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Maximum Frequency Score of a Subarray

Hard

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Topics:

ArraysSliding Windows

You are given a 0-indexed integer array nums.

The frequency score of a subarray nums[l...r] is the frequency of the most frequent element in the subarray.

Return the maximum frequency score of any subarray of nums.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1,2,3,3,3,1,4]
Output: 3
Explanation: The subarray [3,3,3] has a frequency score of 3 since 3 occurs 3 times.
It can be shown that no other subarray has a larger frequency score.

Example 2:

Input: nums = [1,2,1,2,1]
Output: 3
Explanation: The subarray [1,2,1,2,1] has a frequency score of 3 since 1 occurs 3 times.
It can be shown that no other subarray has a larger frequency score.

Constraints:

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

Maximum Frequency Score of a Subarray

Hard

PayPal

1 view

Topics:

ArraysSliding Windows

You are given a 0-indexed integer array nums.

The frequency score of a subarray nums[l...r] is the frequency of the most frequent element in the subarray.

Return the maximum frequency score of any subarray of nums.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1,2,3,3,3,1,4]
Output: 3
Explanation: The subarray [3,3,3] has a frequency score of 3 since 3 occurs 3 times.
It can be shown that no other subarray has a larger frequency score.

Example 2:

Input: nums = [1,2,1,2,1]
Output: 3
Explanation: The subarray [1,2,1,2,1] has a frequency score of 3 since 1 occurs 3 times.
It can be shown that no other subarray has a larger frequency score.

Constraints:

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

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 the range of values within it?
Can the input array `nums` contain negative numbers or zero?
If the array is empty, what should I return?
Could you clarify what is meant by the "score" for a number? Is it simply the frequency of the number within a given subarray?
If multiple subarrays result in the same maximum frequency score, is any one of them acceptable?

Brute Force Solution

Approach

The brute force strategy is all about trying every single possibility. For this problem, we examine every single group of consecutive numbers within the larger set of numbers to find one that matches certain conditions.

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

First, look at just the very first number in the entire set.
Then, look at the first two numbers together.
Next, consider the first three numbers together, and so on, until you've looked at the whole set of numbers starting from the beginning.
Now, shift your focus. Start with the second number alone, then the second and third numbers together, then the second, third, and fourth numbers, and so on.
Keep repeating this process, each time shifting the starting point one number further down the list. Each time you shift, you consider groups of increasing size.
For each of these groups of numbers you consider, check if it satisfies a specific condition related to how often a number occurs within that group.
After checking every possible group, keep track of the best score you found based on how often the most frequent number appears in each group.
The biggest score that you kept track of during that process is the answer.

Code Implementation

def maximum_frequency_score_brute_force(numbers):
    maximum_frequency = 0

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            # Consider each subarray
            subarray = numbers[start_index:end_index + 1]
            if not subarray:
                continue

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

            # Determine the maximum frequency of any number within the subarray
            current_max_frequency = 0
            for number in frequency_map:
                current_max_frequency = max(current_max_frequency, frequency_map[number])

            # Keep track of the largest maximum frequency seen so far.
            maximum_frequency = max(maximum_frequency, current_max_frequency)

    return maximum_frequency

Big(O) Analysis

Time Complexity

O(n²) – The provided brute force approach involves iterating through all possible subarrays of the input array of size n. The outer loop determines the starting index of the subarray, running from 0 to n-1. The inner loop determines the ending index of the subarray, running from the starting index to n-1. Thus, for each starting index, we iterate through the remaining elements. This nested loop structure results in a time complexity proportional to the sum of integers from 1 to n, which is approximately n * (n+1) / 2. This simplifies to O(n²).

Space Complexity

O(N) – The brute force approach, as described, implicitly uses a frequency counter (e.g., a hash map or dictionary) within each subarray to determine the most frequent number. In the worst-case scenario, this frequency counter could store the counts for all unique elements within a subarray of size N. Therefore, the auxiliary space required for each subarray is O(N), and since this occurs repeatedly within the loops, the maximum space used at any given time is proportional to the size of the largest possible subarray which is the entire input array. Thus, the space complexity is O(N).

Optimal Solution

Approach

The key is to efficiently track how often each number appears within a sliding window. Instead of recalculating frequencies from scratch for every possible section, we'll update them as we move the window, letting us quickly find the most frequent number and its count.

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

Imagine a window that can slide across the list of numbers.
Start with a small window, just the first number.
Keep track of how many times each number appears within the current window.
Slide the window one position to the right, including the next number.
As you slide, update the counts of the numbers: increase the count of the new number and decrease the count of the number that left the window.
After each slide, find the number that appears most often in the current window and how many times it appears. This is the current maximum frequency score.
Remember the highest score you've seen so far.
Continue sliding the window and updating the maximum frequency score until you've covered all possible windows.
The highest score you remembered is the answer.

