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Maximum Equal Frequency

Hard

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Topics:

ArraysSliding WindowsGreedy Algorithms

Given an array nums of positive integers, return the longest possible length of an array prefix of nums, such that it is possible to remove exactly one element from this prefix so that every number that has appeared in it will have the same number of occurrences.

If after removing one element there are no remaining elements, it's still considered that every appeared number has the same number of ocurrences (0).

Example 1:

Input: nums = [2,2,1,1,5,3,3,5]
Output: 7
Explanation: For the subarray [2,2,1,1,5,3,3] of length 7, if we remove nums[4] = 5, we will get [2,2,1,1,3,3], so that each number will appear exactly twice.

Example 2:

Input: nums = [1,1,1,2,2,2,3,3,3,4,4,4,5]
Output: 13

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Try coding it in LeetCode

Maximum Equal Frequency

Hard

Asked by:

3 views

Topics:

ArraysSliding WindowsGreedy Algorithms

If after removing one element there are no remaining elements, it's still considered that every appeared number has the same number of ocurrences (0).

Example 1:

Input: nums = [2,2,1,1,5,3,3,5]
Output: 7
Explanation: For the subarray [2,2,1,1,5,3,3] of length 7, if we remove nums[4] = 5, we will get [2,2,1,1,3,3], so that each number will appear exactly twice.

Example 2:

Input: nums = [1,1,1,2,2,2,3,3,3,4,4,4,5]
Output: 13

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

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 are the constraints on the size of the input array `nums`, and what is the range of values within the array?
Can the input array `nums` be empty or null?
If no such prefix exists where removing one element results in equal frequencies, should I return 0 as stated, or is there a different error value I should use?
Could you clarify what happens if there are multiple possible prefixes that satisfy the condition? Should I return the longest one?
Are all elements in the array integers, or could there be other data types like floats?

Brute Force Solution

Approach

The brute force approach to this problem means we're going to try every possible length of the sequence and see if removing just one element can make all the other elements appear with the same frequency. We will check each possibility to see which gives us the longest possible sequence that meets the criteria.

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

Start by considering the entire sequence of numbers.
Check if removing any single number from the entire sequence would result in all the remaining numbers appearing the same number of times.
If removing a number makes all remaining numbers have the same frequency, remember the length of this sequence (before removing the element).
Now, repeat this process, but instead of considering the entire sequence, consider smaller sequences, starting from the beginning.
For each shorter sequence, again try removing each number one by one and check if all remaining numbers have the same frequency.
Keep track of the longest sequence found so far where removing one number makes all the others have equal frequency.
Continue this process until you have checked all possible sequences of numbers from the start, and considered removing each element to achieve equal frequencies.
The longest sequence that meets the condition (after removing one element) is your answer.

Code Implementation

def maximum_equal_frequency_brute_force(numbers):
    maximum_length = 0
    for sublist_end_index in range(1, len(numbers) + 1):
        # Iterate through possible sublist lengths
        for index_to_remove in range(sublist_end_index):
            temp_list = numbers[:sublist_end_index]
            temp_list.pop(index_to_remove)

            if not temp_list:
                maximum_length = max(maximum_length, sublist_end_index)
                continue

            frequency_counts = {}
            for number in temp_list:
                frequency_counts[number] = frequency_counts.get(number, 0) + 1

            # Check if all frequencies are equal after removal
            frequencies = list(frequency_counts.values())
            if len(set(frequencies)) <= 1:
                # Found a valid sublist, update maximum length
                maximum_length = max(maximum_length, sublist_end_index)

    return maximum_length

Big(O) Analysis

Time Complexity

O(n^2) – The outer loop iterates up to n times, considering prefixes of the input array. For each prefix, which has length up to n, an inner loop iterates through each element in that prefix to simulate its removal. Inside the inner loop, checking if all remaining elements have the same frequency takes O(n) time in the worst case. Thus the total time complexity is approximately n * n, resulting in O(n^2).

Space Complexity

O(N) – The brute force approach necessitates checking various sequence lengths and removing elements. This involves creating frequency counts for each subsequence, potentially requiring a hash map (or array) to store the frequency of each number in the subsequence. In the worst-case scenario, the subsequence could be of length N, where N is the length of the input array, leading to a frequency map of size proportional to N. Therefore, the auxiliary space used is O(N).

