What is the time complexity of Cherry Pickup II?

The optimal solution has a time complexity of O(m*n^2).

What is the space complexity of Cherry Pickup II?

The space complexity is O(n^2).

What data structures are used to solve Cherry Pickup II?

This problem uses Array, Dynamic Programming, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Cherry Pickup II?

Insight 1: Dynamic Programming is suitable because we can break down the problem into subproblems: finding the maximum cherries collected up to a certain row for all possible positions of the two robots. Insight 2: We need to maintain a state that captures the current row and the column positions of both robots.

Which companies ask LeetCode 1463. Cherry Pickup II in interviews?

This problem is frequently asked at Flipkart, Rubrik.

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Cherry Pickup II - LeetCode 1463 Solution

Cherry Pickup II - Complete Solution Guide

Cherry Pickup II is LeetCode problem 1463, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the maximum number of cherries two robots can collect in a grid. Robot 1 starts at the top-left corner (0, 0) and Robot 2 starts at the top-right corner (0, cols - 1). Both robots move downwards, one row at a time. From cell (i, j), each robot can move to (i+1, j-1), (i+1, j), or (i+1, j+1). When a robot passes through a cell, it picks up all the cherries, and the cell becomes empty. If both robots are in the same cell, only one of them picks up the cherries. The goal is to maximize the total cherries collected when both robots reach the bottom row.

Solution Approach

The provided code uses a dynamic programming approach. It iterates through each row of the grid, maintaining a 2D DP table `dp` where `dp[c1][c2]` represents the maximum number of cherries that can be collected up to the current row, with Robot 1 at column `c1` and Robot 2 at column `c2`. For each cell in the `dp` table, it considers all possible moves the robots could have made from the previous row and calculates the cherries collected in the current cell, updating the `dp` table accordingly. A temporary `next_dp` table is used to store intermediate values before updating the main `dp` table at the end of each row processing.

Step-by-Step Algorithm

Step 1: Initialize a 2D DP table `dp` of size `cols x cols` with all values set to -1. Set the initial state `dp[0][cols - 1]` to the sum of cherries at (0, 0) and (0, cols - 1) (or just grid[0][0] if cols == 1).
Step 2: Iterate through the rows of the grid from the second row (r = 1) to the last row.
Step 3: For each row, create a temporary DP table `next_dp` to store intermediate results.
Step 4: Iterate through all possible column positions `c1` and `c2` for Robot 1 and Robot 2, respectively.
Step 5: If `dp[c1][c2]` is -1, it means this state is unreachable from the previous row, so skip to the next iteration.
Step 6: For each possible move `d1` and `d2` (-1, 0, or 1) for Robot 1 and Robot 2, calculate the next column positions `nc1 = c1 + d1` and `nc2 = c2 + d2`.
Step 7: Check if `nc1` and `nc2` are within the grid boundaries (0 <= `nc1` < cols and 0 <= `nc2` < cols).
Step 8: If the next positions are valid, calculate the cherries collected in the current row. If `nc1` and `nc2` are the same, the robots are in the same cell, so collect only `grid[r + 1][nc1]`. Otherwise, collect `grid[r + 1][nc1] + grid[r + 1][nc2]`.
Step 9: Update `next_dp[nc1][nc2]` with the maximum of its current value and `dp[c1][c2] + cherries`.
Step 10: After processing all possible moves for the current row, update `dp` with `next_dp`.
Step 11: After processing all rows, find the maximum value in the final `dp` table, which represents the maximum number of cherries that can be collected.

Key Insights

Insight 1: Dynamic Programming is suitable because we can break down the problem into subproblems: finding the maximum cherries collected up to a certain row for all possible positions of the two robots.
Insight 2: We need to maintain a state that captures the current row and the column positions of both robots.
Insight 3: The transition function involves considering all possible moves for both robots from their current positions to the next row, ensuring they stay within the grid boundaries.

Complexity Analysis

Time Complexity: O(m*n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Flipkart, Rubrik.

Cherry Pickup II | LeetCode 1463

Hard

Array Dynamic Programming Matrix

Time: O(m*n^2)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int cherryPickup(int[][] grid) {
        int rows = grid.length;
        int cols = grid[0].length;

        int[][] dp = new int[cols][cols];
        for (int i = 0; i < cols; i++) {
            for (int j = 0; j < cols; j++) {
                dp[i][j] = -1;
            }
        }
        dp[0][cols - 1] = grid[0][0] + grid[0][cols - 1];

        for (int r = 0; r < rows - 1; r++) {
            int[][] next_dp = new int[cols][cols];
            for (int i = 0; i < cols; i++) {
                for (int j = 0; j < cols; j++) {
                    next_dp[i][j] = -1;
                }
            }
            for (int c1 = 0; c1 < cols; c1++) {
                for (int c2 = 0; c2 < cols; c2++) {
                    if (dp[c1][c2] == -1) {
                        continue;
                    }

                    for (int d1 = -1; d1 <= 1; d1++) {
                        for (int d2 = -1; d2 <= 1; d2++) {
                            int nc1 = c1 + d1;
                            int nc2 = c2 + d2;

