Medium
Minimum Subarrays in a Valid Split — C++
Full explanation · Time O(n^2 * logr) · Space O(n)
// Time: O(n^2 * logr), r = max(nums)
// Space: O(n)
// dp
class Solution {
public:
int validSubarraySplit(vector<int>& nums) {
static const int INF = numeric_limits<int>::max();
vector<int> dp(size(nums) + 1, INF); // dp[i]: min number of subarrays in nums[:i]
dp[0] = 0;
for (int i = 1; i <= size(nums); ++i) {
for (int j = 0; j < i; ++j) {
if (gcd(nums[j], nums[i - 1]) != 1) {
if (dp[j] != INF) {
dp[i] = min(dp[i], dp[j] + 1);
}
}
}
}
return dp.back() != INF ? dp.back() : -1;
}
};