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Maximum Frequency After Subarray Operation

Medium

Amazon

33 views

Topics:

Arrays

You are given an array nums of length n. You are also given an integer k.

You perform the following operation on nums once:

Select a subarray nums[i..j] where 0 <= i <= j <= n - 1.
Select an integer x and add x to all the elements in nums[i..j].

Find the maximum frequency of the value k after the operation.

Example 1:

Input: nums = [1,2,3,4,5,6], k = 1

Output: 2

Explanation:

After adding -5 to nums[2..5], 1 has a frequency of 2 in [1, 2, -2, -1, 0, 1].

Example 2:

Input: nums = [10,2,3,4,5,5,4,3,2,2], k = 10

Output: 4

Explanation:

After adding 8 to nums[1..9], 10 has a frequency of 4 in [10, 10, 11, 12, 13, 13, 12, 11, 10, 10].

Constraints:

1 <= n == nums.length <= 10⁵
1 <= nums[i] <= 50
1 <= k <= 50

Maximum Frequency After Subarray Operation

Medium

Amazon

33 views

Topics:

Arrays

You are given an array nums of length n. You are also given an integer k.

You perform the following operation on nums once:

Select a subarray nums[i..j] where 0 <= i <= j <= n - 1.
Select an integer x and add x to all the elements in nums[i..j].

Find the maximum frequency of the value k after the operation.

Example 1:

Input: nums = [1,2,3,4,5,6], k = 1

Output: 2

Explanation:

After adding -5 to nums[2..5], 1 has a frequency of 2 in [1, 2, -2, -1, 0, 1].

Example 2:

Input: nums = [10,2,3,4,5,5,4,3,2,2], k = 10

Output: 4

Explanation:

After adding 8 to nums[1..9], 10 has a frequency of 4 in [10, 10, 11, 12, 13, 13, 12, 11, 10, 10].

Constraints:

1 <= n == nums.length <= 10⁵
1 <= nums[i] <= 50
1 <= k <= 50

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? Can the numbers be negative, zero, or floating-point values?
What is the maximum size of the input array?
If there are multiple subarrays that result in the same maximum frequency after the operation, can I return the length resulting from any of them?
If the input array is empty, what should the return value be?
Could you define more precisely what the 'subarray operation' entails? Is it replacing all the elements inside the chosen subarray with a single, chosen element?

Brute Force Solution

Approach

The brute force method to find the maximum frequency after a subarray operation means we try out every possible combination. We will examine each possible subarray and change its elements to see how it affects the frequency of a single number, and we'll keep track of the best result we see.

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

Consider each possible segment (a continuous portion) of the list of numbers.
For each segment, pick a number that exists in the whole list.
Change all the numbers within that segment to the number you picked.
Count how many times the most frequent number appears in the entire list after this change.
Remember the highest count of the most frequent number you've seen so far.
Repeat steps 2-5 for every possible number in the list.
Repeat steps 1-6 for every possible segment in the list.
The highest count you remembered is the answer.

Code Implementation

def maximum_frequency_after_subarray_operation(numbers):
    maximum_frequency = 0
    list_length = len(numbers)

    for start_index in range(list_length):
        for end_index in range(start_index, list_length):
            # Iterate through all possible subarrays
            for target_number in set(numbers):
                # Iterate through all unique numbers in the input list
                temp_numbers = numbers[:]

                for index in range(start_index, end_index + 1):
                    # Modify the subarray to have the target number
                    temp_numbers[index] = target_number

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

                current_max_frequency = 0
                for number in frequency_counts:
                    current_max_frequency = max(current_max_frequency, frequency_counts[number])

                maximum_frequency = max(maximum_frequency, current_max_frequency)

    return maximum_frequency

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n^3 * m) – Let n be the length of the input list and m be the number of unique elements in the list. First, we iterate through all possible subarrays, which takes O(n^2) time. For each subarray, we iterate through all unique elements (m). For each unique element, we change all elements in the subarray which is O(n). Then, we have to iterate through the entire array to find the maximum frequency after the change, which is O(n). Therefore, the total time complexity is O(n^2 * m * n), which simplifies to O(n^3 * m).

Space Complexity

O(1) – The brute force solution described involves iterating through subarrays and counting frequencies. While frequency counting for each subarray operation technically requires a temporary counter variable, this counter's size is bound by a constant factor (the number of distinct elements in the input list) and doesn't scale with the input size N. No other auxiliary data structures like lists or hash maps of size dependent on N are used. Thus, the auxiliary space remains constant regardless of N.

Optimal Solution

Approach

The key to solving this problem efficiently is to realize we don't need to check every single possible subarray. Instead, we focus on finding the longest possible sequence of the same number that can be made by changing a limited number of other numbers within a sliding window. The sliding window allows us to efficiently find the longest sequence.

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

Think of it as having a 'window' that we slide across the list of numbers.
As we move the window, we keep track of how many times each number appears inside the window.
Our goal is to find the largest possible window where one number appears many times, and the other numbers appear only a few times, and that few number of changes fall under the allowed changes.
For each window, we determine the most frequent number within that window.
We then see how many other numbers are in that window (these are the numbers we'd have to change to all match the most frequent number).
If the number of changes needed is less than or equal to the allowed number of changes, we've found a valid sequence.
We update the maximum length seen so far.
We slide the window across the entire list of numbers, repeating this process, and ultimately return the longest sequence we found.