Optimal Solution

Approach

The core idea is to keep track of how often each number appears and how many numbers appear that many times. By analyzing these counts, we can quickly determine if removing a single number results in all remaining numbers having the same frequency. This approach avoids re-calculating frequencies from scratch each time.

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

Keep track of how many times each number appears as we go through the input.
Also, keep track of how many numbers have each count or frequency. For instance, how many numbers appear once, how many appear twice, and so on.
For each number in the input, consider what happens if we remove it.
Figure out what the 'ideal' frequency would be if all the remaining numbers had the same frequency.
Check if removing the current number would make all the other numbers have the same frequency, using the frequencies we tracked. There are a few possible scenarios where removing a single element can lead to a valid solution: (1) all the numbers have frequency 1, (2) all numbers but one have frequency k, and one number has frequency k+1, (3) one number has frequency 1 and all the others have frequency k, and (4) every number has some frequency k.
Keep track of the longest sequence where removing a number satisfies the 'equal frequency' condition.
Return the length of this longest sequence.

Code Implementation

def maximum_equal_frequency(numbers):
    frequency_of_number = {}
    count_of_frequency = {}
    maximum_length = 0

    for index, number in enumerate(numbers):
        if number in frequency_of_number:
            count_of_frequency[frequency_of_number[number]] -= 1
            if count_of_frequency[frequency_of_number[number]] == 0:
                del count_of_frequency[frequency_of_number[number]]
            frequency_of_number[number] += 1
        else:
            frequency_of_number[number] = 1

        if frequency_of_number[number] in count_of_frequency:
            count_of_frequency[frequency_of_number[number]] += 1
        else:
            count_of_frequency[frequency_of_number[number]] = 1

        current_length = index + 1

        # Checks for the conditions
        if len(count_of_frequency) == 1:
            # Either all elements are 1 or all elements are k
            frequency = next(iter(count_of_frequency))
            if frequency == 1 or count_of_frequency[frequency] == 1:
                maximum_length = current_length
        elif len(count_of_frequency) == 2:
            frequencies = sorted(count_of_frequency.keys())

            # Check if one element has frequency 1 and others have frequency k
            if frequencies[0] == 1 and count_of_frequency[frequencies[0]] == 1:
                maximum_length = current_length
            # Check if one element has frequency k+1 and others have frequency k
            elif frequencies[1] == frequencies[0] + 1 and count_of_frequency[frequencies[1]] == 1:
                maximum_length = current_length

    return maximum_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array nums once, where n is the length of nums. Inside the loop, the frequency counts and the counts of frequencies are updated using hash map operations which take O(1) time on average. The checks to determine if removing an element would result in equal frequencies also take O(1) time, as they involve examining a limited number of frequency counts. Therefore, the overall time complexity is dominated by the single loop through the array, resulting in O(n).

Space Complexity

O(N) – The algorithm uses two main data structures to keep track of frequencies: one to store the frequency of each number (count of each number), and another to store the count of each frequency (count of frequencies). In the worst case, all N numbers in the input array are distinct, leading to a frequency map of size N. Similarly, the count of frequencies map could also potentially store up to N different frequencies. Therefore, the auxiliary space used is proportional to the input size N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as there's no prefix to analyze.
Array with a single element	Return 1 because removing that single element results in all (zero) remaining elements having the same frequency (0).
Array with two identical elements	Return 2 because removing one element leaves a prefix with only one element, thus all elements have the same frequency.
Array with two distinct elements	Return 2 because removing either element leaves a prefix with only one element, thus all elements have the same frequency.
Array with all identical elements	The maximum prefix length will be the array's length, as removing any one element makes all others identical.
Array where removing the last element makes frequencies equal	The solution should correctly identify this as a valid prefix, returning the array's length.
Large input array to test for time complexity issues	The solution should use efficient data structures (e.g., hash maps) to achieve O(n) time complexity.
Input containing negative numbers	Hash map approach works correctly with negative numbers used as keys.

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 are the constraints on the size of the input array `nums`, and what is the range of values within the array?
Can the input array `nums` be empty or null?
If no such prefix exists where removing one element results in equal frequencies, should I return 0 as stated, or is there a different error value I should use?
Could you clarify what happens if there are multiple possible prefixes that satisfy the condition? Should I return the longest one?
Are all elements in the array integers, or could there be other data types like floats?

