How do you solve Jump Game VI on LeetCode?

Jump Game VI (LeetCode #1696) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log k) time complexity and O(n) space complexity. Key insight: Dynamic programming can significantly reduce the number of calculations by storing intermediate results.

What is the time complexity of Jump Game VI?

The optimal solution for Jump Game VI runs in O(n log k) time and uses O(n) space. The time complexity is O(n log k) due to the operations on the max-heap, where we maintain at most k elements at any time.

What is the brute force solution for Jump Game VI?

The brute force approach for Jump Game VI is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach explores all possible paths from the start index to the end index. For each index, we calculate the maximum score by trying every possible jump within the allowed range. The optimal Optimal Solution approach improves this to O(n log k).

What algorithm or data structure is used to solve Jump Game VI?

Jump Game VI uses the following concepts: Array, Dynamic Programming, Queue, Heap (Priority Queue), Monotonic Queue. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Jump Game VI?

Always think about how you can break down the problem into smaller subproblems. Consider edge cases, such as very small arrays or large negative values. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Jump Game VI?

Not considering the bounds of the array when jumping. Failing to optimize the search for the maximum score, leading to unnecessary calculations.

#1696

Jump Game VI

Medium

ArrayDynamic ProgrammingQueueHeap (Priority Queue)Monotonic QueueDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log k)
Space	O(n)	O(n)

💡

Intuition

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

The optimal solution uses dynamic programming and a max-heap to efficiently track the maximum scores from reachable indices. This avoids redundant calculations and speeds up the process.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dp array where dp[i] represents the maximum score to reach the end starting from index i.
2Step 2: Use a max-heap to keep track of the maximum dp values within the last k indices.
3Step 3: Iterate from the end of the array to the beginning, updating dp[i] based on the maximum value from the heap and the current index's value.

solution.py14 lines

1import heapq
2
3def maxScore(nums, k):
4    n = len(nums)
5    dp = [0] * n
6    max_heap = []
7    dp[-1] = nums[-1]
8    heapq.heappush(max_heap, (-dp[-1], n - 1))
9    for i in range(n - 2, -1, -1):
10        while max_heap and max_heap[0][1] > i + k:
11            heapq.heappop(max_heap)
12        dp[i] = nums[i] + (-max_heap[0][0])
13        heapq.heappush(max_heap, (-dp[i], i))
14    return dp[0]

ℹ

Complexity note: The time complexity is O(n log k) due to the operations on the max-heap, where we maintain at most k elements at any time.

1Dynamic programming can significantly reduce the number of calculations by storing intermediate results.
2Using a max-heap allows us to efficiently track the best scores within the allowed jump range.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.