README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/2500-2599/2584.Split%20the%20Array%20to%20Make%20Coprime%20Products/README_EN.md	2159	Weekly Contest 335 Q3	Array Hash Table Math Number Theory

2584. Split the Array to Make Coprime Products

中文文档

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}