What is the time complexity of Jump Game II?

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

What data structures are used to solve Jump Game II?

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 II?

Insight 1: The key insight is to use a greedy approach. At each step, we want to jump to the position that allows us to reach the farthest. Insight 2: We don't need to explicitly simulate every possible jump combination using recursion or backtracking, which would be inefficient. Instead, we can keep track of the farthest reachable index and the current jump's end.

Which companies ask LeetCode 45. Jump Game II in interviews?

This problem is frequently asked at Adobe, Agoda, Amazon, Apple, Atlassian and 25 more companies.

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Jump Game II - LeetCode 45 Solution

Q: How to solve Jump Game II?

The provided code implements a greedy algorithm. It iterates through the array, keeping track of the 'farthest' reachable index and the 'current_end' of the current jump. For each index `i`, it updates the 'farthest' reachable index. When `i` reaches 'current_end', it means we've exhausted the current jump, so we increment the jump count and update 'current_end' to 'farthest'. We stop iterating at `n-1` because we only need to make jumps to reach the second to last element.

Jump Game II - Complete Solution Guide

Jump Game II is LeetCode problem 45, 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 a 0-indexed array of integers nums of length n . You are initially positioned at index 0. Each element nums[i] represents the maximum length of a forward jump from index i . In other words, if you are at index i , you can jump to any index (i + j) where: 0 <= j <= nums[i] and i + j < n Return the minimum number of jumps to reach index n - 1 . The test cases are generated such that you can reach index n - 1 . Example 1: Input: nums = [2,3,1,1,4] Output: 2 Explanation: The minimum nu

Detailed Explanation

The problem "Jump Game II" asks us to find the minimum number of jumps required to reach the last element of an array, given that each element in the array represents the maximum jump distance from that position. We start at index 0, and we need to reach index `n-1` with the fewest jumps possible. It's guaranteed that we can reach the last index.

Solution Approach

The provided code implements a greedy algorithm. It iterates through the array, keeping track of the 'farthest' reachable index and the 'current_end' of the current jump. For each index `i`, it updates the 'farthest' reachable index. When `i` reaches 'current_end', it means we've exhausted the current jump, so we increment the jump count and update 'current_end' to 'farthest'. We stop iterating at `n-1` because we only need to make jumps to reach the second to last element.

Step-by-Step Algorithm

Step 1: Initialize `jumps` to 0, `current_end` to 0, and `farthest` to 0. `jumps` keeps track of the number of jumps, `current_end` marks the end of the current jump range, and `farthest` tracks the maximum reachable index from the current range.
Step 2: Iterate through the array from `i = 0` to `n - 2`. We stop at `n - 2` because reaching `n-1` is the goal and one additional jump is counted when current_end is reached.
Step 3: In each iteration, update `farthest` to the maximum of its current value and `i + nums[i]`. This means from position i, the maximum reachable index is `i + nums[i]`.
Step 4: If `i` is equal to `current_end`, it means we have reached the end of the current jump range. Therefore, we need to make a jump. Increment `jumps` by 1.
Step 5: Update `current_end` to `farthest`. This extends the range of the next jump to the farthest reachable index.
Step 6: After the loop finishes, return `jumps`.

Key Insights

Insight 1: The key insight is to use a greedy approach. At each step, we want to jump to the position that allows us to reach the farthest.
Insight 2: We don't need to explicitly simulate every possible jump combination using recursion or backtracking, which would be inefficient. Instead, we can keep track of the farthest reachable index and the current jump's end.
Insight 3: Instead of finding all possible paths, we keep track of how far we can jump from the current range. If our current position reaches the end of the current range, we make a jump and update our range.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy.

Companies

Asked at: Adobe, Agoda, Amazon, Apple, Atlassian, Bloomberg, ConsultAdd, DE Shaw, DoorDash, Expedia, Flipkart, Goldman Sachs, Google, Groupon, HashedIn, MakeMyTrip, Meta, Microsoft, Mitsogo, Nutanix, Oracle, PayPal, PhonePe, Roblox, Samsung, TikTok, Uber, Walmart Labs, Zepto, Zomato.

