LeetCode #3629 — MEDIUM

Minimum Jumps to Reach End via Prime Teleportation

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math

Example 1

[1,2,4,6]

Example 2

[2,3,4,7,9]

Example 3

[4,6,5,8]

Step 02

Core Insight

What unlocks the optimal approach

Use a breadth-first search.
Precompute prime factors of each <code>nums[i]</code> via a sieve, and build a bucket <code>bucket[p]</code> mapping each prime <code>p</code> to all indices <code>j</code> with <code>nums[j] % p == 0</code>.
During the BFS, when at index <code>i</code>, enqueue its adjacent steps (<code>i+1</code> and <code>i-1</code>) and all indices in <code>bucket[p]</code> for each prime <code>p</code> dividing <code>nums[i]</code>, then clear <code>bucket[p]</code> so each prime's bucket is visited only once.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3629: Minimum Jumps to Reach End via Prime Teleportation
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3629: Minimum Jumps to Reach End via Prime Teleportation
// package main
// 
// // https://space.bilibili.com/206214
// const mx = 1_000_001
// 
// var primeFactors = [mx][]int{}
// 
// func init() {
// 	// 预处理每个数的质因子列表，思路同埃氏筛
// 	for i := 2; i < mx; i++ {
// 		if primeFactors[i] == nil { // i 是质数
// 			for j := i; j < mx; j += i { // i 的倍数有质因子 i
// 				primeFactors[j] = append(primeFactors[j], i)
// 			}
// 		}
// 	}
// }
// 
// func minJumps(nums []int) (ans int) {
// 	n := len(nums)
// 	groups := map[int][]int{}
// 	for i, x := range nums {
// 		if len(primeFactors[x]) == 1 { // x 是质数
// 			groups[x] = append(groups[x], i)
// 		}
// 	}
// 
// 	vis := make([]bool, n)
// 	vis[n-1] = true
// 	q := []int{n - 1}
// 	for {
// 		tmp := q
// 		q = nil
// 		for _, i := range tmp {
// 			if i == 0 {
// 				return
// 			}
// 			if !vis[i-1] {
// 				vis[i-1] = true
// 				q = append(q, i-1)
// 			}
// 			if i < n-1 && !vis[i+1] {
// 				vis[i+1] = true
// 				q = append(q, i+1)
// 			}
// 			// 逆向思维：从 i 倒着跳到 nums[i] 的质因子 p 的下标 j
// 			for _, p := range primeFactors[nums[i]] {
// 				for _, j := range groups[p] {
// 					if !vis[j] {
// 						vis[j] = true
// 						q = append(q, j)
// 					}
// 				}
// 				delete(groups, p) // 避免重复访问下标列表
// 			}
// 		}
// 		ans++
// 	}
// }
// 
// func minJumps1(nums []int) (ans int) {
// 	n := len(nums)
// 	groups := map[int][]int{}
// 	for i, x := range nums {
// 		for _, p := range primeFactors[x] {
// 			groups[p] = append(groups[p], i) // 对于质数 p，可以跳到下标 i
// 		}
// 	}
// 
// 	vis := make([]bool, n)
// 	vis[0] = true
// 	q := []int{0}
// 	for {
// 		tmp := q
// 		q = nil
// 		for _, i := range tmp {
// 			if i == n-1 {
// 				return
// 			}
// 			idx := groups[nums[i]]
// 			idx = append(idx, i+1)
// 			if i > 0 {
// 				idx = append(idx, i-1)
// 			}
// 			for _, j := range idx { // 可以从 i 跳到 j
// 				if !vis[j] {
// 					vis[j] = true
// 					q = append(q, j)
// 				}
// 			}
// 			delete(groups, nums[i]) // 避免重复访问下标列表
// 		}
// 		ans++
// 	}
// }

// Accepted solution for LeetCode #3629: Minimum Jumps to Reach End via Prime Teleportation
package main

// https://space.bilibili.com/206214
const mx = 1_000_001

var primeFactors = [mx][]int{}

func init() {
	// 预处理每个数的质因子列表，思路同埃氏筛
	for i := 2; i < mx; i++ {
		if primeFactors[i] == nil { // i 是质数
			for j := i; j < mx; j += i { // i 的倍数有质因子 i
				primeFactors[j] = append(primeFactors[j], i)
			}
		}
	}
}

func minJumps(nums []int) (ans int) {
	n := len(nums)
	groups := map[int][]int{}
	for i, x := range nums {
		if len(primeFactors[x]) == 1 { // x 是质数
			groups[x] = append(groups[x], i)
		}
	}

