Below is the implementation of the cherry pickup ii:
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202/** * Find the maximum number of cherries collection using both robots. * Dynamic Programming (3D) approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickup(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize 3D DP array const dp = Array(rows).fill().map(() => Array(cols).fill().map(() => Array(cols).fill(-1) ) ); // Define recursive function function dfs(row, col1, col2) { // Base case: reached the bottom row if (row === rows) { return 0; } // Check if already computed if (dp[row][col1][col2] !== -1) { return dp[row][col1][col2]; } // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, dfs(row + 1, newCol1, newCol2)); } } } // Update DP array dp[row][col1][col2] = cherries + maxCherries; return dp[row][col1][col2]; } return dfs(0, 0, cols - 1);} /** * Find the maximum number of cherries collection using both robots. * Dynamic Programming (2D) approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickupOptimized(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize DP arrays let curr = Array(cols).fill().map(() => Array(cols).fill(0)); let next = Array(cols).fill().map(() => Array(cols).fill(0)); // Fill DP arrays from bottom to top for (let row = rows - 1; row >= 0; row--) { // Reset current row for (let col1 = 0; col1 < cols; col1++) { for (let col2 = 0; col2 < cols; col2++) { curr[col1][col2] = 0; } } // Calculate maximum cherries for current row for (let col1 = 0; col1 < cols; col1++) { for (let col2 = 0; col2 < cols; col2++) { // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, next[newCol1][newCol2]); } } } // Update current row curr[col1][col2] = cherries + maxCherries; } } // Swap arrays for next iteration [curr, next] = [next, curr]; } return next[0][cols - 1];} /** * Find the maximum number of cherries collection using both robots. * Dynamic Programming with Memoization approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickupMemoization(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize memoization map const memo = new Map(); // Define recursive function function dp(row, col1, col2) { // Base case: reached the bottom row if (row === rows) { return 0; } // Create a unique key for the memo map const key = `${row},${col1},${col2}`; // Check if already computed if (memo.has(key)) { return memo.get(key); } // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, dp(row + 1, newCol1, newCol2)); } } } // Update memoization map const result = cherries + maxCherries; memo.set(key, result); return result; } return dp(0, 0, cols - 1);} // Test casesconst grid1 = [ [3, 1, 1], [2, 5, 1], [1, 5, 5], [2, 1, 1]];console.log(cherryPickup(grid1)); // 24console.log(cherryPickupOptimized(grid1)); // 24console.log(cherryPickupMemoization(grid1)); // 24 const grid2 = [ [1, 0, 0, 0, 0, 0, 1], [2, 0, 0, 0, 0, 3, 0], [2, 0, 9, 0, 0, 0, 0], [0, 3, 0, 5, 4, 0, 0], [1, 0, 2, 3, 0, 0, 6]];console.log(cherryPickup(grid2)); // 28console.log(cherryPickupOptimized(grid2)); // 28console.log(cherryPickupMemoization(grid2)); // 28
Let's break down the implementation:
Implement the cherry pickup ii solution in different programming languages.
Below is the implementation of the cherry pickup ii in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202/** * Find the maximum number of cherries collection using both robots. * Dynamic Programming (3D) approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickup(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize 3D DP array const dp = Array(rows).fill().map(() => Array(cols).fill().map(() => Array(cols).fill(-1) ) ); // Define recursive function function dfs(row, col1, col2) { // Base case: reached the bottom row if (row === rows) { return 0; } // Check if already computed if (dp[row][col1][col2] !== -1) { return dp[row][col1][col2]; } // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, dfs(row + 1, newCol1, newCol2)); } } } // Update DP array dp[row][col1][col2] = cherries + maxCherries; return dp[row][col1][col2]; } return dfs(0, 0, cols - 1);} /** * Find the maximum number of cherries collection using both robots. * Dynamic Programming (2D) approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickupOptimized(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize DP arrays let curr = Array(cols).fill().map(() => Array(cols).fill(0)); let next = Array(cols).fill().map(() => Array(cols).fill(0)); // Fill DP arrays from bottom to top for (let row = rows - 1; row >= 0; row--) { // Reset current row for (let col1 = 0; col1 < cols; col1++) { for (let col2 = 0; col2 < cols; col2++) { curr[col1][col2] = 0; } } // Calculate maximum cherries for current row for (let col1 = 0; col1 < cols; col1++) { for (let col2 = 0; col2 < cols; col2++) { // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, next[newCol1][newCol2]); } } } // Update current row curr[col1][col2] = cherries + maxCherries; } } // Swap arrays for next iteration [curr, next] = [next, curr]; } return next[0][cols - 1];} /** * Find the maximum number of cherries collection using both robots. * Dynamic Programming with Memoization approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickupMemoization(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize memoization map const memo = new Map(); // Define recursive function function dp(row, col1, col2) { // Base case: reached the bottom row if (row === rows) { return 0; } // Create a unique key for the memo map const key = `${row},${col1},${col2}`; // Check if already computed if (memo.has(key)) { return memo.get(key); } // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, dp(row + 1, newCol1, newCol2)); } } } // Update memoization map const result = cherries + maxCherries; memo.set(key, result); return result; } return dp(0, 0, cols - 1);} // Test casesconst grid1 = [ [3, 1, 1], [2, 5, 1], [1, 5, 5], [2, 1, 1]];console.log(cherryPickup(grid1)); // 24console.log(cherryPickupOptimized(grid1)); // 24console.log(cherryPickupMemoization(grid1)); // 24 const grid2 = [ [1, 0, 0, 0, 0, 0, 1], [2, 0, 0, 0, 0, 3, 0], [2, 0, 9, 0, 0, 0, 0], [0, 3, 0, 5, 4, 0, 0], [1, 0, 2, 3, 0, 0, 6]];console.log(cherryPickup(grid2)); // 28console.log(cherryPickupOptimized(grid2)); // 28console.log(cherryPickupMemoization(grid2)); // 28
Define the state as (row, col1, col2), where row is the current row, col1 is the column of robot 1, and col2 is the column of robot 2.
