What is the time complexity of Jump Game VI?

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

What data structures are used to solve Jump Game VI?

This problem uses Array, Dynamic Programming, Queue, Heap (Priority Queue), Monotonic Queue. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Jump Game VI?

Insight 1: Dynamic Programming (DP) is suitable since the optimal score at index `i` depends on the optimal scores at previous indices within the range `[i-k, i-1]`. Insight 2: Using a naive DP approach (checking all possible previous indices in each step) results in O(n*k) time complexity, which can exceed the time limit for larger inputs. A Monotonic Queue helps optimize the DP transitions to O(n).

Which companies ask LeetCode 1696. Jump Game VI in interviews?

This problem is frequently asked at AQR Capital Management.

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Jump Game VI - LeetCode 1696 Solution

Jump Game VI - Complete Solution Guide

Jump Game VI is LeetCode problem 1696, 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 integer array nums and an integer k . You are initially standing at index 0 . In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive . You want to reach the last index of the array (index n - 1 ). Your score is the sum of all nums[j] for each index j you visited in the array. Return the maximum score you can get . Example 1: Inp

Detailed Explanation

The problem asks us to find the maximum score achievable when jumping from index 0 to index `n-1` in an array `nums`. From any index `i`, you can jump to any index within the range `[i+1, min(n-1, i+k)]`, where `k` is a given integer. The score is the sum of the values of the array elements at the indices you visit. The goal is to maximize this sum.

Solution Approach

The provided code uses a Dynamic Programming approach optimized with a Monotonic Queue (Deque). The `nums` array itself is used as the DP table, where `nums[i]` stores the maximum score achievable when reaching index `i`. The Monotonic Queue efficiently keeps track of the best possible previous indices to jump from, within the allowed jump range `k`.

Step-by-Step Algorithm

Step 1: Initialize a Deque `dq` to store indices. Add the starting index 0 to `dq`.
Step 2: Iterate from index 1 to `n-1` of the `nums` array.
Step 3: Before updating `nums[i]`, check if the index at the front of `dq` is outside the valid jump range `[i-k, i-1]`. If `dq.peekFirst() < i - k`, it means that the index is too far behind, so remove it from the front of `dq`.
Step 4: Update `nums[i]` by adding the maximum score from a previous reachable index, which is `nums[dq.peekFirst()]`. Thus, `nums[i] += nums[dq.peekFirst()]`.
Step 5: Maintain the monotonic property of `dq`. While `dq` is not empty and the score at the last index in `dq` (`nums[dq.peekLast()]`) is less than or equal to the current score (`nums[i]`), remove the last index from `dq`. This ensures that `dq` always contains indices with decreasing scores from front to back.
Step 6: Add the current index `i` to the back of `dq`.
Step 7: After iterating through the array, `nums[n-1]` will contain the maximum score achievable when reaching the last index.
Step 8: Return `nums[n-1]`.

Key Insights

Insight 1: Dynamic Programming (DP) is suitable since the optimal score at index `i` depends on the optimal scores at previous indices within the range `[i-k, i-1]`.
Insight 2: Using a naive DP approach (checking all possible previous indices in each step) results in O(n*k) time complexity, which can exceed the time limit for larger inputs. A Monotonic Queue helps optimize the DP transitions to O(n).
Insight 3: The Monotonic Queue stores indices of potentially optimal 'previous' states (indices) to jump from. It maintains a decreasing order of `nums[index]` values so that the front of the queue always contains the index that yields the highest score when jumping to the current index.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Dynamic Programming, Queue, Heap (Priority Queue), Monotonic Queue.

Companies

Asked at: AQR Capital Management.

