LeetCode 1463. Cherry Pickup II Solution in Java, C++, Python & More | Explanation + Code

Description

You are given a rows x cols matrix grid representing a field of cherries where grid[i][j] represents the number of cherries that you can collect from the (i, j) cell.

You have two robots that can collect cherries for you:

• Robot #1 is located at the top-left corner (0, 0), and
• Robot #2 is located at the top-right corner (0, cols - 1).

Return the maximum number of cherries collection using both robots by following the rules below:

• From a cell (i, j), robots can move to cell (i + 1, j - 1), (i + 1, j), or (i + 1, j + 1).
• When any robot passes through a cell, It picks up all cherries, and the cell becomes an empty cell.
• When both robots stay in the same cell, only one takes the cherries.
• Both robots cannot move outside of the grid at any moment.
• Both robots should reach the bottom row in grid.

 

Example 1:

Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
Output: 24
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
Total of cherries: 12 + 12 = 24.

Example 2:

Input: grid = [[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]]
Output: 28
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17.
Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11.
Total of cherries: 17 + 11 = 28.

 

Constraints:

• rows == grid.length
• cols == grid[i].length
• 2 <= rows, cols <= 70
• 0 <= grid[i][j] <= 100

Solutions

Solution 1: Dynamic Programming

We define f[i][j₁][j₂] as the maximum number of cherries that can be picked when the two robots are at positions j₁ and j₂ in the i-th row. Initially, f[0][0][n-1] = grid[0][0] + grid[0][n-1], and the other values are -1. The answer is max_{0 ≤ j₁, j₂ < n} f[m-1][j₁][j₂].

Consider f[i][j₁][j₂]. If j₁ ≠ j₂, then the number of cherries that the robots can pick in the i-th row is grid[i][j₁] + grid[i][j₂]. If j₁ = j₂, then the number of cherries that the robots can pick in the i-th row is grid[i][j₁]. We can enumerate the previous state of the two robots f[i-1][y1][y2], where y₁, y₂ are the positions of the two robots in the (i-1)-th row, then y₁ ∈ {j₁-1, j₁, j₁+1} and y₂ ∈ {j₂-1, j₂, j₂+1}. The state transition equation is as follows:

f[i][j₁][j₂] = max_{y₁ ∈ {j₁-1, j₁, j₁+1}, y₂ ∈ {j₂-1, j₂, j₂+1}} f[i-1][y₁][y₂] + \begin{cases} grid[i][j₁] + grid[i][j₂], & j₁ ≠ j₂ \ grid[i][j₁], & j₁ = j₂ \end{cases}

Where f[i-1][y₁][y₂] is ignored when it is -1.

The final answer is max_{0 ≤ j₁, j₂ < n} f[m-1][j₁][j₂].

The time complexity is O(m × n²), and the space complexity is O(m × n²). Where m and n are the number of rows and columns of the grid, respectively.

PythonJavaC++GoTypeScript

class Solution: def cherryPickup(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0]) f = [[[-1] * n for _ in range(n)] for _ in range(m)] f[0][0][n - 1] = grid[0][0] + grid[0][n - 1] for i in range(1, m): for j1 in range(n): for j2 in range(n): x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2]) for y1 in range(j1 - 1, j1 + 2): for y2 in range(j2 - 1, j2 + 2): if 0 <= y1 < n and 0 <= y2 < n and f[i - 1][y1][y2] != -1: f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x) return max(f[-1][j1][j2] for j1, j2 in product(range(n), range(n)))(code-box)

class Solution { public int cherryPickup(int[][] grid) { int m = grid.length, n = grid[0].length; int[][][] f = new int[m][n][n]; for (var g : f) { for (var h : g) { Arrays.fill(h, -1); } } f[0][0][n - 1] = grid[0][0] + grid[0][n - 1]; for (int i = 1; i < m; ++i) { for (int j1 = 0; j1 < n; ++j1) { for (int j2 = 0; j2 < n; ++j2) { int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]); for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) { for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) { if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) { f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x); } } } } } } int ans = 0; for (int j1 = 0; j1 < n; ++j1) { for (int j2 = 0; j2 < n; ++j2) { ans = Math.max(ans, f[m - 1][j1][j2]); } } return ans; } }(code-box)

class Solution { public: int cherryPickup(vector<vector<int>>& grid) { int m = grid.size(), n = grid[0].size(); int f[m][n][n]; memset(f, -1, sizeof(f)); f[0][0][n - 1] = grid[0][0] + grid[0][n - 1]; for (int i = 1; i < m; ++i) { for (int j1 = 0; j1 < n; ++j1) { for (int j2 = 0; j2 < n; ++j2) { int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]); for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) { for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) { if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) { f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x); } } } } } } int ans = 0; for (int j1 = 0; j1 < n; ++j1) { for (int j2 = 0; j2 < n; ++j2) { ans = max(ans, f[m - 1][j1][j2]); } } return ans; } };(code-box)

func cherryPickup(grid [][]int) (ans int) { m, n := len(grid), len(grid[0]) f := make([][][]int, m) for i := range f { f[i] = make([][]int, n) for j := range f[i] { f[i][j] = make([]int, n) for k := range f[i][j] { f[i][j][k] = -1 } } } f[0][0][n-1] = grid[0][0] + grid[0][n-1] for i := 1; i < m; i++ { for j1 := 0; j1 < n; j1++ { for j2 := 0; j2 < n; j2++ { x := grid[i][j1] if j1 != j2 { x += grid[i][j2] } for y1 := j1 - 1; y1 <= j1+1; y1++ { for y2 := j2 - 1; y2 <= j2+1; y2++ { if y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i-1][y1][y2] != -1 { f[i][j1][j2] = max(f[i][j1][j2], f[i-1][y1][y2]+x) } } } } } } for j1 := 0; j1 < n; j1++ { ans = max(ans, slices.Max(f[m-1][j1])) } return }(code-box)

