What is the time complexity of Jump Game?

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

What data structures are used to solve Jump Game?

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

What are the key insights for solving Jump Game?

Insight 1: The problem can be solved greedily by tracking the maximum reachable index at each step. Insight 2: If at any point the current index `i` exceeds the maximum reachable index, it means we cannot reach that index and thus cannot reach the end.

Which companies ask LeetCode 55. Jump Game in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Cisco and 29 more companies.

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Jump Game - LeetCode 55 Solution

Jump Game - Complete Solution Guide

Jump Game is LeetCode problem 55, a Medium 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 an integer array nums . You are initially positioned at the array's first index , and each element in the array represents your maximum jump length at that position. Return true if you can reach the last index, or false otherwise . Example 1: Input: nums = [2,3,1,1,4] Output: true Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index. Example 2: Input: nums = [3,2,1,0,4] Output: false Explanation: You will always arrive at index 3 no matter what. Its maximum ju

Detailed Explanation

The Jump Game problem asks whether you can reach the last index of an integer array `nums`, starting from the first index. Each element `nums[i]` represents the maximum jump length you can take from that position `i`. The goal is to determine if there exists a sequence of jumps that allows you to arrive at the last index.

Solution Approach

The solution uses a greedy approach to determine if the last index can be reached. It iterates through the array, maintaining a variable `max_reach` representing the furthest index reachable from the indices visited so far. At each index `i`, it checks if `i` is greater than `max_reach`. If it is, it means we cannot reach index `i`, and therefore cannot reach the end, so we return `false`. Otherwise, we update `max_reach` to be the maximum of its current value and `i + nums[i]` (the furthest we can reach from index `i`). If we successfully iterate through the entire array without returning `false`, it means we can reach the last index, so we return `true`.

Step-by-Step Algorithm

Step 1: Initialize `max_reach` to 0, representing the furthest index reachable from the starting position (index 0).
Step 2: Iterate through the `nums` array using a loop, with `i` as the current index.
Step 3: Inside the loop, check if the current index `i` is greater than `max_reach`. If it is, it means we cannot reach this index, so return `false`.
Step 4: Update `max_reach` to be the maximum of its current value and `i + nums[i]`. This updates the furthest reachable index based on the current index's jump length.
Step 5: After the loop finishes, if no `false` has been returned, return `true`, indicating that the last index can be reached.

Key Insights

Insight 1: The problem can be solved greedily by tracking the maximum reachable index at each step.
Insight 2: If at any point the current index `i` exceeds the maximum reachable index, it means we cannot reach that index and thus cannot reach the end.
Insight 3: We only need to keep track of the furthest reachable index, not the exact jump sequence.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Cisco, DoorDash, Flipkart, Goldman Sachs, Google, HashedIn, Informatica, Infosys, Karat, MakeMyTrip, Media.net, Meta, Microsoft, Morgan Stanley, Oracle, PayPal, PhonePe, Salesforce, ServiceNow, Shopee, Tekion, TikTok, Turing, Uber, Verily, Wipro, Yahoo, ZScaler, Zoho, tcs.

Jump Game | LeetCode 55

Medium

Array Dynamic Programming Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean canJump(int[] nums) {
        int max_reach = 0;
        for (int i = 0; i < nums.length; i++) {
            int jump_length = nums[i];
            if (i > max_reach) {
                return false;
            }
            max_reach = Math.max(max_reach, i + jump_length);
        }
        return true;
    }
}

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Problem Info

Description

Problem 55: Jump Game - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy

Companies

AdobeAmazonAppleBloombergCiscoDoorDashFlipkartGoldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:34

Problem #:55

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

5 steps

1
Step 1: Initialize `max_reach` to 0, representing the furthest index reachable from the starting position (index 0).
2
Step 2: Iterate through the `nums` array using a loop, with `i` as the current index.
3
Step 3: Inside the loop, check if the current index `i` is greater than `max_reach`. If it is, it means we cannot reach this index, so return `false`.
4
Step 4: Update `max_reach` to be the maximum of its current value and `i + nums[i]`. This updates the furthest reachable index based on the current index's jump length.
5
Step 5: After the loop finishes, if no `false` has been returned, return `true`, indicating that the last index can be reached.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because we visit each element of the array once. The space complexity is constant because we only use a single variable to store the maximum reachable index, regardless of the input array size.

📖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

Dynamic Programming (Top-Down with Memoization)

Use a recursive function with memoization to store whether each index can reach the end. Start from the beginning and explore all possible jump combinations. This approach is less efficient because of recursive calls and memoization overhead.

Time: O(n^2) in the worst case without memoization, O(n) with memoization.Space: O(n) due to the recursion stack and memoization table.

Trade-offs: This approach is less efficient than the greedy approach. The greedy approach is preferred due to lower time and space complexities.

Dynamic Programming (Bottom-Up)

Create a boolean array `dp` of the same size as `nums`. `dp[i]` is `true` if index `i` can reach the end. Start from the end and work backwards, filling in the `dp` array. This approach is more structured but requires more space.

Time: O(n^2) in the worst case, since for each index, we have to check all reachable indices if they can reach to the last index.Space: O(n) due to the `dp` array.

Trade-offs: Requires O(n) space. The greedy approach is preferred for better space efficiency.

✨

Key Insights

1Insight 1: The problem can be solved greedily by tracking the maximum reachable index at each step.
2Insight 2: If at any point the current index `i` exceeds the maximum reachable index, it means we cannot reach that index and thus cannot reach the end.
3Insight 3: We only need to keep track of the furthest reachable index, not the exact jump sequence.

📚

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.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `nums = [0]`. The code will correctly return `true` because the loop will not execute (or will execute once where i=0 which doesn't invalidate the condition) and the function returns true.
Edge case 2: `nums = [3,2,1,0,4]`. The code will correctly return `false` because when `i = 3`, `max_reach` will likely be 3 (depending on previous values), and `3 > 3` is false, and the updated `max_reach` will be `3`, the next `i` will be 4, where 4 > 3, which satisfies the if condition and returns false.
Edge case 3: `nums = [2,0,0]`. The code correctly returns true. max_reach becomes 2. i becomes 1, which is less than max_reach. max_reach remains 2. i becomes 2, which is less than max_reach. max_reach remains 2. Return true.
Edge case 4: `nums = [2,5,0,0]`. The code correctly returns true.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to update `max_reach` correctly. If `max_reach` is not updated appropriately, the algorithm might incorrectly determine that the end cannot be reached.

✗Common mistake 2: Confusing the jump length with the actual index to jump to. The jump length is the *maximum* jump possible; it is not mandatory to take the full jump.

✗Common mistake 3: Thinking that you have to find the *optimal* jump sequence. The problem only asks if it's *possible* to reach the end, not to find the shortest path.

✅

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: Resource allocation. Imagine you have a limited budget to invest in different projects, where each project has a potential return. The Jump Game analogy can be used to determine if you can achieve a specific financial goal by strategically allocating your resources.

🏦Real-world scenario 2: Network routing. In a network, nodes can be considered as array elements, and the jump length can be interpreted as the capacity of a node to transmit data to other nodes. The Jump Game algorithm can be used to determine if it is possible to transmit data from a source node to a destination node.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 26, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Jump Game"?

For this medium-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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