# 2584. Split the Array to Make Coprime Products

## Description

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 <= 104
• 1 <= nums[i] <= 106

## Solutions

• class Solution {
public int findValidSplit(int[] nums) {
Map<Integer, Integer> first = new HashMap<>();
int n = nums.length;
int[] last = new int[n];
for (int i = 0; i < n; ++i) {
last[i] = i;
}
for (int i = 0; i < n; ++i) {
int x = nums[i];
for (int j = 2; j <= x / j; ++j) {
if (x % j == 0) {
if (first.containsKey(j)) {
last[first.get(j)] = i;
} else {
first.put(j, i);
}
while (x % j == 0) {
x /= j;
}
}
}
if (x > 1) {
if (first.containsKey(x)) {
last[first.get(x)] = i;
} else {
first.put(x, i);
}
}
}
int mx = last[0];
for (int i = 0; i < n; ++i) {
if (mx < i) {
return mx;
}
mx = Math.max(mx, last[i]);
}
return -1;
}
}

• class Solution {
public:
int findValidSplit(vector<int>& nums) {
unordered_map<int, int> first;
int n = nums.size();
vector<int> last(n);
iota(last.begin(), last.end(), 0);
for (int i = 0; i < n; ++i) {
int x = nums[i];
for (int j = 2; j <= x / j; ++j) {
if (x % j == 0) {
if (first.count(j)) {
last[first[j]] = i;
} else {
first[j] = i;
}
while (x % j == 0) {
x /= j;
}
}
}
if (x > 1) {
if (first.count(x)) {
last[first[x]] = i;
} else {
first[x] = i;
}
}
}
int mx = last[0];
for (int i = 0; i < n; ++i) {
if (mx < i) {
return mx;
}
mx = max(mx, last[i]);
}
return -1;
}
};

• class Solution:
def findValidSplit(self, nums: List[int]) -> int:
first = {}
n = len(nums)
last = list(range(n))
for i, x in enumerate(nums):
j = 2
while j <= x // j:
if x % j == 0:
if j in first:
last[first[j]] = i
else:
first[j] = i
while x % j == 0:
x //= j
j += 1
if x > 1:
if x in first:
last[first[x]] = i
else:
first[x] = i
mx = last[0]
for i, x in enumerate(last):
if mx < i:
return mx
mx = max(mx, x)
return -1


• func findValidSplit(nums []int) int {
first := map[int]int{}
n := len(nums)
last := make([]int, n)
for i := range last {
last[i] = i
}
for i, x := range nums {
for j := 2; j <= x/j; j++ {
if x%j == 0 {
if k, ok := first[j]; ok {
last[k] = i
} else {
first[j] = i
}
for x%j == 0 {
x /= j
}
}
}
if x > 1 {
if k, ok := first[x]; ok {
last[k] = i
} else {
first[x] = i
}
}
}
mx := last[0]
for i, x := range last {
if mx < i {
return mx
}
mx = max(mx, x)
}
return -1
}