	vis := make([]bool, n)
	vis[n-1] = true
	q := []int{n - 1}
	for {
		tmp := q
		q = nil
		for _, i := range tmp {
			if i == 0 {
				return
			}
			if !vis[i-1] {
				vis[i-1] = true
				q = append(q, i-1)
			}
			if i < n-1 && !vis[i+1] {
				vis[i+1] = true
				q = append(q, i+1)
			}
			// 逆向思维：从 i 倒着跳到 nums[i] 的质因子 p 的下标 j
			for _, p := range primeFactors[nums[i]] {
				for _, j := range groups[p] {
					if !vis[j] {
						vis[j] = true
						q = append(q, j)
					}
				}
				delete(groups, p) // 避免重复访问下标列表
			}
		}
		ans++
	}
}

func minJumps1(nums []int) (ans int) {
	n := len(nums)
	groups := map[int][]int{}
	for i, x := range nums {
		for _, p := range primeFactors[x] {
			groups[p] = append(groups[p], i) // 对于质数 p，可以跳到下标 i
		}
	}

	vis := make([]bool, n)
	vis[0] = true
	q := []int{0}
	for {
		tmp := q
		q = nil
		for _, i := range tmp {
			if i == n-1 {
				return
			}
			idx := groups[nums[i]]
			idx = append(idx, i+1)
			if i > 0 {
				idx = append(idx, i-1)
			}
			for _, j := range idx { // 可以从 i 跳到 j
				if !vis[j] {
					vis[j] = true
					q = append(q, j)
				}
			}
			delete(groups, nums[i]) // 避免重复访问下标列表
		}
		ans++
	}
}

# Accepted solution for LeetCode #3629: Minimum Jumps to Reach End via Prime Teleportation
# Time:  precompute: O(r)
#        runtime:    O(nlogr)
# Space: O(r + nlogr)

import collections


# number theory, bfs
def linear_sieve_of_eratosthenes(n):  # Time: O(n), Space: O(n)
    primes = []
    spf = [-1]*(n+1)  # the smallest prime factor
    for i in xrange(2, n+1):
        if spf[i] == -1:
            spf[i] = i
            primes.append(i)
        for p in primes:
            if i*p > n or p > spf[i]:
                break
            spf[i*p] = p
    return spf


MAX_NUM = 10**6
SPF = linear_sieve_of_eratosthenes(MAX_NUM)
class Solution(object):
    def minJumps(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        adj = collections.defaultdict(list)
        for i, x in enumerate(nums):
            while x != 1:
                p = SPF[x]
                while x%p == 0:
                    x //= p
                adj[p].append(i)
        dist = [-1]*len(nums)
        dist[0] = 0
        q = [0]
        while q:
            new_q = []
            for i in q:
                if i == len(nums)-1:
                    return dist[-1]
                for di in (-1, +1):
                    ni = i+di
                    if 0 <= ni < len(nums) and dist[ni] == -1:
                        dist[ni] = dist[i]+1
                        new_q.append(ni)
                p = nums[i]
                if SPF[p] != p or p not in adj:
                    continue
                for ni in adj[p]:
                    if dist[ni] != -1:
                        continue
                    dist[ni] = dist[i]+1
                    new_q.append(ni)
                del adj[p]
            q = new_q
        return -1