Set up a dynamic programming array to keep track of the maximum cherries that can be collected for each state.
For each state, calculate the cherries collected by both robots, being careful not to double-count when both robots are in the same cell.
For each state, try all possible moves for both robots (9 combinations in total) and find the maximum cherries that can be collected.
Optimize the space complexity by using a 2D array instead of a 3D array, as we only need the values from the next row.
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the cherry pickup ii problem using a different approach than shown above.
Handle the case where the grid has the minimum size (2 × 2).
Handle the case where some cells are empty (have 0 cherries).
Handle the case where all cells have the maximum number of cherries.
Handle large grid sizes efficiently to avoid timeout issues.
Below is the implementation of the cherry pickup ii:
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202/** * Find the maximum number of cherries collection using both robots. * Dynamic Programming (3D) approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickup(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize 3D DP array const dp = Array(rows).fill().map(() => Array(cols).fill().map(() => Array(cols).fill(-1) ) ); // Define recursive function function dfs(row, col1, col2) { // Base case: reached the bottom row if (row === rows) { return 0; } // Check if already computed if (dp[row][col1][col2] !== -1) { return dp[row][col1][col2]; } // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, dfs(row + 1, newCol1, newCol2)); } } } // Update DP array dp[row][col1][col2] = cherries + maxCherries; return dp[row][col1][col2]; } return dfs(0, 0, cols - 1);} /** * Find the maximum number of cherries collection using both robots. * Dynamic Programming (2D) approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickupOptimized(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize DP arrays let curr = Array(cols).fill().map(() => Array(cols).fill(0)); let next = Array(cols).fill().map(() => Array(cols).fill(0)); // Fill DP arrays from bottom to top for (let row = rows - 1; row >= 0; row--) { // Reset current row for (let col1 = 0; col1 < cols; col1++) { for (let col2 = 0; col2 < cols; col2++) { curr[col1][col2] = 0; } } // Calculate maximum cherries for current row for (let col1 = 0; col1 < cols; col1++) { for (let col2 = 0; col2 < cols; col2++) { // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, next[newCol1][newCol2]); } } } // Update current row curr[col1][col2] = cherries + maxCherries; } } // Swap arrays for next iteration [curr, next] = [next, curr]; } return next[0][cols - 1];} /** * Find the maximum number of cherries collection using both robots. * Dynamic Programming with Memoization approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickupMemoization(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize memoization map const memo = new Map(); // Define recursive function function dp(row, col1, col2) { // Base case: reached the bottom row if (row === rows) { return 0; } // Create a unique key for the memo map const key = `${row},${col1},${col2}`; // Check if already computed if (memo.has(key)) { return memo.get(key); } // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, dp(row + 1, newCol1, newCol2)); } } } // Update memoization map const result = cherries + maxCherries; memo.set(key, result); return result; } return dp(0, 0, cols - 1);} // Test casesconst grid1 = [ [3, 1, 1], [2, 5, 1], [1, 5, 5], [2, 1, 1]];console.log(cherryPickup(grid1)); // 24console.log(cherryPickupOptimized(grid1)); // 24console.log(cherryPickupMemoization(grid1)); // 24 const grid2 = [ [1, 0, 0, 0, 0, 0, 1], [2, 0, 0, 0, 0, 3, 0], [2, 0, 9, 0, 0, 0, 0], [0, 3, 0, 5, 4, 0, 0], [1, 0, 2, 3, 0, 0, 6]];console.log(cherryPickup(grid2)); // 28console.log(cherryPickupOptimized(grid2)); // 28console.log(cherryPickupMemoization(grid2)); // 28
Let's break down the implementation:
Implement the cherry pickup ii solution in different programming languages.
