Leetcode #2464: Minimum Subarrays in a Valid Split
In this guide, we solve Leetcode #2464 Minimum Subarrays in a Valid Split in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer array nums. Splitting of an integer array nums into subarrays is valid if: the greatest common divisor of the first and last elements of each subarray is greater than 1, and each element of nums belongs to exactly one subarray.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, Math, Dynamic Programming, Number Theory
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: nums = [2,6,3,4,3]
Output: 2
Explanation: We can create a valid split in the following way: [2,6] | [3,4,3].
- The starting element of the 1st subarray is 2 and the ending is 6. Their greatest common divisor is 2, which is greater than 1.
- The starting element of the 2nd subarray is 3 and the ending is 3. Their greatest common divisor is 3, which is greater than 1.
It can be proved that 2 is the minimum number of subarrays that we can obtain in a valid split.
Python Solution
class Solution:
def validSubarraySplit(self, nums: List[int]) -> int:
def dfs(i):
if i >= n:
return 0
ans = inf
for j in range(i, n):
if gcd(nums[i], nums[j]) > 1:
ans = min(ans, 1 + dfs(j + 1))
return ans
n = len(nums)
ans = dfs(0)
dfs.cache_clear()
return ans if ans < inf else -1
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.