Code Implementation

def maximum_frequency_score(numbers):
    maximum_score = 0
    for window_start in range(len(numbers)):
        frequency_map = {}
        current_maximum_frequency = 0

        for window_end in range(window_start, len(numbers)):
            right_element = numbers[window_end]
            #Update the frequency of the current element.
            if right_element in frequency_map:
                frequency_map[right_element] += 1
            else:
                frequency_map[right_element] = 1

            #Find the most frequent element.
            current_maximum_frequency = max(frequency_map.values(), default = 0)

            #Update the overall maximum score.
            maximum_score = max(maximum_score, current_maximum_frequency)

    return maximum_score

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once using a sliding window. Within each iteration, frequency counts are updated, and the maximum frequency is determined. Assuming the frequency updates and maximum frequency determination within the window take constant time on average (e.g., using a hash map or other efficient data structure), the overall time complexity is dominated by the single pass through the array. Therefore, the runtime is directly proportional to n, resulting in O(n) time complexity.

Space Complexity

O(N) – The solution uses a sliding window approach, tracking the frequency of each number within the current window. This requires a hash map (or dictionary) to store the counts of each number. In the worst case, all N numbers in the input array could be distinct, leading to N entries in the frequency map. Therefore, the auxiliary space required grows linearly with the input size N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0, as there are no subarrays to consider.
Array with a single element	Return 1, as the score is 1 for that single element subarray.
Array with all identical values (e.g., [5, 5, 5, 5])	The frequency score will simply be the length of the array, as each element contributes 1 to its own score.
Array with highly skewed distribution (e.g., [1, 1, 1, 1, 1000])	The algorithm should correctly calculate frequencies even with outlier values.
Array containing only negative numbers (e.g., [-1, -2, -3])	The algorithm should work correctly with negative numbers without modification.
Array containing zero (e.g., [0, 1, 2])	Zero values should be treated the same as any other number when calculating frequencies.
Large array size (e.g., nums.length close to Integer.MAX_VALUE)	The space complexity of frequency counters (if used) might become an issue; optimize for space or use a more efficient data structure if necessary.
Integer overflow in frequency counts (if numbers repeat many times)	Use long data type for frequency counts to avoid potential overflow issues.

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 the range of values within it?
Can the input array `nums` contain negative numbers or zero?
If the array is empty, what should I return?
Could you clarify what is meant by the "score" for a number? Is it simply the frequency of the number within a given subarray?
If multiple subarrays result in the same maximum frequency score, is any one of them acceptable?

Brute Force Solution

Approach

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

First, look at just the very first number in the entire set.
Then, look at the first two numbers together.
Next, consider the first three numbers together, and so on, until you've looked at the whole set of numbers starting from the beginning.
Now, shift your focus. Start with the second number alone, then the second and third numbers together, then the second, third, and fourth numbers, and so on.
Keep repeating this process, each time shifting the starting point one number further down the list. Each time you shift, you consider groups of increasing size.
For each of these groups of numbers you consider, check if it satisfies a specific condition related to how often a number occurs within that group.
After checking every possible group, keep track of the best score you found based on how often the most frequent number appears in each group.
The biggest score that you kept track of during that process is the answer.

Code Implementation

def maximum_frequency_score_brute_force(numbers):
    maximum_frequency = 0

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            # Consider each subarray
            subarray = numbers[start_index:end_index + 1]
            if not subarray:
                continue

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

            # Determine the maximum frequency of any number within the subarray
            current_max_frequency = 0
            for number in frequency_map:
                current_max_frequency = max(current_max_frequency, frequency_map[number])

            # Keep track of the largest maximum frequency seen so far.
            maximum_frequency = max(maximum_frequency, current_max_frequency)

    return maximum_frequency

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine a window that can slide across the list of numbers.
Start with a small window, just the first number.
Keep track of how many times each number appears within the current window.
Slide the window one position to the right, including the next number.
As you slide, update the counts of the numbers: increase the count of the new number and decrease the count of the number that left the window.
After each slide, find the number that appears most often in the current window and how many times it appears. This is the current maximum frequency score.
Remember the highest score you've seen so far.
Continue sliding the window and updating the maximum frequency score until you've covered all possible windows.
The highest score you remembered is the answer.

Code Implementation

def maximum_frequency_score(numbers):
    maximum_score = 0
    for window_start in range(len(numbers)):
        frequency_map = {}
        current_maximum_frequency = 0

        for window_end in range(window_start, len(numbers)):
            right_element = numbers[window_end]
            #Update the frequency of the current element.
            if right_element in frequency_map:
                frequency_map[right_element] += 1
            else:
                frequency_map[right_element] = 1

            #Find the most frequent element.
            current_maximum_frequency = max(frequency_map.values(), default = 0)

            #Update the overall maximum score.
            maximum_score = max(maximum_score, current_maximum_frequency)

    return maximum_score

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0, as there are no subarrays to consider.
Array with a single element	Return 1, as the score is 1 for that single element subarray.
Array with all identical values (e.g., [5, 5, 5, 5])	The frequency score will simply be the length of the array, as each element contributes 1 to its own score.
Array with highly skewed distribution (e.g., [1, 1, 1, 1, 1000])	The algorithm should correctly calculate frequencies even with outlier values.
Array containing only negative numbers (e.g., [-1, -2, -3])	The algorithm should work correctly with negative numbers without modification.
Array containing zero (e.g., [0, 1, 2])	Zero values should be treated the same as any other number when calculating frequencies.
Large array size (e.g., nums.length close to Integer.MAX_VALUE)	The space complexity of frequency counters (if used) might become an issue; optimize for space or use a more efficient data structure if necessary.
Integer overflow in frequency counts (if numbers repeat many times)	Use long data type for frequency counts to avoid potential overflow issues.