(() => Array(m).fill().map(() => Array(n).fill(0) ) ); // Initialize the starting position dp[0][startRow][startColumn] = 1; // Initialize the result let result = 0; // Define the four directions: up, down, left, right const directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]; // Iterate through all moves for (let k = 1; k <= maxMove; k++) { // Iterate through all positions in the grid for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { // 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 (const [di, dj] of directions) { const ni = i + di; const 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;} // Test casesconsole.log(findPaths(2, 2, 2, 0, 0)); // 6console.log(findPaths(1, 3, 3, 0, 1)); // 12 console.log(findPathsIterative(2, 2, 2, 0, 0)); // 6console.log(findPathsIterative(1, 3, 3, 0, 1)); // 12
Let's break down the implementation:
Implement the out of boundary paths solution in different programming languages.
Below is the implementation of the out of boundary paths in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132/** * Find the number of paths to move the ball out of the grid boundary. * Dynamic Programming (3D) approach with recursion and memoization. * * @param {number} m - Number of rows in the grid * @param {number} n - Number of columns in the grid * @param {number} maxMove - Maximum number of moves allowed * @param {number} startRow - Starting row position * @param {number} startColumn - Starting column position * @return {number} - Number of paths to move the ball out of the grid boundary */function findPaths(m, n, maxMove, startRow, startColumn) { // Define the modulo value const MOD = 1000000007; // Initialize the DP array const dp = Array(m).fill().map(() => Array(n).fill().map(() => Array(maxMove + 1).fill(-1) ) ); /** * Recursive function to find the number of paths. * * @param {number} i - Current row position * @param {number} j - Current column position * @param {number} k - Number of moves remaining * @return {number} - Number of paths to move the ball out of the grid boundary */ 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 let 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);} /** * Find the number of paths to move the ball out of the grid boundary. * Dynamic Programming (Iterative) approach. * * @param {number} m - Number of rows in the grid * @param {number} n - Number of columns in the grid * @param {number} maxMove - Maximum number of moves allowed * @param {number} startRow - Starting row position * @param {number} startColumn - Starting column position * @return {number} - Number of paths to move the ball out of the grid boundary */function findPathsIterative(m, n, maxMove, startRow, startColumn) { // Define the modulo value const MOD = 1000000007; // Initialize the DP array const dp = Array(maxMove + 1).fill().map(() => Array(m).fill().map(() => Array(n).fill(0) ) ); // Initialize the starting position dp[0][startRow][startColumn] = 1; // Initialize the result let result = 0; // Define the four directions: up, down, left, right const directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]; // Iterate through all moves for (let k = 1; k <= maxMove; k++) { // Iterate through all positions in the grid for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { // 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 (const [di, dj] of directions) { const ni = i + di; const 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;} // Test casesconsole.log(findPaths(2, 2, 2, 0, 0)); // 6console.log(findPaths(1, 3, 3, 0, 1)); // 12 console.log(findPathsIterative(2, 2, 2, 0, 0)); // 6console.log(findPathsIterative(1, 3, 3, 0, 1)); // 12
Understand that we need to find the number of paths to move the ball out of the grid boundary within a maximum number of moves.
Initialize a 3D DP array to store the number of paths for each position and remaining moves.
Define a recursive function that calculates the number of paths for each state (position and remaining moves).
If the ball is out of the grid boundary, return 1. If there are no moves remaining, return 0.
For each state, consider all four possible directions to move the ball and sum up the results.
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the out of boundary paths problem using a different approach than shown above.
Handle the case where maxMove is 0 (return 0, as the ball cannot move out of the grid boundary).
Handle the case where the ball starts at a position adjacent to the grid boundary (it can move out in 1 step).
Handle the case where the grid is large and the maximum number of moves is small (the ball may not be able to reach the boundary).