Problem Description

In this problem, you're given an array of integers nums and an integer k. The task is to find the maximum length of a subarray where all elements are equal, after optionally deleting up to k elements from the array.

A subarray is defined as a contiguous part of the array which could be as small as an empty array or as large as the entire array. The concept of 'equality' here means that every item in the subarray is the same.

The challenge, therefore, is to figure out the strategy to remove up to k elements to maximize the length of this uniform subarray.

Intuition

The intuition behind the solution is to use a sliding window approach. The key insight is that the maximum length of equal subarray can be found by maintaining the count of elements within a given window and adjusting its size (using two pointers) to remove elements when necessary.

To implement this, iterate over the array while keeping track of the frequency of each element in the current window using a counter. The variable mx is used to keep track of the maximum frequency of any element that we've seen in our current window. This represents the largest potential equal subarray if we were to remove other elements.

We can have a window that exceeds this maximum frequency by k elements, as we're allowed to delete at most k elements. Whenever the size of our current window exceeds mx + k, this implies that we need to remove some elements to bring it back within the allowed size. We do this by shrinking the window from the left.

Iteration continues until we've processed all elements, and the length of the largest possible equal subarray is returned.

This approach works because it continually adjusts the window size to accommodate the highest frequency element while keeping count of the deletions within the limit of k. Whenever the limit is exceeded, the size of the window is reduced to restore the balance.

Learn more about Binary Search and Sliding Window patterns.

Solution Approach

The provided solution code employs a sliding window technique along with a counter to efficiently keep track of the number of occurrences of each element within the current window.

Here's a step-by-step analysis of the implementation:

1. A counter named cnt is initialized using Counter from the collections module. This counter will keep track of the frequency of each element within the current window.
2. Two pointers, l (left) and r (right), are used to define the boundaries of the sliding window. l starts at 0, and r is incremented in each iteration of the for loop.
3. A variable mx, initialized at 0, is used to store the maximum frequency of any element within the current window throughout the iterations.
4. The main loop iterates over the indices and values from the nums array. As r moves to the right, the corresponding value x in nums is added to the cnt counter.
5. After each new element is added to cnt, the mx variable is updated to the maximum value between itself and the updated count for that element, effectively tracking the highest frequency element in the window.
6. At any point, if the size of the current window (given by r - l + 1) minus the maximum frequency (mx) exceeds k, it indicates that we cannot delete enough elements to make the subarray equal. Thus, it's necessary to shrink the window from the left by incrementing l and decrementing the frequency of the lth element in cnt.
7. This process continues until the end of the array is reached, ensuring that at each step, the window size exceeds mx + k by no more than one element.
8. Finally, the length of the longest possible equal subarray (mx) that satisfies the condition after deleting at most k elements is returned.

This solution efficiently finds the longest equal subarray with a complexity of O(n), since it only requires a single pass over the input array, and operations within the loop are O(1) on average due to the use of the counter.

Example Walkthrough

Let's consider the array nums = [1, 1, 2, 2, 4, 4, 4, 2] and k = 2.

The aim is to find the maximum length of a subarray where all elements are equal after optionally deleting up to 2 elements.

Now, here's a step-by-step walkthrough using the sliding window approach:

1. Initialize the counter cnt and pointers l = 0, r = 0. Also, initialize mx as 0.

2. Start with the right pointer at the first element, i.e., nums[0] which is 1.

3. Move to nums[1], which is also 1. Update cnt[1] to 2 and mx = 2.

4. Next, we encounter nums[2] which is 2. Add to cnt and mx remains = 2.

5. Continue to nums[3], another 2, update cnt and still mx = 2.

6. At nums[4], the element is 4. Add to cnt and mx remains 2.

Now, the window size is r - l + 1 = 5 - 0 + 1 = 6, but mx + k = 2 + 2 = 4. Hence, we must shrink the window from the left.

7. Increment l to point at nums[1], decrement the count of nums[0] which is 1, and window size is now 5, which is again greater than mx + k. Therefore, increment l again.

8. Continue the process for the rest of the elements, adjusting l and r accordingly and maintaining cnt and mx.

At nums[6], the element is 4; we add it to cnt, resulting in cnt[4] being 3 and mx being updated to 3.

When the entire array has been checked with this approach, the maximum length of the possible equal subarray at any point considering at most k deletions will be retained in mx.

For the given example, after moving through the entire array with the process mentioned, you would find the maximum length to be 5, using the subarray [4, 4, 4, 2 (deleted), 2 (deleted)].

In the end, the implementation will thus return 5, the maximum length of an equal subarray, given the constraints, for array nums and integer k = 2.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is primarily determined by the for loop which iterates through each element of the nums array once. Therefore, the complexity is dependent on the number of elements n in the nums array. Within the loop, various operations are performed in constant time, including updating the Counter, comparing and assigning the maximum value, and possibly decrementing a count in the Counter. However, as the Counter operations could in the worst case take O(n) when the Counter has grown to the size of the distinct elements in the array and we're decrementing, the dominant operation is still the for loop.

Hence, the time complexity of the code is O(n), where n is the length of the nums array.

Space Complexity

Space complexity is influenced by the additional data structures used in the algorithm. The use of a Counter to store the frequency of each distinct number results in a space complexity proportional to the number of distinct numbers in the nums array. In the worst case scenario, all numbers are distinct, leading to a space complexity equal to the number of distinct elements, which is O(n). However, if numbers repeat often, the space complexity could be much less.

Thus, the space complexity is also O(n), where n is the length of the nums array, representing the worst-case scenario where all elements are unique.

Learn more about how to find time and space complexity quickly using problem constraints.