992-subarrays-with-k-different-integers

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Description

Given an integer array nums and an integer k, return the number of good subarrays of nums.

A good array is an array where the number of different integers in that array is exactly k.

• For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.

A subarray is a contiguous part of an array.

 

Example 1:

Input: nums = [1,2,1,2,3], k = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2]

Example 2:

Input: nums = [1,2,1,3,4], k = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].

 

Constraints:

• 1 <= nums.length <= 2 * 104
• 1 <= nums[i], k <= nums.length

Solution(Python)

class Solution:
    def subarraysWithKDistinct(self, nums: List[int], k: int) -> int:
        return self.subarraysWithAtmostKDistinct(
            nums, k
        ) - self.subarraysWithAtmostKDistinct(nums, k - 1)

    def subarraysWithAtmostKDistinct(self, nums: List[int], k: int) -> int:
        cnt = 0
        n = len(nums)

        # for left in range(n):
        #     hashmap = set()
        #     for right in range(left,n):
        #         hashmap.add(nums[right])
        #         if len(hashmap) > k:
        #             break
        #         cnt +=1
        #
        # Time Complexity : O(n^2)

        left, right = 0, 0
        size = 0
        hashmap = {}
        while right < n:
            if nums[right] not in hashmap:
                hashmap[nums[right]] = 1
            else:
                hashmap[nums[right]] += 1

            if hashmap[nums[right]] == 1:
                size += 1

            while size > k:
                hashmap[nums[left]] -= 1
                if hashmap[nums[left]] == 0:
                    size -= 1
                left += 1

            cnt += right - left + 1
            right += 1

        return cnt
        # Time complexity : O(N)