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Formatted question description: https://leetcode.ca/all/2464.html

2464. Minimum Subarrays in a Valid Split

Description

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.

Return the minimum number of subarrays in a valid subarray splitting of nums. If a valid subarray splitting is not possible, return -1.

Note that:

  • The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.
  • A subarray is a contiguous non-empty part of an array.

 

Example 1:

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.

Example 2:

Input: nums = [3,5]
Output: 2
Explanation: We can create a valid split in the following way: [3] | [5].
- The starting element of the 1st subarray is 3 and the ending is 3. Their greatest common divisor is 3, which is greater than 1.
- The starting element of the 2nd subarray is 5 and the ending is 5. Their greatest common divisor is 5, 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.

Example 3:

Input: nums = [1,2,1]
Output: -1
Explanation: It is impossible to create valid split.

 

Constraints:

  • 1 <= nums.length <= 1000
  • 1 <= nums[i] <= 105

Solutions

Solution 1: Memoization Search

We design a function $dfs(i)$ to represent the minimum number of partitions starting from index $i$. For index $i$, we can enumerate all partition points $j$, i.e., $i \leq j < n$, where $n$ is the length of the array. For each partition point $j$, we need to determine whether the greatest common divisor of $nums[i]$ and $nums[j]$ is greater than $1$. If it is greater than $1$, we can partition, and the number of partitions is $1 + dfs(j + 1)$; otherwise, the number of partitions is $+\infty$. Finally, we take the minimum of all partition numbers.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

  • class Solution {
    private int n;
    private int[] f;
    private int[] nums;
    private int inf = 0x3f3f3f3f;

    public int validSubarraySplit(int[] nums) {
        n = nums.length;
        f = new int[n];
        this.nums = nums;
        int ans = dfs(0);
        return ans < inf ? ans : -1;
    }

    private int dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i] > 0) {
            return f[i];
        }
        int ans = inf;
        for (int j = i; j < n; ++j) {
            if (gcd(nums[i], nums[j]) > 1) {
                ans = Math.min(ans, 1 + dfs(j + 1));
            }
        }
        f[i] = ans;
        return ans;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

  • class Solution {
public:
    const int inf = 0x3f3f3f3f;
    int validSubarraySplit(vector<int>& nums) {
        int n = nums.size();
        vector<int> f(n);
        function<int(int)> dfs = [&](int i) -> int {
            if (i >= n) return 0;
            if (f[i]) return f[i];
            int ans = inf;
            for (int j = i; j < n; ++j) {
                if (__gcd(nums[i], nums[j]) > 1) {
                    ans = min(ans, 1 + dfs(j + 1));
                }
            }
            f[i] = ans;
            return ans;
        };
        int ans = dfs(0);
        return ans < inf ? ans : -1;
    }
};

  • class Solution:
    def validSubarraySplit(self, nums: List[int]) -> int:
        @cache
        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


  • func validSubarraySplit(nums []int) int {
	n := len(nums)
	f := make([]int, n)
	var dfs func(int) int
	const inf int = 0x3f3f3f3f
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if f[i] > 0 {
			return f[i]
		}
		ans := inf
		for j := i; j < n; j++ {
			if gcd(nums[i], nums[j]) > 1 {
				ans = min(ans, 1+dfs(j+1))
			}
		}
		f[i] = ans
		return ans
	}
	ans := dfs(0)
	if ans < inf {
		return ans
	}
	return -1
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

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