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

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