Is Jump Game VII easy or hard?

Jump Game VII is rated Medium but often feels closer to hard for many candidates. The challenge is recognizing that naive range checks are too slow and applying a sliding window optimization to reduce the complexity to O(n).

How to solve Jump Game VII in O(n)?

Use a DP array where dp[i] indicates if index i is reachable. Maintain a sliding window count of reachable indices between i - maxJump and i - minJump. If the count is positive and s[i] equals '0', mark dp[i] as reachable. Updating the window incrementally keeps the entire algorithm linear.

What is the best approach for Jump Game VII?

The most efficient approach uses sliding window dynamic programming. Track whether each index is reachable and maintain a running count of reachable indices within the window [i - maxJump, i - minJump]. This avoids scanning the entire range for every position and achieves O(n) time complexity.

Is Jump Game VII asked at Google/Amazon/Meta?

Jump Game style problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variants test reachability, dynamic programming, and sliding window optimization. Interviewers often expect candidates to move from a BFS idea to an O(n) optimized solution.

What data structure is used in Jump Game VII?

Common implementations use a queue for the BFS approach and an array for dynamic programming. The optimized solution also uses a sliding window counter, similar to prefix sum techniques, to track reachable positions efficiently.

What is the time complexity of Jump Game VII?

The optimal solution runs in O(n) time where n is the length of the string. Each index is processed once while maintaining a sliding window of reachable positions. Space complexity is O(n) for the DP or visited array.

Jump Game VII Python or Java solution approach?

Both Python and Java implementations typically follow the sliding window DP method. Maintain a dp array and update a running window count as you iterate through the string. This keeps the complexity O(n) and works efficiently even for large input sizes.

Description Hints 2 Solution Video 12:28

1871. Jump Game VII

Medium35.4% acceptance4 companies

You are given a 0-indexed binary string s and two integers minJump and maxJump. In the beginning, you are standing at index 0, which is equal to '0'. You can move from index i to index j if the following conditions are fulfilled:

i + minJump <= j <= min(i + maxJump, s.length - 1), and
s[j] == '0'.

Return true if you can reach index s.length - 1 in s, or false otherwise.

Example 1:

Input: s = "011010", minJump = 2, maxJump = 3
Output: true
Explanation:
In the first step, move from index 0 to index 3. 
In the second step, move from index 3 to index 5.

Example 2:

Input: s = "01101110", minJump = 2, maxJump = 3
Output: false

Constraints:

2 <= s.length <= 10⁵
s[i] is either '0' or '1'.
s[0] == '0'
1 <= minJump <= maxJump < s.length

Asked at 4 companies

Bloomberg Meta

Google Amazon

Topics

Hints open one at a time

Hint 1

Consider for each reachable index i the interval [i + a, i + b].

Hint 2

Use partial sums to mark the intervals as reachable.

12:28

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