LeetCode 2831: Find the Longest Equal Subarray Solution

Master LeetCode problem 2831 (Find the Longest Equal Subarray), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2831. Find the Longest Equal Subarray

Medium

Array Hash Table Binary Search Sliding Window

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We will maintain a window such that the number of elements to delete is less than or equal to k. We will track the frequency of the elements in the current window and update the maximum length of the equal subarray accordingly. If at any point the number of elements to delete exceeds k, we will shrink the window from the left side.

Algorithm:

Initialize variables: left pointing to the start of the window, maxLen to store the maximum equal subarray length, freq to map element frequencies, and maxFreq to track the most frequent element in the window.
Iterate over the array using a right pointer:
- Update the frequency of the current element.
- Update maxFreq if the frequency of the current element is higher.
- Check if the number of elements to delete is greater than k, shrink the window from the left side and update frequencies accordingly.
- Update maxLen with the length of the current window.
Return maxLen.

Time Complexity:

The time complexity of this algorithm is O(n) where n is the size of the input array.

Space Complexity:

The space complexity of this algorithm is O(n) where n is the size of the input array.

Solution Code

class Solution {
    public int equalSubarray(int[] nums, int k) {
        int left = 0, maxLen = 0, maxFreq = 0;
        Map<Integer, Integer> freq = new HashMap<>();
        
        for (int right = 0; right < nums.length; right++) {
            freq.put(nums[right], freq.getOrDefault(nums[right], 0) + 1);
            maxFreq = Math.max(maxFreq, freq.get(nums[right]));
            
            while (right - left + 1 - maxFreq > k) {
                freq.put(nums[left], freq.get(nums[left]) - 1);
                left++;
            }
            
            maxLen = Math.max(maxLen, right - left + 1);
        }
        
        return maxLen;
    }
}

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