function cherryPickup(grid: number[][]): number { const m = grid.length; const n = grid[0].length; const f = Array.from({ length: m }, () => Array.from({ length: n }, () => Array.from({ length: n }, () => -1)), ); f[0][0][n - 1] = grid[0][0] + grid[0][n - 1]; for (let i = 1; i < m; ++i) { for (let j1 = 0; j1 < n; ++j1) { for (let j2 = 0; j2 < n; ++j2) { const x = grid[i][j1] + (j1 === j2 ? 0 : grid[i][j2]); for (let y1 = j1 - 1; y1 <= j1 + 1; ++y1) { for (let y2 = j2 - 1; y2 <= j2 + 1; ++y2) { if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] !== -1) { f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x); } } } } } } return Math.max(...f[m - 1].flat()); }(code-box)

Solution 2: Dynamic Programming (Space Optimization)

Notice that the calculation of f[i][j₁][j₂] is only related to f[i-1][y₁][y₂]. Therefore, we can use a rolling array to optimize the space complexity. After optimizing the space complexity, the time complexity is O(n²).

PythonJavaC++GoTypeScript

class Solution: def cherryPickup(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0]) f = [[-1] * n for _ in range(n)] g = [[-1] * n for _ in range(n)] f[0][n - 1] = grid[0][0] + grid[0][n - 1] for i in range(1, m): for j1 in range(n): for j2 in range(n): x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2]) for y1 in range(j1 - 1, j1 + 2): for y2 in range(j2 - 1, j2 + 2): if 0 <= y1 < n and 0 <= y2 < n and f[y1][y2] != -1: g[j1][j2] = max(g[j1][j2], f[y1][y2] + x) f, g = g, f return max(f[j1][j2] for j1, j2 in product(range(n), range(n)))(code-box)

class Solution { public int cherryPickup(int[][] grid) { int m = grid.length, n = grid[0].length; int[][] f = new int[n][n]; int[][] g = new int[n][n]; for (int i = 0; i < n; ++i) { Arrays.fill(f[i], -1); Arrays.fill(g[i], -1); } f[0][n - 1] = grid[0][0] + grid[0][n - 1]; for (int i = 1; i < m; ++i) { for (int j1 = 0; j1 < n; ++j1) { for (int j2 = 0; j2 < n; ++j2) { int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]); for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) { for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) { if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1) { g[j1][j2] = Math.max(g[j1][j2], f[y1][y2] + x); } } } } } int[][] t = f; f = g; g = t; } int ans = 0; for (int j1 = 0; j1 < n; ++j1) { for (int j2 = 0; j2 < n; ++j2) { ans = Math.max(ans, f[j1][j2]); } } return ans; } }(code-box)

class Solution { public: int cherryPickup(vector<vector<int>>& grid) { int m = grid.size(), n = grid[0].size(); vector<vector<int>> f(n, vector<int>(n, -1)); vector<vector<int>> g(n, vector<int>(n, -1)); f[0][n - 1] = grid[0][0] + grid[0][n - 1]; for (int i = 1; i < m; ++i) { for (int j1 = 0; j1 < n; ++j1) { for (int j2 = 0; j2 < n; ++j2) { int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]); for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) { for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) { if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1) { g[j1][j2] = max(g[j1][j2], f[y1][y2] + x); } } } } } swap(f, g); } int ans = 0; for (int j1 = 0; j1 < n; ++j1) { for (int j2 = 0; j2 < n; ++j2) { ans = max(ans, f[j1][j2]); } } return ans; } };(code-box)

func cherryPickup(grid [][]int) (ans int) { m, n := len(grid), len(grid[0]) f := make([][]int, n) g := make([][]int, n) for i := range f { f[i] = make([]int, n) g[i] = make([]int, n) for j := range f[i] { f[i][j] = -1 g[i][j] = -1 } } f[0][n-1] = grid[0][0] + grid[0][n-1] for i := 1; i < m; i++ { for j1 := 0; j1 < n; j1++ { for j2 := 0; j2 < n; j2++ { x := grid[i][j1] if j1 != j2 { x += grid[i][j2] } for y1 := j1 - 1; y1 <= j1+1; y1++ { for y2 := j2 - 1; y2 <= j2+1; y2++ { if y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1 { g[j1][j2] = max(g[j1][j2], f[y1][y2]+x) } } } } } f, g = g, f } for j1 := 0; j1 < n; j1++ { ans = max(ans, slices.Max(f[j1])) } return }(code-box)

function cherryPickup(grid: number[][]): number { const m = grid.length; const n = grid[0].length; let f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => -1)); let g: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => -1)); f[0][n - 1] = grid[0][0] + grid[0][n - 1]; for (let i = 1; i < m; ++i) { for (let j1 = 0; j1 < n; ++j1) { for (let j2 = 0; j2 < n; ++j2) { const x = grid[i][j1] + (j1 === j2 ? 0 : grid[i][j2]); for (let y1 = j1 - 1; y1 <= j1 + 1; ++y1) { for (let y2 = j2 - 1; y2 <= j2 + 1; ++y2) { if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] !== -1) { g[j1][j2] = Math.max(g[j1][j2], f[y1][y2] + x); } } } } } [f, g] = [g, f]; } return Math.max(...f.flat()); }(code-box)