LeetCode #1463 — HARD

Cherry Pickup II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[[3,1,1],[2,5,1],[1,5,5],[2,1,1]]

Example 2

[[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]]

Step 02

Core Insight

What unlocks the optimal approach

Use dynamic programming, define DP[i][j][k]: The maximum cherries that both robots can take starting on the ith row, and column j and k of Robot 1 and 2 respectively.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1463: Cherry Pickup II
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;
    }
}

// Accepted solution for LeetCode #1463: Cherry Pickup II
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
}

# Accepted solution for LeetCode #1463: Cherry Pickup II
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)))

// Accepted solution for LeetCode #1463: Cherry Pickup II
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn cherry_pickup(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let m = grid[0].len();
        let mut memo: HashMap<(usize, usize, usize), i32> = HashMap::new();
        Self::dp(0, 0, m - 1, &mut memo, &grid, n, m)
    }
    fn dp(
        i: usize,
        j: usize,
        k: usize,
        memo: &mut HashMap<(usize, usize, usize), i32>,
        grid: &[Vec<i32>],
        n: usize,
        m: usize,
    ) -> i32 {
        if let Some(&res) = memo.get(&(i, j, k)) {
            return res;
        }

        let mut res = if k != j {
            grid[i][j] + grid[i][k]
        } else {
            grid[i][j]
        };

        if i + 1 < n {
            let mut max = 0;
            if j > 0 && k > 0 {
                max = max.max(Self::dp(i + 1, j - 1, k - 1, memo, grid, n, m));
            }
            if k > 0 {
                max = max.max(Self::dp(i + 1, j, k - 1, memo, grid, n, m));
            }
            if j + 1 < m && k > 0 {
                max = max.max(Self::dp(i + 1, j + 1, k - 1, memo, grid, n, m));
            }
            if j > 0 {
                max = max.max(Self::dp(i + 1, j - 1, k, memo, grid, n, m));
            }
            max = max.max(Self::dp(i + 1, j, k, memo, grid, n, m));
            if j + 1 < m {
                max = max.max(Self::dp(i + 1, j + 1, k, memo, grid, n, m));
            }
            if j > 0 && k + 1 < m {
                max = max.max(Self::dp(i + 1, j - 1, k + 1, memo, grid, n, m));
            }
            if k + 1 < m {
                max = max.max(Self::dp(i + 1, j, k + 1, memo, grid, n, m));
            }
            if j + 1 < m && k + 1 < m {
                max = max.max(Self::dp(i + 1, j + 1, k + 1, memo, grid, n, m));
            }
            res += max;
        }
        memo.insert((i, j, k), res);
        res
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[3, 1, 1], [2, 5, 1], [1, 5, 5], [2, 1, 1]];
    let res = 24;
    assert_eq!(Solution::cherry_pickup(grid), res);
    let grid = vec_vec_i32![
        [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]
    ];
    let res = 28;
    assert_eq!(Solution::cherry_pickup(grid), res);
}

// Accepted solution for LeetCode #1463: Cherry Pickup II
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());
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n^2)

Space

O(m × n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.