Brute Force Solution

Approach

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

Start by considering the entire sequence of numbers.
Check if removing any single number from the entire sequence would result in all the remaining numbers appearing the same number of times.
If removing a number makes all remaining numbers have the same frequency, remember the length of this sequence (before removing the element).
Now, repeat this process, but instead of considering the entire sequence, consider smaller sequences, starting from the beginning.
For each shorter sequence, again try removing each number one by one and check if all remaining numbers have the same frequency.
Keep track of the longest sequence found so far where removing one number makes all the others have equal frequency.
Continue this process until you have checked all possible sequences of numbers from the start, and considered removing each element to achieve equal frequencies.
The longest sequence that meets the condition (after removing one element) is your answer.

Code Implementation

def maximum_equal_frequency_brute_force(numbers):
    maximum_length = 0
    for sublist_end_index in range(1, len(numbers) + 1):
        # Iterate through possible sublist lengths
        for index_to_remove in range(sublist_end_index):
            temp_list = numbers[:sublist_end_index]
            temp_list.pop(index_to_remove)

            if not temp_list:
                maximum_length = max(maximum_length, sublist_end_index)
                continue

            frequency_counts = {}
            for number in temp_list:
                frequency_counts[number] = frequency_counts.get(number, 0) + 1

            # Check if all frequencies are equal after removal
            frequencies = list(frequency_counts.values())
            if len(set(frequencies)) <= 1:
                # Found a valid sublist, update maximum length
                maximum_length = max(maximum_length, sublist_end_index)

    return maximum_length

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Keep track of how many times each number appears as we go through the input.
Also, keep track of how many numbers have each count or frequency. For instance, how many numbers appear once, how many appear twice, and so on.
For each number in the input, consider what happens if we remove it.
Figure out what the 'ideal' frequency would be if all the remaining numbers had the same frequency.
Check if removing the current number would make all the other numbers have the same frequency, using the frequencies we tracked. There are a few possible scenarios where removing a single element can lead to a valid solution: (1) all the numbers have frequency 1, (2) all numbers but one have frequency k, and one number has frequency k+1, (3) one number has frequency 1 and all the others have frequency k, and (4) every number has some frequency k.
Keep track of the longest sequence where removing a number satisfies the 'equal frequency' condition.
Return the length of this longest sequence.

Code Implementation

def maximum_equal_frequency(numbers):
    frequency_of_number = {}
    count_of_frequency = {}
    maximum_length = 0

    for index, number in enumerate(numbers):
        if number in frequency_of_number:
            count_of_frequency[frequency_of_number[number]] -= 1
            if count_of_frequency[frequency_of_number[number]] == 0:
                del count_of_frequency[frequency_of_number[number]]
            frequency_of_number[number] += 1
        else:
            frequency_of_number[number] = 1

        if frequency_of_number[number] in count_of_frequency:
            count_of_frequency[frequency_of_number[number]] += 1
        else:
            count_of_frequency[frequency_of_number[number]] = 1

        current_length = index + 1

        # Checks for the conditions
        if len(count_of_frequency) == 1:
            # Either all elements are 1 or all elements are k
            frequency = next(iter(count_of_frequency))
            if frequency == 1 or count_of_frequency[frequency] == 1:
                maximum_length = current_length
        elif len(count_of_frequency) == 2:
            frequencies = sorted(count_of_frequency.keys())

            # Check if one element has frequency 1 and others have frequency k
            if frequencies[0] == 1 and count_of_frequency[frequencies[0]] == 1:
                maximum_length = current_length
            # Check if one element has frequency k+1 and others have frequency k
            elif frequencies[1] == frequencies[0] + 1 and count_of_frequency[frequencies[1]] == 1:
                maximum_length = current_length

    return maximum_length

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as there's no prefix to analyze.
Array with a single element	Return 1 because removing that single element results in all (zero) remaining elements having the same frequency (0).
Array with two identical elements	Return 2 because removing one element leaves a prefix with only one element, thus all elements have the same frequency.
Array with two distinct elements	Return 2 because removing either element leaves a prefix with only one element, thus all elements have the same frequency.
Array with all identical elements	The maximum prefix length will be the array's length, as removing any one element makes all others identical.
Array where removing the last element makes frequencies equal	The solution should correctly identify this as a valid prefix, returning the array's length.
Large input array to test for time complexity issues	The solution should use efficient data structures (e.g., hash maps) to achieve O(n) time complexity.
Input containing negative numbers	Hash map approach works correctly with negative numbers used as keys.