Question

2464. Minimum Subarrays in a Valid Split

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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:
  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

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;
  }
};

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)
}