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992. Subarrays with K Different Integers

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

Solutions

Solution 1

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class Solution:
    def subarraysWithKDistinct(self, nums: List[int], k: int) -> int:
        def f(k):
            pos = [0] * len(nums)
            cnt = Counter()
            j = 0
            for i, x in enumerate(nums):
                cnt[x] += 1
                while len(cnt) > k:
                    cnt[nums[j]] -= 1
                    if cnt[nums[j]] == 0:
                        cnt.pop(nums[j])
                    j += 1
                pos[i] = j
            return pos

        return sum(a - b for a, b in zip(f(k - 1), f(k)))

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class Solution {
    public int subarraysWithKDistinct(int[] nums, int k) {
        int[] left = f(nums, k);
        int[] right = f(nums, k - 1);
        int ans = 0;
        for (int i = 0; i < nums.length; ++i) {
            ans += right[i] - left[i];
        }
        return ans;
    }

    private int[] f(int[] nums, int k) {
        int n = nums.length;
        int[] cnt = new int[n + 1];
        int[] pos = new int[n];
        int s = 0;
        for (int i = 0, j = 0; i < n; ++i) {
            if (++cnt[nums[i]] == 1) {
                ++s;
            }
            for (; s > k; ++j) {
                if (--cnt[nums[j]] == 0) {
                    --s;
                }
            }
            pos[i] = j;
        }
        return pos;
    }
}

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class Solution {
public:
    int subarraysWithKDistinct(vector<int>& nums, int k) {
        vector<int> left = f(nums, k);
        vector<int> right = f(nums, k - 1);
        int ans = 0;
        for (int i = 0; i < nums.size(); ++i) {
            ans += right[i] - left[i];
        }
        return ans;
    }

    vector<int> f(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> pos(n);
        int cnt[n + 1];
        memset(cnt, 0, sizeof(cnt));
        int s = 0;
        for (int i = 0, j = 0; i < n; ++i) {
            if (++cnt[nums[i]] == 1) {
                ++s;
            }
            for (; s > k; ++j) {
                if (--cnt[nums[j]] == 0) {
                    --s;
                }
            }
            pos[i] = j;
        }
        return pos;
    }
};

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func subarraysWithKDistinct(nums []int, k int) (ans int) {
    f := func(k int) []int {
        n := len(nums)
        pos := make([]int, n)
        cnt := make([]int, n+1)
        s, j := 0, 0
        for i, x := range nums {
            cnt[x]++
            if cnt[x] == 1 {
                s++
            }
            for ; s > k; j++ {
                cnt[nums[j]]--
                if cnt[nums[j]] == 0 {
                    s--
                }
            }
            pos[i] = j
        }
        return pos
    }
    left, right := f(k), f(k-1)
    for i := range left {
        ans += right[i] - left[i]
    }
    return
}

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