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Maximum Frequency of an Element After Performing Operations II

Hard

Google

2 views

Topics:

ArraysSliding Windows

You are given an integer array nums and two integers k and numOperations.

You must perform an operation numOperations times on nums, where in each operation you:

Select an index i that was not selected in any previous operations.
Add an integer in the range [-k, k] to nums[i].

Return the maximum possible frequency of any element in nums after performing the operations.

For example:

nums = [1,4,5], k = 1, numOperations = 2

We can achieve a maximum frequency of two by:

Adding 0 to nums[1], after which nums becomes [1, 4, 5]. (nums[1] is the second element)
Adding -1 to nums[2], after which nums becomes [1, 4, 4]. (nums[2] is the third element)

Another example:

nums = [5,11,20,20], k = 5, numOperations = 1

We can achieve a maximum frequency of two by adding 0 to nums[1].

Maximum Frequency of an Element After Performing Operations II

Hard

Google

2 views

Topics:

ArraysSliding Windows

You are given an integer array nums and two integers k and numOperations.

You must perform an operation numOperations times on nums, where in each operation you:

Select an index i that was not selected in any previous operations.
Add an integer in the range [-k, k] to nums[i].

Return the maximum possible frequency of any element in nums after performing the operations.

For example:

nums = [1,4,5], k = 1, numOperations = 2

We can achieve a maximum frequency of two by:

Adding 0 to nums[1], after which nums becomes [1, 4, 5]. (nums[1] is the second element)
Adding -1 to nums[2], after which nums becomes [1, 4, 4]. (nums[2] is the third element)

Another example:

nums = [5,11,20,20], k = 5, numOperations = 1

We can achieve a maximum frequency of two by adding 0 to nums[1].

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 `nums` array and the value of `k`? Are they within a specific range?
Can the numbers in the `nums` array be negative, zero, or only positive integers?
If it's impossible to achieve a frequency greater than 1 even after using all `k` operations, what value should I return?
Are we allowed to modify the original `nums` array, or should I consider creating a copy to avoid side effects?
If multiple elements can achieve the same maximum frequency after applying operations, is any one of them acceptable?

Brute Force Solution

Approach

The brute force method tackles this problem by trying every single possible combination of operations. It checks, for each number, how many times it can become the most frequent after applying the maximum allowed number of actions to other numbers. It then picks the best outcome.

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

Pick a number from the list.
For every other number in the list, determine the number of actions needed to make that number equal to the number you picked in the first step.
Add up all the actions needed to make the appropriate numbers equal to the number you picked in the first step.
If the total actions needed is less than or equal to the maximum allowed actions, count how many numbers are now equal to the number you picked in the first step.
If the total actions needed exceeds the maximum allowed actions, skip and proceed to the next possible number.
Repeat steps 1-5 for every number in the original list.
Keep track of the maximum count obtained in step 4 across all these attempts.
The largest count found is the answer.

Code Implementation

def max_frequency_brute_force(numbers, max_operations):
    max_frequency = 0

    for i in range(len(numbers)):
        target_number = numbers[i]
        current_frequency = 0
        total_operations_needed = 0

        # Iterate through all numbers to check how many can be made equal to the target.
        for j in range(len(numbers)):
            if numbers[j] != target_number:
                operations_needed = abs(numbers[j] - target_number)
                total_operations_needed += operations_needed

        # Only counts the element as part of the frequency if the total operations allow it.
        if total_operations_needed <= max_operations:
            for k in range(len(numbers)):
                operations_needed_to_equal_target = abs(numbers[k] - target_number)
                if operations_needed_to_equal_target == 0:
                    current_frequency += 1
        else:
            continue

        # Keep track of the largest frequency so far.
        if total_operations_needed <= max_operations:
            max_frequency = max(max_frequency, current_frequency)

    return max_frequency

Big(O) Analysis

Time Complexity

O(n²) – The outer loop iterates through each of the n elements in the input array. For each element in the outer loop, the inner loop iterates through all the other n-1 elements to calculate the cost of making them equal to the selected element. The cost calculation and comparison happen within the inner loop, leading to a nested loop structure. This results in approximately n * (n-1) operations which simplifies to n² - n and consequently, O(n²).

Space Complexity

O(1) – The brute force algorithm described iterates through the input list and calculates the actions needed for each number. It primarily uses a few variables to store the current number being considered, the actions needed, and the maximum count found so far. Since the space used by these variables is independent of the size of the input list (N), the auxiliary space complexity is constant, O(1).

Optimal Solution

Approach

To find the maximum frequency of an element after operations, it's crucial to efficiently explore the possibilities without wasting time. We'll focus on maintaining a relevant section of the data and strategically adjusting it to discover the highest possible frequency.