Code Implementation

def max_frequency(numbers, allowed_changes):
    window_start = 0
    maximum_length = 0
    frequency_map = {}

    for window_end in range(len(numbers)):
        right_number = numbers[window_end]
        if right_number not in frequency_map:
            frequency_map[right_number] = 0
        frequency_map[right_number] += 1

        # Find the most frequent number's count in current window.
        most_frequent_number_count = max(frequency_map.values())

        # Shrink window if changes exceed allowed changes.
        while (window_end - window_start + 1) - most_frequent_number_count > allowed_changes:
            left_number = numbers[window_start]
            frequency_map[left_number] -= 1
            window_start += 1

        # Update maximum length after each window adjustment.
        maximum_length = max(maximum_length, window_end - window_start + 1)

    return maximum_length

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The described sliding window approach iterates through the input array of size n once, adjusting the window boundaries as it progresses. Inside the window, operations like counting frequency of elements and determining the maximum frequency take constant time since they involve a limited number of elements within the window. Thus, the overall time complexity is determined by the single pass through the array, resulting in O(n) time complexity.

Space Complexity

O(N) – The algorithm uses a sliding window approach and keeps track of the frequency of numbers within that window. This frequency count is maintained using a hash map or dictionary. In the worst-case scenario, all numbers in the input list are distinct, leading to the hash map storing the count for each of the N numbers. Thus, the space required for the hash map grows linearly with the input size N. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no elements, thus no frequency.
Array with a single element	Return 1, as the frequency of that single element is 1.
Array with all identical elements	The frequency of the identical element is the array length, which should be returned.
Large array with values close to the maximum integer value, potentially leading to overflow during calculations.	Use a data type with a larger range (e.g., long) for intermediate calculations to prevent integer overflow.
Array contains negative numbers	The sliding window approach should work correctly regardless of negative number existence because we track the difference between window sum and target * window size
The array is already sorted in ascending or descending order.	The sliding window approach handles pre-sorted arrays correctly, as it iterates through the array regardless of the order.
Target is much larger than array elements	The window size will increase until it reaches the array size or it is limited by k and a valid maximum frequency will be returned.
k is zero	The algorithm should return the frequency of the most frequent element without any operations (k=0).

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? Can the numbers be negative, zero, or floating-point values?
What is the maximum size of the input array?
If there are multiple subarrays that result in the same maximum frequency after the operation, can I return the length resulting from any of them?
If the input array is empty, what should the return value be?
Could you define more precisely what the 'subarray operation' entails? Is it replacing all the elements inside the chosen subarray with a single, chosen element?

Brute Force Solution

Approach

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

Consider each possible segment (a continuous portion) of the list of numbers.
For each segment, pick a number that exists in the whole list.
Change all the numbers within that segment to the number you picked.
Count how many times the most frequent number appears in the entire list after this change.
Remember the highest count of the most frequent number you've seen so far.
Repeat steps 2-5 for every possible number in the list.
Repeat steps 1-6 for every possible segment in the list.
The highest count you remembered is the answer.

Code Implementation

def maximum_frequency_after_subarray_operation(numbers):
    maximum_frequency = 0
    list_length = len(numbers)

    for start_index in range(list_length):
        for end_index in range(start_index, list_length):
            # Iterate through all possible subarrays
            for target_number in set(numbers):
                # Iterate through all unique numbers in the input list
                temp_numbers = numbers[:]

                for index in range(start_index, end_index + 1):
                    # Modify the subarray to have the target number
                    temp_numbers[index] = target_number

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

                current_max_frequency = 0
                for number in frequency_counts:
                    current_max_frequency = max(current_max_frequency, frequency_counts[number])

                maximum_frequency = max(maximum_frequency, current_max_frequency)

    return maximum_frequency

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Think of it as having a 'window' that we slide across the list of numbers.
As we move the window, we keep track of how many times each number appears inside the window.
Our goal is to find the largest possible window where one number appears many times, and the other numbers appear only a few times, and that few number of changes fall under the allowed changes.
For each window, we determine the most frequent number within that window.
We then see how many other numbers are in that window (these are the numbers we'd have to change to all match the most frequent number).
If the number of changes needed is less than or equal to the allowed number of changes, we've found a valid sequence.
We update the maximum length seen so far.
We slide the window across the entire list of numbers, repeating this process, and ultimately return the longest sequence we found.

Code Implementation

def max_frequency(numbers, allowed_changes):
    window_start = 0
    maximum_length = 0
    frequency_map = {}

    for window_end in range(len(numbers)):
        right_number = numbers[window_end]
        if right_number not in frequency_map:
            frequency_map[right_number] = 0
        frequency_map[right_number] += 1

        # Find the most frequent number's count in current window.
        most_frequent_number_count = max(frequency_map.values())

        # Shrink window if changes exceed allowed changes.
        while (window_end - window_start + 1) - most_frequent_number_count > allowed_changes:
            left_number = numbers[window_start]
            frequency_map[left_number] -= 1
            window_start += 1

        # Update maximum length after each window adjustment.
        maximum_length = max(maximum_length, window_end - window_start + 1)

    return maximum_length

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no elements, thus no frequency.
Array with a single element	Return 1, as the frequency of that single element is 1.
Array with all identical elements	The frequency of the identical element is the array length, which should be returned.
Large array with values close to the maximum integer value, potentially leading to overflow during calculations.	Use a data type with a larger range (e.g., long) for intermediate calculations to prevent integer overflow.
Array contains negative numbers	The sliding window approach should work correctly regardless of negative number existence because we track the difference between window sum and target * window size
The array is already sorted in ascending or descending order.	The sliding window approach handles pre-sorted arrays correctly, as it iterates through the array regardless of the order.
Target is much larger than array elements	The window size will increase until it reaches the array size or it is limited by k and a valid maximum frequency will be returned.
k is zero	The algorithm should return the frequency of the most frequent element without any operations (k=0).