Find the Longest Equal Subarray

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ArraysTwo PointersSliding Windows

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

Try coding it in LeetCode

Find the Longest Equal Subarray

Medium

Asked by:

8 views

Topics:

ArraysTwo PointersSliding Windows

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the input array? Are they all positive integers?
What should be returned if the input array is empty or null?
Are there any restrictions on the maximum size of the array?
If multiple equal-length subarrays exist, should I return the first one found, or is there a specific criteria for which one to return?
Could you define more formally what qualifies as an 'equal subarray'? Specifically, do elements outside the subarray impact whether the subarray is deemed equal?

Brute Force Solution

Approach

The brute force approach means we're going to check every single possible group of numbers to see if it fits our rules. We'll look at the shortest groups first, then bigger ones, until we've tried everything. Then we pick the longest one that matches the rules.

Here's how the algorithm would work step-by-step:

First, consider only the first number in the whole list.
Then, consider the first two numbers together, then the first three, and so on, up to using the entire list.
Each time you consider a group of numbers, check if it's an equal subarray.
An equal subarray is one where, if you were to count how many times the most frequent number appears, and compare that to a target number, it's either equal to or less than our target number.
Remember the longest equal subarray you find so far.
Next, move on by starting at the second number in the list and repeat the same process: consider the second number by itself, then the second and third numbers together, then the second, third, and fourth numbers together, and so on.
Continue this process, shifting the starting point one number at a time until you've checked every possible group of numbers in the entire list.
In the end, the longest equal subarray you kept track of is your answer.

Code Implementation

def find_longest_equal_subarray(numbers, max_frequency):
    longest_subarray_length = 0

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            subarray = numbers[start_index:end_index + 1]

            # Count frequency of each number in the subarray.
            frequency_map = {}
            for number in subarray:
                frequency_map[number] = frequency_map.get(number, 0) + 1

            # Find the maximum frequency in the subarray.
            max_subarray_frequency = 0
            for number in frequency_map:
                max_subarray_frequency = max(max_subarray_frequency, frequency_map[number])

            # Check if the current subarray is an equal subarray
            if max_subarray_frequency <= max_frequency:

                # Update the longest equal subarray length if needed
                longest_subarray_length = max(longest_subarray_length, len(subarray))

    return longest_subarray_length

Big(O) Analysis

Time Complexity

O(n³) – The algorithm iterates through all possible subarrays of the input array of size n. For each starting index, it considers subarrays of increasing length. This nested iteration contributes O(n²). For each subarray, it counts the frequency of each element to determine the most frequent one, which takes O(n) time in the worst case. Multiplying the cost of subarray generation by the cost of the frequency counting gives a total time complexity of O(n² * n) which simplifies to O(n³).

Space Complexity

O(N) – The provided brute force approach, as described, requires us to keep track of the longest equal subarray found so far. This implies storing the subarray itself or at least its start and end indices. In addition, for each subarray being considered, we need to count the frequency of each number to determine if it is an equal subarray. The frequency counting could be done using a hash map, which, in the worst case (where all numbers in the subarray are unique), could store up to N distinct elements. Therefore, the space used for the frequency map can grow linearly with the size of the input array, N. Additionally, storing the longest equal subarray could potentially take up to O(N) space in the worst-case scenario where the entire input array is an equal subarray. Consequently, the space complexity is O(N).

Optimal Solution

Approach

To efficiently find the longest equal subarray, the best approach is to track the positions of each number and then efficiently calculate the length of equal subarrays based on those positions. This way, we avoid repeatedly checking overlapping subarrays and quickly find the largest one that satisfies the conditions.

Here's how the algorithm would work step-by-step:

First, record where each number appears in the list.
For each different number in the list, consider it as the number that we want to make an equal subarray with.
Using the positions of that number, imagine trying to build an equal subarray by picking a starting point and an ending point.
For a given start and end point, calculate how many of the other numbers will have to be removed to form a fully equal subarray.
If the number of elements that need to be removed is less than or equal to the allowed number of removals, calculate the length of this potential equal subarray. Keep track of the longest one you find.
Repeat this process for every possible starting and ending position for each number.
The largest length calculated is the answer.

Code Implementation

def find_longest_equal_subarray(numbers, allowed_removals):
    number_positions = {}
    for index, number in enumerate(numbers):
        if number not in number_positions:
            number_positions[number] = []
        number_positions[number].append(index)

    longest_equal_subarray_length = 0

    for target_number in number_positions:
        positions = number_positions[target_number]
        for start_index in range(len(positions)):
            for end_index in range(start_index, len(positions)):
                # Calculate potential subarray length
                subarray_length = end_index - start_index + 1

                start_position = positions[start_index]
                end_position = positions[end_index]

                # Calculate elements to remove.
                elements_to_remove = (end_position - start_position + 1)\
                    - subarray_length

                # Check if removal count is within limit
                if elements_to_remove <= allowed_removals:

                    # Update if longer subarray found.
                    longest_equal_subarray_length = max(\
                        longest_equal_subarray_length, \
                        end_position - start_position + 1
                    )

    return longest_equal_subarray_length

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each distinct number in the input array of size n. For each number, it examines all possible subarrays defined by its positions. In the worst case, this requires iterating through all possible start and end positions of the number, which can be O(n^2) in the extreme case where a number appears many times. Calculating the number of removals for each subarray is O(1). Thus, the dominant cost comes from checking all possible subarrays, leading to a time complexity of O(n²).

Space Complexity

O(N) – The algorithm uses a hash map to store the positions of each number in the list. In the worst case, all numbers in the list are distinct, resulting in the hash map storing N numbers, where N is the number of elements in the input list. Therefore, the auxiliary space used by the hash map is proportional to N. No other significant auxiliary space is used, so the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0, as there's no subarray to consider.
Array with a single element	Return 1 since a single element is a subarray of length 1.
Array with all identical elements	Return the length of the array, as the entire array is an equal subarray.
Array with no equal subarrays	Return 1, as each individual element forms a subarray of length 1.
Large input array causing potential memory issues	Use an efficient sliding window approach to avoid storing the entire array content in memory.
Array containing only negative numbers	The algorithm should correctly identify the longest equal subarray regardless of the number's sign.
Array containing large positive and negative numbers (potential integer overflow)	Use a data type with sufficient range (e.g., long) for calculations involving element counts.
Array with multiple equal subarrays of the same maximum length	The algorithm should return one of the longest equal subarrays, and the specific choice is acceptable unless explicitly required.