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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) {
          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;
      }
  };
  

• 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
  
  

• 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
      }
  }
  
  

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