2831. Find the Longest Equal Subarray

Problem Statement

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

Intuition

Maintain a window keeping max freq of elements in that interval of Sliding window

Like for eg-
1,3,2,3,1,3
0 1 2 3 4 5

In index 0 to 3
len = 1 , max freq of any element is 2, element = 3
therefore 4-2 = 2 , which is permitted as k=3

This means that if we delete 1 and 2 in that interval, we get equal subarray

Optimal answer between interval 1 and 5
Max freq = 3 of 3
len5-freq3 = 2 < 3 Permitted

Which implies that after 2 deletions 2 and 1 , We get equal subarrray of size 3

Links

https://leetcode.com/problems/find-the-longest-equal-subarray/description/

Video Links

Approach 1:

Sliding Window

C++

class Solution {
public:
    int longestEqualSubarray(vector<int>& arr, int k) {
        unordered_map<int,int> freq;
        int maxf = 0;
        int start = 0;

        for(int end=0; end<arr.size(); end++){
            maxf = max(maxf, ++freq[arr[end]]);

            if(end-start+1 - maxf > k){
                freq[arr[start]]--;
                start++;
            }
        }

        return maxf;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++