How do you solve Jump Game VII on LeetCode?

Jump Game VII (LeetCode #1871) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the jump constraints is crucial for determining reachable indices.

What is the brute force solution for Jump Game VII?

The brute force approach for Jump Game VII is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach explores all possible jumps from the starting index to see if we can reach the end. It checks each index to see if it can be reached based on the jump constraints. The optimal Optimal Solution approach improves this to O(n).

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

Jump Game VII uses the following concepts: String, Dynamic Programming, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Breadth-First Search (BFS).

What are the interview tips for Jump Game VII?

Always clarify the constraints and edge cases with the interviewer. Think about how you can optimize your solution after the brute force approach. Practice explaining your thought process clearly as you code.

What are common mistakes when solving Jump Game VII?

Not considering the case where you might skip over reachable indices. Failing to properly update the farthest reachable index can lead to incorrect results.

#1871

Jump Game VII

Medium

StringDynamic ProgrammingSliding WindowPrefix SumSliding WindowBreadth-First Search (BFS)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach to keep track of the farthest index we can reach. This reduces unnecessary checks and efficiently determines if we can reach the end.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to track the farthest index we can reach.
2Step 2: Iterate through the string, updating the farthest index based on the current index and the jump limits.
3Step 3: If at any point the farthest index reaches or exceeds the last index, return true.
4Step 4: If the loop completes without reaching the last index, return false.

solution.py9 lines

1def canReach(s, minJump, maxJump):
2    n = len(s)
3    farthest = 0
4    for i in range(n):
5        if i > farthest:
6            return False
7        if s[i] == '0' and i + minJump <= n - 1:
8            farthest = max(farthest, i + maxJump)
9    return farthest >= n - 1

ℹ

Complexity note: This complexity is linear because we only traverse the string once, updating the farthest reachable index without nested loops.

1Understanding the jump constraints is crucial for determining reachable indices.
2Using a sliding window approach can significantly reduce the number of checks needed.

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