minimum-subarrays-in-a-valid-split.cpp

// 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;
    }
};