Jump Game VI | LeetCode 1696

Medium

Array Dynamic Programming Queue Heap (Priority Queue)Monotonic Queue

Time: O(n)Space: O(k)

Solution Code

Available in:

import java.util.Deque;
import java.util.LinkedList;

class Solution {
    public int maxResult(int[] nums, int k) {
        int n = nums.length;
        Deque<Integer> dq = new LinkedList<>();
        dq.offer(0);

        for (int i = 1; i < n; i++) {
            if (!dq.isEmpty() && dq.peekFirst() < i - k) {
                dq.pollFirst();
            }

            nums[i] += nums[dq.peekFirst()];

            while (!dq.isEmpty() && nums[dq.peekLast()] <= nums[i]) {
                dq.pollLast();
            }

            dq.offer(i);
        }

        return nums[n - 1];
    }
}

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Problem Info

Description

Problem 1696: Jump Game VI - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Queue Heap (Priority Queue)Monotonic Queue

Companies

AQR Capital Management

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1696

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 Deque `dq` to store indices. Add the starting index 0 to `dq`.
2
Step 2: Iterate from index 1 to `n-1` of the `nums` array.
3
Step 3: Before updating `nums[i]`, check if the index at the front of `dq` is outside the valid jump range `[i-k, i-1]`. If `dq.peekFirst() < i - k`, it means that the index is too far behind, so remove it from the front of `dq`.
4
Step 4: Update `nums[i]` by adding the maximum score from a previous reachable index, which is `nums[dq.peekFirst()]`. Thus, `nums[i] += nums[dq.peekFirst()]`.
5
Step 5: Maintain the monotonic property of `dq`. While `dq` is not empty and the score at the last index in `dq` (`nums[dq.peekLast()]`) is less than or equal to the current score (`nums[i]`), remove the last index from `dq`. This ensures that `dq` always contains indices with decreasing scores from front to back.
6
Step 6: Add the current index `i` to the back of `dq`.
7
Step 7: After iterating through the array, `nums[n-1]` will contain the maximum score achievable when reaching the last index.
8
Step 8: Return `nums[n-1]`.

💡

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(k)

📝

Why?

The time complexity is dominated by the single loop that iterates through the `nums` array. The monotonic queue operations take amortized O(1) time per element because each element is added and removed at most once. The space complexity is determined by the maximum size of the monotonic queue, which is bounded by `k`.

📖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

Priority Queue (Heap)

A priority queue (max heap) could be used to store (score, index) pairs. For each index `i`, extract the maximum score from the heap whose index is within `k` steps behind `i`. Insert the new score at `i` into the heap.

Time: O(n log k)Space: O(k)

Trade-offs: Priority queue has O(log k) time for insertion and extraction, which results in O(n log k) overall time complexity. While it might be easier to implement, Monotonic Queue provides better time complexity.

✨

Key Insights

1Insight 1: Dynamic Programming (DP) is suitable since the optimal score at index `i` depends on the optimal scores at previous indices within the range `[i-k, i-1]`.
2Insight 2: Using a naive DP approach (checking all possible previous indices in each step) results in O(n*k) time complexity, which can exceed the time limit for larger inputs. A Monotonic Queue helps optimize the DP transitions to O(n).
3Insight 3: The Monotonic Queue stores indices of potentially optimal 'previous' states (indices) to jump from. It maintains a decreasing order of `nums[index]` values so that the front of the queue always contains the index that yields the highest score when jumping to the current index.

📚

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.

Queue

FIFO structure - First In, First Out. Like a line at a store!

Edge Cases

Edge case 1: `nums` array with only one element. The loop does not execute, and the initial value of `nums[0]` is returned, which is correct.
Edge case 2: All negative numbers in `nums`. The algorithm correctly handles negative numbers. The monotonic queue ensures that the index with the least negative (or most positive if any are positive) is always considered. If all numbers are negative it picks the least negative sequence.
Edge case 3: `k` is equal to or greater than the length of the array. The algorithm correctly accounts for this since the index at the front of the deque is checked whether it exceeds the max index.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not maintaining the monotonic property of the queue. If the queue isn't monotonic, you won't be able to guarantee that the index at the front of the queue always leads to the optimal score.

✗Common mistake 2: Forgetting to remove indices from the front of the queue that are out of the jump range. This can lead to incorrect scores and exceeding the time limit.

✗Common mistake 3: Using brute force DP, leading to O(n*k) complexity, which can result in TLE (Time Limit Exceeded).

✅

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 where you want to maximize a certain utility function over a period, with a constraint on the number of resources you can allocate in each step.

🏦Real-world scenario 2: Stock trading where you want to maximize profit over a series of days, with a limit on how many days you can hold a stock before selling it.

🎮Real-world scenario 3: Game playing AI where the goal is to maximize the score given the constraint of the number of moves possible in a turn.

💡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.

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: August 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Jump Game VI"?

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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