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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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  };
  

• 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
  
  

• 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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