2958. Length of Longest Subarray With at Most K Frequency

class Solution {
    public int maxSubarrayLength(int[] nums, int k) {
        Map<Integer, Integer> freq = new HashMap<>();
        int left = 0;
        int maxLen = 0;

        for (int right = 0; right < nums.length; right++) {
            freq.put(nums[right], freq.getOrDefault(nums[right], 0) + 1);

            while (freq.get(nums[right]) > k) {
                freq.put(nums[left], freq.get(nums[left]) - 1);
                left++;
            }

            maxLen = Math.max(maxLen, right - left + 1);
        }

        return maxLen;
    }
}

class Solution:
    def maxSubarrayLength(self, nums, k):
        freq = {}
        left = 0
        max_len = 0

        for right in range(len(nums)):
            freq[nums[right]] = freq.get(nums[right], 0) + 1

            while freq[nums[right]] > k:
                freq[nums[left]] -= 1
                left += 1

            max_len = max(max_len, right - left + 1)

        return max_len

class Solution {
public:
    int maxSubarrayLength(vector<int>& nums, int k) {
        unordered_map<int, int> freq;
        int left = 0;
        int maxLen = 0;

        for (int right = 0; right < nums.size(); right++) {
            freq[nums[right]]++;

            while (freq[nums[right]] > k) {
                freq[nums[left]]--;
                left++;
            }

            maxLen = max(maxLen, right - left + 1);
        }

        return maxLen;
    }
};

class Solution {
    maxSubarrayLength(nums, k) {
        const freq = new Map();
        let left = 0;
        let maxLen = 0;

        for (let right = 0; right < nums.length; right++) {
            freq.set(nums[right], (freq.get(nums[right]) || 0) + 1);

            while (freq.get(nums[right]) > k) {
                freq.set(nums[left], freq.get(nums[left]) - 1);
                left++;
            }

            maxLen = Math.max(maxLen, right - left + 1);
        }

        return maxLen;
    }
}