Formatted question description: https://leetcode.ca/all/992.html

992. Subarrays with K Different Integers (Hard)

Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.

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

Return the number of good subarrays of A.

 

Example 1:

Input: A = [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: A = [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].

 

Note:

  1. 1 <= A.length <= 20000
  2. 1 <= A[i] <= A.length
  3. 1 <= K <= A.length

Related Topics:
Hash Table, Two Pointers, Sliding Window

Similar Questions:

Solution 1. Sliding Window

Use [i, j) as a sliding window to find the maximum window which contains no more than K unique elements.

To achieve this, we use a map m to store the last position of each number in the current window.

When m.size() > K, we should move forward i to shrink the window until it become valid again.

When m.size() == K, [i, j) is the maximum window we are looking for.

Within this maximum window [i, j), there is a minimum window [k, j) containing no more than K unique elements. k is the minimal index in m.

Now, the number of valid subarrays in this window is k - i + 1.

Since k is monotonically increasing and must be no less than i, we can use k as a global pointer just like i and j so that the overall time complexity of moving k is O(N).

Whenever m.size() == K, we can move k forward until m[A[k] - '0'] == k, and add k - i + 1 to the answer.

// OJ: https://leetcode.com/problems/subarrays-with-k-different-integers/

// Time: O(N)
// Space: O(N)
class Solution {
public:
    int subarraysWithKDistinct(vector<int>& A, int K) {
        int ans = 0, i = 0, j = 0, k = 0, N = A.size();
        unordered_map<int, int> m;
        while (j < N) {
            m[A[j] - '0'] = j;
            ++j;
            while (m.size() > K) {
                int d = A[i++] - '0';
                if (m[d] < i) m.erase(d);
            }
            if (m.size() == K) {
                k = max(i, k);
                while (m[A[k] - '0'] != k) ++k;
                ans += k - i + 1;
            }
        }
        return ans;
    }
};

Java

class Solution {
    public int subarraysWithKDistinct(int[] A, int K) {
        if (A == null || A.length == 0 || K == 0)
            return 0;
        int subarrays = 0;
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        int left = 0, right = 0;
        int length = A.length;
        while (right < length) {
            int rightNum = A[right];
            int rightCount = map.getOrDefault(rightNum, 0) + 1;
            map.put(rightNum, rightCount);
            right++;
            while (map.size() > K) {
                int leftNum = A[left];
                int leftCount = map.getOrDefault(leftNum, 0) - 1;
                if (leftCount > 0)
                    map.put(leftNum, leftCount);
                else
                    map.remove(leftNum);
                left++;
            }
            int pointer = left;
            while (map.size() == K) {
                subarrays++;
                int curNum = A[pointer];
                int curCount = map.getOrDefault(curNum, 0) - 1;
                if (curCount > 0)
                    map.put(curNum, curCount);
                else
                    map.remove(curNum);
                pointer++;
            }
            pointer--;
            while (pointer >= left) {
                int curNum = A[pointer];
                int curCount = map.getOrDefault(curNum, 0) + 1;
                map.put(curNum, curCount);
                pointer--;
            }
        }
        return subarrays;
    }
}

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