How do you solve Jump Game IX on LeetCode?

Jump Game IX (LeetCode #3660) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem can be visualized as a directed graph of jumps.

What is the time complexity of Jump Game IX?

The optimal solution for Jump Game IX runs in O(n) time and uses O(n) space. We traverse the array once and use a stack to store potential maximums, leading to linear complexity.

What is the brute force solution for Jump Game IX?

The brute force approach for Jump Game IX is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible jump from each index to find the maximum reachable value. This involves exploring all valid paths, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

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

Jump Game IX uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Graph Traversal, Dynamic Programming.

What are the interview tips for Jump Game IX?

Clarify the jump rules before coding. Discuss the time complexity of your approach. Think about edge cases and test them.

What are common mistakes when solving Jump Game IX?

Not considering both forward and backward jumps. Overlooking edge cases where no jumps are possible.

#3660

Jump Game IX

Medium

ArrayDynamic ProgrammingGraph TraversalDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use a stack to efficiently track reachable values. By processing indices in reverse, we can ensure that we always find the maximum value reachable from each index.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack and an answer array filled with zeros.
2Step 2: Traverse the array from right to left, using the stack to track potential maximum values.
3Step 3: For each index, pop from the stack until the top value is valid, then update the answer.

solution.py10 lines

1def jump_game_ix(nums):
2    n = len(nums)
3    ans = [0] * n
4    stack = []
5    for i in range(n - 1, -1, -1):
6        while stack and (stack[-1] >= nums[i] or (i < stack[-1] and nums[stack[-1]] < nums[i])):
7            stack.pop()
8        ans[i] = max(nums[i], stack[-1] if stack else nums[i])
9        stack.append(nums[i])
10    return ans

ℹ

Complexity note: We traverse the array once and use a stack to store potential maximums, leading to linear complexity.

1The problem can be visualized as a directed graph of jumps.
2Using a stack allows efficient tracking of maximum reachable values.

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