                            if (0 <= nc1 && nc1 < cols && 0 <= nc2 && nc2 < cols) {
                                int cherries = grid[r + 1][nc1];
                                if (nc1 != nc2) {
                                    cherries += grid[r + 1][nc2];
                                }

                                int new_val = dp[c1][c2] + cherries;
                                next_dp[nc1][nc2] = Math.max(next_dp[nc1][nc2], new_val);
                            }
                        }
                    }
                }
            }
            dp = next_dp;
        }

        int max_cherries = 0;
        for (int[] row : dp) {
            for (int val : row) {
                max_cherries = Math.max(max_cherries, val);
            }
        }

        return max_cherries;
    }
}

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Problem Info

Description

Problem 1463: Cherry Pickup II - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

FlipkartRubrik

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1463

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize a 2D DP table `dp` of size `cols x cols` with all values set to -1. Set the initial state `dp[0][cols - 1]` to the sum of cherries at (0, 0) and (0, cols - 1) (or just grid[0][0] if cols == 1).
2
Step 2: Iterate through the rows of the grid from the second row (r = 1) to the last row.
3
Step 3: For each row, create a temporary DP table `next_dp` to store intermediate results.
4
Step 4: Iterate through all possible column positions `c1` and `c2` for Robot 1 and Robot 2, respectively.
5
Step 5: If `dp[c1][c2]` is -1, it means this state is unreachable from the previous row, so skip to the next iteration.
6
Step 6: For each possible move `d1` and `d2` (-1, 0, or 1) for Robot 1 and Robot 2, calculate the next column positions `nc1 = c1 + d1` and `nc2 = c2 + d2`.
7
Step 7: Check if `nc1` and `nc2` are within the grid boundaries (0 <= `nc1` < cols and 0 <= `nc2` < cols).
8
Step 8: If the next positions are valid, calculate the cherries collected in the current row. If `nc1` and `nc2` are the same, the robots are in the same cell, so collect only `grid[r + 1][nc1]`. Otherwise, collect `grid[r + 1][nc1] + grid[r + 1][nc2]`.
9
Step 9: Update `next_dp[nc1][nc2]` with the maximum of its current value and `dp[c1][c2] + cherries`.
10
Step 10: After processing all possible moves for the current row, update `dp` with `next_dp`.
11
Step 11: After processing all rows, find the maximum value in the final `dp` table, which represents the maximum number of cherries that can be collected.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n^2)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is derived from the three nested loops: one for iterating through rows (m), and two for iterating through possible column positions of the robots (n x n). The space complexity is determined by the size of the DP table, which is n x n.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Memoization (Top-Down Dynamic Programming)

A recursive approach with memoization can also be used. Define a recursive function that takes the current row and the column positions of both robots as input. Memoize the results to avoid redundant computations.

Time: O(m*n^2)Space: O(m*n^2) in worst case (if all states are visited), but can be lower if some states are not needed. Stack space due to recursion should also be considered.

Trade-offs: Memoization might be slightly easier to understand conceptually, but it can have a larger memory footprint due to the recursion stack. Bottom-up DP can sometimes be more efficient due to better cache locality.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because we can break down the problem into subproblems: finding the maximum cherries collected up to a certain row for all possible positions of the two robots.
2Insight 2: We need to maintain a state that captures the current row and the column positions of both robots.
3Insight 3: The transition function involves considering all possible moves for both robots from their current positions to the next row, ensuring they stay within the grid boundaries.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Grid with only one row. The algorithm correctly initializes the dp table and proceeds to return the sum of cherries collected by both robots.
Edge case 2: Robots ending up in the same column. The algorithm handles this case by only adding the cherry value of that single column once.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where both robots are in the same cell. The code must ensure that the cherries are only counted once in this scenario.

✗Common mistake 2: Incorrectly handling boundary conditions. Robots cannot move outside the grid, so the code needs to check for valid column positions after each move.

✗Common mistake 3: Forgetting to initialize DP array with a default value that indicates that the state has not been calculated yet. In this case -1 is used.

✗Common mistake 4: Not considering the order of loops. Outer loop should be rows, then robot positions for each row.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Path planning for multiple agents in a constrained environment. Similar logic can be applied to optimize the paths of multiple robots in a warehouse to maximize efficiency and minimize collisions.

🏦Real-world scenario 2: Resource allocation problems where multiple resources (analogous to robots) are used to collect items (analogous to cherries) in different locations (analogous to grid cells).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 24, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Cherry Pickup II"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?