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2799. Count Complete Subarrays in an Array

Description

You are given an array nums consisting of positive integers.

We call a subarray of an array complete if the following condition is satisfied:

  • The number of distinct elements in the subarray is equal to the number of distinct elements in the whole array.

Return the number of complete subarrays.

A subarray is a contiguous non-empty part of an array.

 

Example 1:

Input: nums = [1,3,1,2,2]
Output: 4
Explanation: The complete subarrays are the following: [1,3,1,2], [1,3,1,2,2], [3,1,2] and [3,1,2,2].

Example 2:

Input: nums = [5,5,5,5]
Output: 10
Explanation: The array consists only of the integer 5, so any subarray is complete. The number of subarrays that we can choose is 10.

 

Constraints:

  • 1 <= nums.length <= 1000
  • 1 <= nums[i] <= 2000

Solutions

Solution 1

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class Solution:
    def countCompleteSubarrays(self, nums: List[int]) -> int:
        cnt = len(set(nums))
        ans, n = 0, len(nums)
        for i in range(n):
            s = set()
            for x in nums[i:]:
                s.add(x)
                if len(s) == cnt:
                    ans += 1
        return ans

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class Solution {
    public int countCompleteSubarrays(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int x : nums) {
            s.add(x);
        }
        int cnt = s.size();
        int ans = 0, n = nums.length;
        for (int i = 0; i < n; ++i) {
            s.clear();
            for (int j = i; j < n; ++j) {
                s.add(nums[j]);
                if (s.size() == cnt) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countCompleteSubarrays(vector<int>& nums) {
        unordered_set<int> s(nums.begin(), nums.end());
        int cnt = s.size();
        int ans = 0, n = nums.size();
        for (int i = 0; i < n; ++i) {
            s.clear();
            for (int j = i; j < n; ++j) {
                s.insert(nums[j]);
                if (s.size() == cnt) {
                    ++ans;
                }
            }
        }
        return ans;
    }
};

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func countCompleteSubarrays(nums []int) (ans int) {
    s := map[int]bool{}
    for _, x := range nums {
        s[x] = true
    }
    cnt := len(s)
    for i := range nums {
        s = map[int]bool{}
        for _, x := range nums[i:] {
            s[x] = true
            if len(s) == cnt {
                ans++
            }
        }
    }
    return
}

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function countCompleteSubarrays(nums: number[]): number {
    const s: Set<number> = new Set(nums);
    const cnt = s.size;
    const n = nums.length;
    let ans = 0;
    for (let i =