Subarrays with K Different Integers

Jun 20, 2026madhavcodes25

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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].

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Approaches:

1. Sliding Window

To find the number of subarrays with exactly $K$ different integers directly can be complex. Instead, we can use a clever sliding window technique: The number of subarrays with exactly $K$ different integers is equal to the number of subarrays with at most $K$ different integers minus the number of subarrays with at most $K - 1$ different integers.

$Exact(K) = AtMost(K) - AtMost(K-1)$

1. Create a helper function atMost(k) that calculates the number of subarrays with at most k distinct elements.
2. In the helper function, maintain a sliding window [left, right] and a frequency map to keep track of the count of each element in the current window.
3. Expand the window by moving right and adding elements to the frequency map.
4. If the number of distinct elements exceeds k, shrink the window from the left by moving left and updating the frequency map until the number of distinct elements is valid again.
5. For every valid window ending at right, the number of valid subarrays ending at right is right - left + 1. Add this to the total count.
6. Finally, return the result of atMost(k) - atMost(k - 1).

• Time Complexity: $O(N)$ where $N$ is the number of elements in nums. Both the left and right pointers traverse the array at most once in the atMost helper function.
• Space Complexity: $O(N)$ in the worst-case scenario to store the frequencies of the elements in a hash map.

Solutions:

C++

class Solution {
public:
    int subarraysWithKDistinct(vector<int>& nums, int k) {
        return atMost(nums, k) - atMost(nums, k - 1);
    }

private:
    int atMost(vector<int>& nums, int k) {
        unordered_map<int, int> count;
        int left = 0, res = 0;
        for (int right = 0; right < nums.size(); ++right) {
            if (count[nums[right]]++ == 0) {
                k--;
            }
            while (k < 0) {
                if (--count[nums[left]] == 0) {
                    k++;
                }
                left++;
            }
            res += right - left + 1;
        }
        return res;
    }
};

Java

class Solution {
    public int subarraysWithKDistinct(int[] nums, int k) {
        return atMost(nums, k) - atMost(nums, k - 1);
    }

    private int atMost(int[] nums, int k) {
        Map<Integer, Integer> count = new HashMap<>();
        int left = 0, res = 0;
        for (int right = 0; right < nums.length; right++) {
            if (count.getOrDefault(nums[right], 0) == 0) {
                k--;
            }
            count.put(nums[right], count.getOrDefault(nums[right], 0) + 1);
            while (k < 0) {
                count.put(nums[left], count.get(nums[left]) - 1);
                if (count.get(nums[left]) == 0) {
                    k++;
                }
                left++;
            }
            res += right - left + 1;
        }
        return res;
    }
}

Python

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

    def atMost(self, nums: List[int], k: int) -> int:
        count = collections.Counter()
        left = 0
        res = 0
        for right in range(len(nums)):
            if count[nums[right]] == 0:
                k -= 1
            count[nums[right]] += 1
            
            while k < 0:
                count[nums[left]] -= 1
                if count[nums[left]] == 0:
                    k += 1
                left += 1
            res += right - left + 1
        return res

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
const subarraysWithKDistinct = function(nums, k) {
    const atMost = (k) => {
        const count = new Map();
        let left = 0;
        let res = 0;

        for (let right = 0; right < nums.length; right++) {
            if (!count.has(nums[right]) || count.get(nums[right]) === 0) {
                k--;
            }
            count.set(nums[right], (count.get(nums[right]) || 0) + 1);

            while (k < 0) {
                count.set(nums[left], count.get(nums[left]) - 1);
                if (count.get(nums[left]) === 0) {
                    k++;
                }
                left++;
            }
            res += right - left + 1;
        }
        return res;
    };

    return atMost(k) - atMost(k - 1);
};