题目 #

You are given a 0-indexed integer array nums of length n.

A split at an index i where 0 <= i <= n - 2 is called valid if the product of the first i + 1 elements and the product of the remaining elements are coprime.

• For example, if nums = [2, 3, 3], then a split at the index i = 0 is valid because 2 and 9 are coprime, while a split at the index i = 1 is not valid because 6 and 3 are not coprime. A split at the index i = 2 is not valid because i == n - 1.

Return the smallest index i at which the array can be split validly or -1 if there is no such split.

Two values val1 and val2 are coprime if gcd(val1, val2) == 1 where gcd(val1, val2) is the greatest common divisor of val1 and val2.

Example 1:

Input: nums = [4,7,8,15,3,5]
Output: 2
Explanation: The table above shows the values of the product of the first i + 1 elements, the remaining elements, and their gcd at each index i.
The only valid split is at index 2.

Example 2:

Input: nums = [4,7,15,8,3,5]
Output: -1
Explanation: The table above shows the values of the product of the first i + 1 elements, the remaining elements, and their gcd at each index i.
There is no valid split.

Constraints:

• n == nums.length
• $1 <= n <= 10^4$
• $1 <= nums[i] <= 10^6$

思路1 合并区间 #

分析 #

• 找一个点，让前后互质。就是找一个点，前面的质因子和后面的质因子都不相同
• 转化一下，好像可以用区间来抽象这道题。每个质因子是一个区间，有最左边和最右边。
• 分割点就是找到一个点正好没有切到某个区间，就是说某个左端点大于其左侧的最大的右端点

代码 #

• 需要的数据只有左端点对应的最大的右端点，并不需要每个质数的区间，所以代码上来了个区间合并
• 只需要统计某个点为左端点时，右端点最大的值就好。map存每个质数对应的左端点，数组存每个左端点对应的最大的右端点

 1func getPrimeFactor(n int, handle func(prime int)) {
 2	max := int(math.Sqrt(float64(n))) + 1
 3	for i := 2; i < max && i < n; i++ {
 4		if n%i == 0 {
 5			handle(i)
 6			n /= i
 7			i = 1
 8		}
 9	}
10	if n > 1 {
11		handle(n)
12	}
13}
14
15func findValidSplit(nums []int) int {
16	left := make(map[int]int)
17	right := make([]int, len(nums)) // 左端点为i的右端点最大值
18	for i, v := range nums {
19		getPrimeFactor(v, func(prime int) {
20			if l, ok := left[prime]; ok {
21				right[l] = i
22			} else {
23				left[prime] = i
24			}
25		})
26	}
27	maxR := 0
28	for l, r := range right {
29		if l > maxR {
30			return maxR
31		}
32		if r > maxR {
33			maxR = r
34		}
35	}
36	return -1
37}