// Accepted solution for LeetCode #3629: Minimum Jumps to Reach End via Prime Teleportation
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3629: Minimum Jumps to Reach End via Prime Teleportation
// package main
// 
// // https://space.bilibili.com/206214
// const mx = 1_000_001
// 
// var primeFactors = [mx][]int{}
// 
// func init() {
// 	// 预处理每个数的质因子列表，思路同埃氏筛
// 	for i := 2; i < mx; i++ {
// 		if primeFactors[i] == nil { // i 是质数
// 			for j := i; j < mx; j += i { // i 的倍数有质因子 i
// 				primeFactors[j] = append(primeFactors[j], i)
// 			}
// 		}
// 	}
// }
// 
// func minJumps(nums []int) (ans int) {
// 	n := len(nums)
// 	groups := map[int][]int{}
// 	for i, x := range nums {
// 		if len(primeFactors[x]) == 1 { // x 是质数
// 			groups[x] = append(groups[x], i)
// 		}
// 	}
// 
// 	vis := make([]bool, n)
// 	vis[n-1] = true
// 	q := []int{n - 1}
// 	for {
// 		tmp := q
// 		q = nil
// 		for _, i := range tmp {
// 			if i == 0 {
// 				return
// 			}
// 			if !vis[i-1] {
// 				vis[i-1] = true
// 				q = append(q, i-1)
// 			}
// 			if i < n-1 && !vis[i+1] {
// 				vis[i+1] = true
// 				q = append(q, i+1)
// 			}
// 			// 逆向思维：从 i 倒着跳到 nums[i] 的质因子 p 的下标 j
// 			for _, p := range primeFactors[nums[i]] {
// 				for _, j := range groups[p] {
// 					if !vis[j] {
// 						vis[j] = true
// 						q = append(q, j)
// 					}
// 				}
// 				delete(groups, p) // 避免重复访问下标列表
// 			}
// 		}
// 		ans++
// 	}
// }
// 
// func minJumps1(nums []int) (ans int) {
// 	n := len(nums)
// 	groups := map[int][]int{}
// 	for i, x := range nums {
// 		for _, p := range primeFactors[x] {
// 			groups[p] = append(groups[p], i) // 对于质数 p，可以跳到下标 i
// 		}
// 	}
// 
// 	vis := make([]bool, n)
// 	vis[0] = true
// 	q := []int{0}
// 	for {
// 		tmp := q
// 		q = nil
// 		for _, i := range tmp {
// 			if i == n-1 {
// 				return
// 			}
// 			idx := groups[nums[i]]
// 			idx = append(idx, i+1)
// 			if i > 0 {
// 				idx = append(idx, i-1)
// 			}
// 			for _, j := range idx { // 可以从 i 跳到 j
// 				if !vis[j] {
// 					vis[j] = true
// 					q = append(q, j)
// 				}
// 			}
// 			delete(groups, nums[i]) // 避免重复访问下标列表
// 		}
// 		ans++
// 	}
// }

// Accepted solution for LeetCode #3629: Minimum Jumps to Reach End via Prime Teleportation
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3629: Minimum Jumps to Reach End via Prime Teleportation
// package main
// 
// // https://space.bilibili.com/206214
// const mx = 1_000_001
// 
// var primeFactors = [mx][]int{}
// 
// func init() {
// 	// 预处理每个数的质因子列表，思路同埃氏筛
// 	for i := 2; i < mx; i++ {
// 		if primeFactors[i] == nil { // i 是质数
// 			for j := i; j < mx; j += i { // i 的倍数有质因子 i
// 				primeFactors[j] = append(primeFactors[j], i)
// 			}
// 		}
// 	}
// }
// 
// func minJumps(nums []int) (ans int) {
// 	n := len(nums)
// 	groups := map[int][]int{}
// 	for i, x := range nums {
// 		if len(primeFactors[x]) == 1 { // x 是质数
// 			groups[x] = append(groups[x], i)
// 		}
// 	}
// 
// 	vis := make([]bool, n)
// 	vis[n-1] = true
// 	q := []int{n - 1}
// 	for {
// 		tmp := q
// 		q = nil
// 		for _, i := range tmp {
// 			if i == 0 {
// 				return
// 			}
// 			if !vis[i-1] {
// 				vis[i-1] = true
// 				q = append(q, i-1)
// 			}
// 			if i < n-1 && !vis[i+1] {
// 				vis[i+1] = true
// 				q = append(q, i+1)
// 			}
// 			// 逆向思维：从 i 倒着跳到 nums[i] 的质因子 p 的下标 j
// 			for _, p := range primeFactors[nums[i]] {
// 				for _, j := range groups[p] {
// 					if !vis[j] {
// 						vis[j] = true
// 						q = append(q, j)
// 					}
// 				}
// 				delete(groups, p) // 避免重复访问下标列表
// 			}
// 		}
// 		ans++
// 	}
// }
// 
// func minJumps1(nums []int) (ans int) {
// 	n := len(nums)
// 	groups := map[int][]int{}
// 	for i, x := range nums {
// 		for _, p := range primeFactors[x] {
// 			groups[p] = append(groups[p], i) // 对于质数 p，可以跳到下标 i
// 		}
// 	}
// 
// 	vis := make([]bool, n)
// 	vis[0] = true
// 	q := []int{0}
// 	for {
// 		tmp := q
// 		q = nil
// 		for _, i := range tmp {
// 			if i == n-1 {
// 				return
// 			}
// 			idx := groups[nums[i]]
// 			idx = append(idx, i+1)
// 			if i > 0 {
// 				idx = append(idx, i-1)
// 			}
// 			for _, j := range idx { // 可以从 i 跳到 j
// 				if !vis[j] {
// 					vis[j] = true
// 					q = append(q, j)
// 				}
// 			}
// 			delete(groups, nums[i]) // 避免重复访问下标列表
// 		}
// 		ans++
// 	}
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.