2831. Find the Longest Equal Subarray

Problem

You are given a 0-indexed integer array nums and an integer k.

A subarray is called equal if all of its elements are equal. Note that the empty subarray is an equal subarray.

Return **the length of the *longest* possible equal subarray after deleting **at most** *k* elements from **nums.

A subarray is a contiguous, possibly empty sequence of elements within an array.

Example 1:

Input: nums = [1,3,2,3,1,3], k = 3
Output: 3
Explanation: It's optimal to delete the elements at index 2 and index 4.
After deleting them, nums becomes equal to [1, 3, 3, 3].
The longest equal subarray starts at i = 1 and ends at j = 3 with length equal to 3.
It can be proven that no longer equal subarrays can be created.

Example 2:

Input: nums = [1,1,2,2,1,1], k = 2
Output: 4
Explanation: It's optimal to delete the elements at index 2 and index 3.
After deleting them, nums becomes equal to [1, 1, 1, 1].
The array itself is an equal subarray, so the answer is 4.
It can be proven that no longer equal subarrays can be created.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= nums.length
• 0 <= k <= nums.length

Solution (Java)

class Solution {
    public int longestEqualSubarray(List<Integer> A, int k) {
        int maxf = 0, i = 0, n = A.size();
        Map<Integer, Integer> count = new HashMap<>();
        for (int j = 0; j < n; j++) {
            count.put(A.get(j), count.getOrDefault(A.get(j), 0) + 1);
            maxf = Math.max(maxf, count.get(A.get(j)));
            if (j - i + 1 - maxf > k) {
                count.put(A.get(i), count.get(A.get(i)) - 1);
                i++;
            }
        }
        return maxf;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).