Is Minimum Jumps to Reach End via Prime Teleportation easy or hard?

The problem is typically rated Medium. The BFS portion is straightforward, but recognizing that teleport edges can be grouped by prime factors and optimized with hashing and number theory is the key challenge.

Minimum Jumps to Reach End via Prime Teleportation Python/Java solution

Implement BFS starting from index 0. Factorize each array value and store indices grouped by their prime factors in a hash map. During traversal, push adjacent indices and teleport indices into the queue, marking visited nodes and clearing processed prime groups for efficiency.

How to solve Minimum Jumps to Reach End via Prime Teleportation in O(n)?

Model the array as a graph and run BFS from index 0. Precompute prime factors for each value and store indices in a hash map keyed by prime factor. During BFS, jump to neighbors i-1, i+1, and all indices sharing a prime factor, then clear that factor's list to avoid repeated processing.

What is the best approach for Minimum Jumps to Reach End via Prime Teleportation?

The most efficient approach uses Breadth-First Search combined with a hash map from prime factors to array indices. Each index is treated as a graph node, and teleport edges are created between indices sharing a prime factor. By processing each prime group only once, the algorithm runs in about O(n log V) time.

Is Minimum Jumps to Reach End via Prime Teleportation asked at Google/Amazon/Meta?

Problems combining BFS with number theory and hash maps appear frequently in interviews at companies like Google, Amazon, and Meta. Variants of graph traversal with teleport rules are common because they test graph modeling, optimization, and data structure design.

What data structure is used in Minimum Jumps to Reach End via Prime Teleportation?

The solution primarily uses a queue for Breadth-First Search, a hash table mapping prime factors to indices, and a visited array to prevent revisiting nodes. Prime factorization or sieve preprocessing is also used to identify teleport connections.

What is the time complexity of Minimum Jumps to Reach End via Prime Teleportation?

The optimized solution runs in O(n log V) time where n is the array length and V is the maximum value in the array. The log V factor comes from prime factorization. Space complexity is O(n + P) for the BFS queue, visited array, and prime-to-indices hash map.

Description Hints 3 Solution Video 28:54

3629. Minimum Jumps to Reach End via Prime Teleportation

Medium44.6% acceptance4 companies

You are given an integer array nums of length n.

You start at index 0, and your goal is to reach index n - 1.

From any index i, you may perform one of the following operations:

Adjacent Step: Jump to index i + 1 or i - 1, if the index is within bounds.
Prime Teleportation: If nums[i] is a prime number p, you may instantly jump to any index j != i such that nums[j] % p == 0.

Return the minimum number of jumps required to reach index n - 1.

Example 1:

Input: nums = [1,2,4,6]

Output: 2

Explanation:

One optimal sequence of jumps is:

Start at index i = 0. Take an adjacent step to index 1.
At index i = 1, nums[1] = 2 is a prime number. Therefore, we teleport to index i = 3 as nums[3] = 6 is divisible by 2.

Thus, the answer is 2.

Example 2:

Input: nums = [2,3,4,7,9]

Output: 2

Explanation:

One optimal sequence of jumps is:

Start at index i = 0. Take an adjacent step to index i = 1.
At index i = 1, nums[1] = 3 is a prime number. Therefore, we teleport to index i = 4 since nums[4] = 9 is divisible by 3.

Thus, the answer is 2.

Example 3:

Input: nums = [4,6,5,8]

Output: 3

Explanation:

Since no teleportation is possible, we move through 0 → 1 → 2 → 3. Thus, the answer is 3.

Constraints:

1 <= n == nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Asked at 4 companies

Uber

Google Amazon Meta

Topics

Hints open one at a time

Hint 1

Use a breadth-first search.

Hint 2

Precompute prime factors of each nums[i] via a sieve, and build a bucket bucket[p] mapping each prime p to all indices j with nums[j] % p == 0.

Hint 3

During the BFS, when at index i, enqueue its adjacent steps (i+1 and i-1) and all indices in bucket[p] for each prime p dividing nums[i], then clear bucket[p] so each prime's bucket is visited only once.

28:54

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1class Solution:
2    def minJumps(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[1,2,4,6]

Input

[2,3,4,7,9]

Input

[4,6,5,8]

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