Medium
Minimum Subarrays in a Valid Split — Python
Full explanation · Time O(n^2 * logr) · Space O(n)
# Time: O(n^2 * logr), r = max(nums)
# Space: O(n)
# dp
class Solution(object):
def validSubarraySplit(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
def gcd(a, b):
while b:
a, b = b, a%b
return a
dp = [float("inf")]*(len(nums)+1) # dp[i]: min number of subarrays in nums[:i]
dp[0] = 0
for i in xrange(1, len(nums)+1):
for j in xrange(i):
if gcd(nums[j], nums[i-1]) != 1:
dp[i] = min(dp[i], dp[j]+1)
return dp[-1] if dp[-1] != float("inf") else -1