# 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

• class Solution {
public int countCompleteSubarrays(int[] nums) {
Set<Integer> s = new HashSet<>();
for (int x : nums) {
}
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) {
if (s.size() == cnt) {
++ans;
}
}
}
return ans;
}
}

• 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;
}
};

• 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:]:
if len(s) == cnt:
ans += 1
return ans


• 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
}

• 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 = 0; i < n; ++i) {
s.clear();
for (let j = i; j < n; ++j) {
if (s.size === cnt) {
++ans;
}
}
}
return ans;
}


• use std::collections::HashSet;
impl Solution {
pub fn count_complete_subarrays(nums: Vec<i32>) -> i32 {
let mut set: HashSet<&i32> = nums.iter().collect();
let n = nums.len();
let m = set.len();
let mut ans = 0;
for i in 0..n {
set.clear();
for j in i..n {
set.insert(&nums[j]);
if set.len() == m {
ans += n - j;
break;
}
}
}
ans as i32
}
}