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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

Solution 1

Thinking

Find the leftmost \(i\) such that the prefix product is coprime to the suffix product. The products overflow; coprimality means the two sides share no prime.

The first and last occurrences of each prime must lie on the same side of the cut. Record the first index of every prime and extend that index's cover to the last occurrence. Scan these covers: if the running right end ends before index \(i\), \(i-1\) is a valid split.

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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

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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;
    }
}

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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;
    }
};

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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
}

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