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

Python3JavaC++GoTypeScriptRust

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

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;
    }
}

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;
    }
};

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) {
            s.add(nums[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
    }
}

Solution 2

Python3JavaC++GoTypeScriptRust

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

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;
    }
}

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;
    }
};

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
}

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;
}

use std::collections::HashMap;
use std::collections::HashSet;
impl Solution {
    pub fn count_complete_subarrays(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let m = nums.iter().collect::<HashSet<&i32>>().len();
        let mut map = HashMap::new();
        let mut ans = 0;
        let mut i = 0;
        for j in 0..n {
            *map.entry(nums[j]).or_insert(0) += 1;
            while map.len() == m {
                ans += n - j;
                let v = map.entry(nums[i]).or_default();
                *v -= 1;
                if *v == 0 {
                    map.remove(&nums[i]);
                }
                i += 1;
            }
        }
        ans as i32
    }
}

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