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 <= 10001 <= nums[i] <= 2000
Solutions
Solution 1: Hash Table + Enumeration
First, we use a hash table to count the number of distinct elements in the array, denoted as $cnt$.
Next, we enumerate the left endpoint index $i$ of the subarray and maintain a set $s$ to store the elements in the subarray. Each time we move the right endpoint index $j$ to the right, we add $nums[j]$ to the set $s$ and check whether the size of the set $s$ equals $cnt$. If it equals $cnt$, it means the current subarray is a complete subarray, and we increment the answer by $1$.
After the enumeration ends, we return the answer.
Time complexity: $O(n^2)$, Space complexity: $O(n)$, where $n$ is the length of the array.
Python3
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
Java
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;
}
}
C++
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;
}
};
Go
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
}
TypeScript
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) {
s.add(nums[j]);
if (s.size === cnt) {
++ans;
}
}
}
return ans;
}
Rust
use std::collections::HashSet;
impl Solution {
pub fn count_complete_subarrays(nums: Vec<i32>) -> i32 {
let mut s = HashSet::new();
for &x in &nums {
s.insert(x);
}
let cnt = s.len();
let n = nums.len();
let mut ans = 0;
for i in 0..n {
s.clear();
for j in i..n {
s.insert(nums[j]);
if s.len() == cnt {
ans += 1;
}
}
}
ans
}
}
Solution 2: Hash Table + Two Pointers
Similar to Solution 1, we can use a hash table to count the number of distinct elements in the array, denoted as $cnt$.
Next, we use two pointers to maintain a sliding window, where the right endpoint index is $j$ and the left endpoint index is $i$.
Each time we fix the left endpoint index $i$, we move the right endpoint index $j$ to the right. When the number of distinct elements in the sliding window equals $cnt$, it means that all subarrays from the left endpoint index $i$ to the right endpoint index $j$ and beyond are complete subarrays. We then increment the answer by $n - j$, where $n$ is the length of the array. Afterward, we move the left endpoint index $i$ one step to the right and repeat the process.
Time complexity: $O(n)$, Space complexity: $O(n)$, where $n$ is the length of the array.
Python3
class Solution:
def countCompleteSubarrays(self, nums: List[int]) -> int:
cnt = len(set(nums))
d = Counter()
ans, n = 0, len(nums)
i = 0
for j, x in enumerate(nums):
d[x] += 1
while len(d) == cnt:
ans += n - j
d[nums[i]] -= 1
if d[nums[i]] == 0:
d.pop(nums[i])
i += 1
return ans
Java
class Solution {
public int countCompleteSubarrays(int[] nums) {
Map<Integer, Integer> d = new HashMap<>();
for (int x : nums) {
d.put(x, 1);
}
int cnt = d.size();
int ans = 0, n = nums.length;
d.clear();
for (int i = 0, j = 0; j < n; ++j) {
d.merge(nums[j], 1, Integer::sum);
while (d.size() == cnt) {
ans += n - j;
if (d.merge(nums[i], -1, Integer::sum) == 0) {
d.remove(nums[i]);
}
++i;
}
}
return ans;
}
}
C++
class Solution {
public:
int countCompleteSubarrays(vector<int>& nums) {
unordered_map<int, int> d;
for (int x : nums) {
d[x] = 1;
}
int cnt = d.size();
d.clear();
int ans = 0, n = nums.size();
for (int i = 0, j = 0; j < n; ++j) {
d[nums[j]]++;
while (d.size() == cnt) {
ans += n - j;
if (--d[nums[i]] == 0) {
d.erase(nums[i]);
}
++i;
}
}
return ans;
}
};
Go
func countCompleteSubarrays(nums []int) (ans int) {
d := map[int]int{}
for _, x := range nums {
d[x] = 1
}
cnt := len(d)
i, n := 0, len(nums)
d = map[int]int{}
for j, x := range nums {
d[x]++
for len(d) == cnt {
ans += n - j
d[nums[i]]--
if d[nums[i]] == 0 {
delete(d, nums[i])
}
i++
}
}
return
}
TypeScript
function countCompleteSubarrays(nums: number[]): number {
const d: Map<number, number> = new Map();
for (const x of nums) {
d.set(x, (d.get(x) ?? 0) + 1);
}
const cnt = d.size;
d.clear();
const n = nums.length;
let ans = 0;
let i = 0;
for (let j = 0; j < n; ++j) {
d.set(nums[j], (d.get(nums[j]) ?? 0) + 1);
while (d.size === cnt) {
ans += n - j;
d.set(nums[i], d.get(nums[i])! - 1);
if (d.get(nums[i]) === 0) {
d.delete(nums[i]);
}
++i;
}
}
return ans;
}
Rust
use std::collections::HashMap;
impl Solution {
pub fn count_complete_subarrays(nums: Vec<i32>) -> i32 {
let mut d = HashMap::new();
for &x in &nums {
d.insert(x, 1);
}
let cnt = d.len();
let mut ans = 0;
let n = nums.len();
d.clear();
let (mut i, mut j) = (0, 0);
while j < n {
*d.entry(nums[j]).or_insert(0) += 1;
while d.len() == cnt {
ans += (n - j) as i32;
let e = d.get_mut(&nums[i]).unwrap();
*e -= 1;
if *e == 0 {
d.remove(&nums[i]);
}
i += 1;
}
j += 1;
}
ans
}
}