Below is the implementation of the cherry pickup ii in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202/** * Find the maximum number of cherries collection using both robots. * Dynamic Programming (3D) approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickup(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize 3D DP array const dp = Array(rows).fill().map(() => Array(cols).fill().map(() => Array(cols).fill(-1) ) ); // Define recursive function function dfs(row, col1, col2) { // Base case: reached the bottom row if (row === rows) { return 0; } // Check if already computed if (dp[row][col1][col2] !== -1) { return dp[row][col1][col2]; } // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, dfs(row + 1, newCol1, newCol2)); } } } // Update DP array dp[row][col1][col2] = cherries + maxCherries; return dp[row][col1][col2]; } return dfs(0, 0, cols - 1);} /** * Find the maximum number of cherries collection using both robots. * Dynamic Programming (2D) approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickupOptimized(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize DP arrays let curr = Array(cols).fill().map(() => Array(cols).fill(0)); let next = Array(cols).fill().map(() => Array(cols).fill(0)); // Fill DP arrays from bottom to top for (let row = rows - 1; row >= 0; row--) { // Reset current row for (let col1 = 0; col1 < cols; col1++) { for (let col2 = 0; col2 < cols; col2++) { curr[col1][col2] = 0; } } // Calculate maximum cherries for current row for (let col1 = 0; col1 < cols; col1++) { for (let col2 = 0; col2 < cols; col2++) { // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, next[newCol1][newCol2]); } } } // Update current row curr[col1][col2] = cherries + maxCherries; } } // Swap arrays for next iteration [curr, next] = [next, curr]; } return next[0][cols - 1];} /** * Find the maximum number of cherries collection using both robots. * Dynamic Programming with Memoization approach. * * @param {number[][]} grid - Matrix representing a field of cherries * @return {number} - Maximum number of cherries */function cherryPickupMemoization(grid) { const rows = grid.length; const cols = grid[0].length; // Initialize memoization map const memo = new Map(); // Define recursive function function dp(row, col1, col2) { // Base case: reached the bottom row if (row === rows) { return 0; } // Create a unique key for the memo map const key = `${row},${col1},${col2}`; // Check if already computed if (memo.has(key)) { return memo.get(key); } // Calculate cherries collected at current state let cherries = grid[row][col1]; if (col1 !== col2) { cherries += grid[row][col2]; } // Initialize maximum cherries let maxCherries = 0; // Try all possible moves for both robots for (let d1 = -1; d1 <= 1; d1++) { for (let d2 = -1; d2 <= 1; d2++) { const newCol1 = col1 + d1; const newCol2 = col2 + d2; // Check if new positions are valid if (newCol1 >= 0 && newCol1 < cols && newCol2 >= 0 && newCol2 < cols) { maxCherries = Math.max(maxCherries, dp(row + 1, newCol1, newCol2)); } } } // Update memoization map const result = cherries + maxCherries; memo.set(key, result); return result; } return dp(0, 0, cols - 1);} // Test casesconst grid1 = [ [3, 1, 1], [2, 5, 1], [1, 5, 5], [2, 1, 1]];console.log(cherryPickup(grid1)); // 24console.log(cherryPickupOptimized(grid1)); // 24console.log(cherryPickupMemoization(grid1)); // 24 const grid2 = [ [1, 0, 0, 0, 0, 0, 1], [2, 0, 0, 0, 0, 3, 0], [2, 0, 9, 0, 0, 0, 0], [0, 3, 0, 5, 4, 0, 0], [1, 0, 2, 3, 0, 0, 6]];console.log(cherryPickup(grid2)); // 28console.log(cherryPickupOptimized(grid2)); // 28console.log(cherryPickupMemoization(grid2)); // 28
Define the state as (row, col1, col2), where row is the current row, col1 is the column of robot 1, and col2 is the column of robot 2.
Set up a dynamic programming array to keep track of the maximum cherries that can be collected for each state.
For each state, calculate the cherries collected by both robots, being careful not to double-count when both robots are in the same cell.
For each state, try all possible moves for both robots (9 combinations in total) and find the maximum cherries that can be collected.
Optimize the space complexity by using a 2D array instead of a 3D array, as we only need the values from the next row.
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the cherry pickup ii problem using a different approach than shown above.
Handle the case where the grid has the minimum size (2 × 2).
Handle the case where some cells are empty (have 0 cherries).
Handle the case where all cells have the maximum number of cherries.
Handle large grid sizes efficiently to avoid timeout issues.