What is the time complexity of Frog Jump?

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

What is the space complexity of Frog Jump?

The space complexity is O(n^2).

What data structures are used to solve Frog Jump?

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

What are the key insights for solving Frog Jump?

Insight 1: Dynamic programming is crucial. We need to keep track of which jump sizes can reach each stone, not just whether a stone can be reached. Insight 2: Using a hash map (or dictionary) is effective for storing the mapping between stone positions and the possible jump sizes that can reach them.

Which companies ask LeetCode 403. Frog Jump in interviews?

This problem is frequently asked at LINE, Otter.ai, PhonePe, Snap, Zomato.

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Frog Jump - LeetCode 403 Solution

Frog Jump - Complete Solution Guide

Frog Jump is LeetCode problem 403, 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

A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones positions (in units) in sorted ascending order , determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit. If the frog's last jump was k units, its next jump must be either k - 1 ,

Detailed Explanation

The Frog Jump problem asks us to determine if a frog can cross a river by jumping on stones. The river is divided into units, and stones are placed at certain units. The frog starts at the first stone (position 0) and must jump 1 unit to the second stone. Subsequent jumps can be of size k-1, k, or k+1, where k is the size of the previous jump. The goal is to determine if the frog can reach the last stone.

Solution Approach

The solution uses dynamic programming to track the possible jump sizes that can lead the frog to each stone. A hash map (dictionary in Python) is used where the key is the stone position and the value is a set of jump sizes that can reach that stone. The algorithm iterates through the stones. For each stone and each jump size that can reach it, it explores the next possible jumps (k-1, k, k+1) to see if they lead to other stones. If a jump lands on another stone, the corresponding jump size is added to the set of possible jump sizes for that stone. If the frog reaches the last stone, the function returns `true`. Otherwise, it returns `false` after exploring all possibilities.

Step-by-Step Algorithm

Step 1: Initialize a hash map (or dictionary) called `dp` to store the possible jump sizes for each stone. The key is the stone's position, and the value is a set of jump sizes.
Step 2: Initialize the jump sizes for the first stone (position 0) to be {0}, since frog starts with a 'jump size' of zero at first position.
Step 3: Iterate through the stones array.
Step 4: For each stone, iterate through the set of jump sizes that can reach it.
Step 5: For each jump size `k`, calculate the next possible positions by jumping `k-1`, `k`, and `k+1` units forward.
Step 6: Check if these new positions exist as stones. If they do, add `k-1`, `k`, or `k+1` to the set of jump sizes for that stone (only if the jump size is > 0).
Step 7: If any jump leads to the last stone, return `true`.
Step 8: If, after iterating through all stones and jump sizes, the last stone is not reached, return `false`.

Key Insights

Insight 1: Dynamic programming is crucial. We need to keep track of which jump sizes can reach each stone, not just whether a stone can be reached.
Insight 2: Using a hash map (or dictionary) is effective for storing the mapping between stone positions and the possible jump sizes that can reach them.
Insight 3: Early exit condition. Check if the second stone exists and equals 1, because the frog's first jump must be 1. If the second stone is not 1 or doesn't exist, return false

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: LINE, Otter.ai, PhonePe, Snap, Zomato.

Frog Jump | LeetCode 403

Hard

Array Dynamic Programming

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public boolean canCross(int[] stones) {
        int n = stones.length;
        if (n > 1 && stones[1] != 1) {
            return false;
        }

        // dp maps a stone's position to a set of jump sizes that can reach it.
        HashMap<Integer, Set<Integer>> dp = new HashMap<>();
        for (int stone : stones) {
            dp.put(stone, new HashSet<>());
        }
        dp.get(0).add(0);

        for (int stone : stones) {
            // For each jump size 'k' that can reach the current stone
            for (int k : dp.get(stone)) {
                // Explore next possible jump sizes: k-1, k, k+1
                for (int step : new int[]{k - 1, k, k + 1}) {
                    if (step > 0) {
                        int nextPos = stone + step;
                        if (nextPos == stones[n - 1]) {
                            return true;
                        }
                        if (dp.containsKey(nextPos)) {
                            dp.get(nextPos).add(step);
                        }
                    }
                }
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 403: Frog Jump - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

LINEOtter.aiPhonePeSnapZomato

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:403

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

8 steps

1
Step 1: Initialize a hash map (or dictionary) called `dp` to store the possible jump sizes for each stone. The key is the stone's position, and the value is a set of jump sizes.
2
Step 2: Initialize the jump sizes for the first stone (position 0) to be {0}, since frog starts with a 'jump size' of zero at first position.
3
Step 3: Iterate through the stones array.
4
Step 4: For each stone, iterate through the set of jump sizes that can reach it.
5
Step 5: For each jump size `k`, calculate the next possible positions by jumping `k-1`, `k`, and `k+1` units forward.
6
Step 6: Check if these new positions exist as stones. If they do, add `k-1`, `k`, or `k+1` to the set of jump sizes for that stone (only if the jump size is > 0).
7
Step 7: If any jump leads to the last stone, return `true`.
8
Step 8: If, after iterating through all stones and jump sizes, the last stone is not reached, return `false`.

💡

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(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The nested loops dominate the time complexity. The space complexity arises from storing jump sizes for each stone, potentially leading to O(n^2) space usage in the worst-case scenario where each stone can be reached by many different jump sizes.

📖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

Recursion with Memoization

A recursive solution can be implemented with memoization to avoid redundant calculations. The recursive function would explore the possible jumps from each stone. Memoization would store the results of already-computed states (stone position and jump size).

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

Trade-offs: This approach is generally less efficient than the iterative DP solution because of the overhead of recursive calls, although both have the same asymptotic complexity. It might be easier to understand for some, but iterative DP generally performs better in practice due to avoiding function call overhead.

✨

Key Insights

1Insight 1: Dynamic programming is crucial. We need to keep track of which jump sizes can reach each stone, not just whether a stone can be reached.
2Insight 2: Using a hash map (or dictionary) is effective for storing the mapping between stone positions and the possible jump sizes that can reach them.
3Insight 3: Early exit condition. Check if the second stone exists and equals 1, because the frog's first jump must be 1. If the second stone is not 1 or doesn't exist, return false

📚

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: Empty input or very small input. The provided constraints specify at least two stones. The code handles this implicitly.
Edge case 2: The frog cannot make the first jump (if the second stone is not at position 1). The code explicitly checks `if n > 1 and stones[1] != 1: return False`.
Edge case 3: Stones not sorted. The problem states the stones are sorted, so there is no need to handle that case.
Edge case 4: Large gaps between stones that make it impossible to reach the next stone. This is the main scenario that the algorithm addresses by checking the reachable positions and jump sizes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case where the second stone is not at position 1. This would cause the frog to be unable to start the journey.

✗Common mistake 2: Forgetting to check if the next jump leads to an existing stone. The frog can only jump on stones, not into the water.

✗Common mistake 3: Using a boolean DP array instead of a hash map to store jump sizes. This won't be able to keep track of the possible jump sizes, which is necessary to determine future jump distances.

✗Common mistake 4: Not checking step > 0. The jump size should always be greater than zero.

✅

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: Game development. This problem is similar to pathfinding in games, where an agent needs to navigate a terrain with certain movement restrictions.

🏦Real-world scenario 2: Robotics. Similar to game development, robot path planning involves navigating an environment with obstacles and constraints on movement. The frog jumping problem shares similarities with this general navigation task.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: July 14, 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 "Frog Jump"?

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.

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