Jump Game II | LeetCode 45

Medium

Array Dynamic Programming Greedy

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

Solution Code

Available in:

class Solution {
    public int jump(int[] nums) {
        int n = nums.length;
        if (n <= 1) {
            return 0;
        }
        
        int jumps = 0;
        int current_end = 0;
        int farthest = 0;
        
        for (int i = 0; i < n - 1; i++) {
            farthest = Math.max(farthest, i + nums[i]);
            
            if (i == current_end) {
                jumps++;
                current_end = farthest;
            }
        }
        
        return jumps;
    }
}

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Problem Info

Description

Problem 45: Jump Game II - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy

Companies

AdobeAgodaAmazonAppleAtlassianBloombergConsultAddDE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:30

Problem #:45

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

6 steps

1
Step 1: Initialize `jumps` to 0, `current_end` to 0, and `farthest` to 0. `jumps` keeps track of the number of jumps, `current_end` marks the end of the current jump range, and `farthest` tracks the maximum reachable index from the current range.
2
Step 2: Iterate through the array from `i = 0` to `n - 2`. We stop at `n - 2` because reaching `n-1` is the goal and one additional jump is counted when current_end is reached.
3
Step 3: In each iteration, update `farthest` to the maximum of its current value and `i + nums[i]`. This means from position i, the maximum reachable index is `i + nums[i]`.
4
Step 4: If `i` is equal to `current_end`, it means we have reached the end of the current jump range. Therefore, we need to make a jump. Increment `jumps` by 1.
5
Step 5: Update `current_end` to `farthest`. This extends the range of the next jump to the farthest reachable index.
6
Step 6: After the loop finishes, return `jumps`.

💡

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 algorithm iterates through the array elements once in a single for loop. Therefore, the time complexity is linear. The algorithm uses a fixed number of variables irrespective of the size of the input. Hence, the space complexity is constant.

📖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

We can use dynamic programming to solve this problem. Create a `dp` array where `dp[i]` represents the minimum jumps to reach index `i`. The base case is `dp[0] = 0`. For each index `i`, iterate through the previous indices `j` and check if we can reach `i` from `j`. If yes, update `dp[i] = min(dp[i], dp[j] + 1)`.

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

Trade-offs: While conceptually simpler to understand for beginners, dynamic programming has a higher time complexity than the greedy approach. The greedy approach is more efficient for larger input arrays. Dynamic programming could be helpful if variations require storing intermediate states. The guarantee of reaching the end enables the greedy optimization.

✨

Key Insights

1Insight 1: The key insight is to use a greedy approach. At each step, we want to jump to the position that allows us to reach the farthest.
2Insight 2: We don't need to explicitly simulate every possible jump combination using recursion or backtracking, which would be inefficient. Instead, we can keep track of the farthest reachable index and the current jump's end.
3Insight 3: Instead of finding all possible paths, we keep track of how far we can jump from the current range. If our current position reaches the end of the current range, we make a jump and update our range.

📚

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.length <= 1`. If the array has only one element or is empty, we don't need any jumps, so the code returns 0.
Edge case 2: The problem guarantees that we can always reach the end. So there is no need to explicitly check for unsolvable cases within the main loop.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the base case where the array length is 1 or less. This can lead to errors or incorrect output.

✗Common mistake 2: Using a recursive approach without memoization or dynamic programming. This can lead to exponential time complexity and Time Limit Exceeded errors.

✗Common mistake 3: Iterating to the last element of the array (n-1) in the loop instead of stopping at n-2. If the loop goes to n-1, there can be incorrect incrementing of the 'jumps' counter. The final jump is counted when i == current_end occurs before i reaches n-1.

✅

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 different resources with different ranges of effect, and you want to achieve a goal with the fewest resources. The problem can be modeled as a jump game.

🏦Real-world scenario 2: Route planning with fuel constraints. If each location provides a certain amount of fuel that allows you to travel a certain distance, determining the minimum number of refueling stops to reach a destination can be modeled as this problem.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 16, 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 II"?

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.

🤖 Need help with this problem?