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

First, arrange the numbers in increasing order, as this will simplify figuring out how many operations are needed.
Imagine a 'window' that slides across the ordered numbers. This window represents the group of numbers we are considering for making equal.
As the window slides, we calculate how many operations it would take to make all the numbers inside the window equal to the largest number within that window.
If the operations needed are within the allowed limit, we record the size of the window. This size represents a potential maximum frequency.
If the operations needed exceed the limit, we shrink the window from the beginning (left side) until the operations are back within the allowed limit.
We continue sliding and adjusting the window, always checking if the current window size is the largest we've seen so far.
The largest window size recorded during this process is the maximum frequency of an element that can be achieved within the given operation limit.

Code Implementation

def maximum_frequency(numbers, allowed_operations):
    numbers.sort()
    maximum_frequency_so_far = 0
    window_start = 0
    current_sum = 0

    for window_end in range(len(numbers)):
        current_sum += numbers[window_end]

        # Shrink the window if operations exceed limit
        while (window_end - window_start + 1) * numbers[window_end] - current_sum > allowed_operations:
            current_sum -= numbers[window_start]
            window_start += 1

        # Track the largest valid window size
        maximum_frequency_so_far = max(maximum_frequency_so_far, window_end - window_start + 1)

    return maximum_frequency_so_far

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the input array of size n, which takes O(n log n) time. The sliding window approach then iterates through the sorted array once. Inside the sliding window, we calculate the cost of making all elements equal, but this calculation is done incrementally as the window slides and shrinks from the left side. The window adjustment (shrinking) is done with a while loop but the total number of iterations of this loop is at most n since the left pointer never goes backwards. Therefore the time complexity of the sliding window is O(n). The total time complexity is the sum of sorting and sliding window which is O(n log n) + O(n). Thus the overall time complexity is O(n log n).

Space Complexity

O(1) – The algorithm sorts the input array in-place, meaning it modifies the original array directly and does not create a new copy. It primarily uses a sliding window approach with a few variables to keep track of the window's boundaries and the number of operations. These variables consume a constant amount of extra space, irrespective of the input array's size, N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty or null array	Return 0 if the input array is null or empty since no elements exist.
Array with a single element	Return 1 if the array has only one element as that is the maximum frequency.
All elements are identical	The sliding window will expand to the entire array, and the result will be the size of the array.
Large input array with large numbers and large k	Ensure integer overflow doesn't occur when calculating the sum within the sliding window by using long data type.
All elements are zero	The sliding window will expand to the entire array and the result will be the size of the array.
Sorted array with a single large value at the end	The sliding window will expand until the operation budget is exhausted, demonstrating algorithm efficiency.
Input array contains negative numbers	The solution should correctly handle negative numbers as the goal is to minimize the difference to a common number with the given operation cost.
k is zero	Return the frequency of the most frequent number in the array directly, as no operations can be performed.

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 `nums` array and the value of `k`? Are they within a specific range?
Can the numbers in the `nums` array be negative, zero, or only positive integers?
If it's impossible to achieve a frequency greater than 1 even after using all `k` operations, what value should I return?
Are we allowed to modify the original `nums` array, or should I consider creating a copy to avoid side effects?
If multiple elements can achieve the same maximum frequency after applying operations, is any one of them acceptable?

Brute Force Solution

Approach

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

Pick a number from the list.
For every other number in the list, determine the number of actions needed to make that number equal to the number you picked in the first step.
Add up all the actions needed to make the appropriate numbers equal to the number you picked in the first step.
If the total actions needed is less than or equal to the maximum allowed actions, count how many numbers are now equal to the number you picked in the first step.
If the total actions needed exceeds the maximum allowed actions, skip and proceed to the next possible number.
Repeat steps 1-5 for every number in the original list.
Keep track of the maximum count obtained in step 4 across all these attempts.
The largest count found is the answer.

Code Implementation

def max_frequency_brute_force(numbers, max_operations):
    max_frequency = 0

    for i in range(len(numbers)):
        target_number = numbers[i]
        current_frequency = 0
        total_operations_needed = 0

        # Iterate through all numbers to check how many can be made equal to the target.
        for j in range(len(numbers)):
            if numbers[j] != target_number:
                operations_needed = abs(numbers[j] - target_number)
                total_operations_needed += operations_needed

        # Only counts the element as part of the frequency if the total operations allow it.
        if total_operations_needed <= max_operations:
            for k in range(len(numbers)):
                operations_needed_to_equal_target = abs(numbers[k] - target_number)
                if operations_needed_to_equal_target == 0:
                    current_frequency += 1
        else:
            continue

        # Keep track of the largest frequency so far.
        if total_operations_needed <= max_operations:
            max_frequency = max(max_frequency, current_frequency)

    return max_frequency

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, arrange the numbers in increasing order, as this will simplify figuring out how many operations are needed.
Imagine a 'window' that slides across the ordered numbers. This window represents the group of numbers we are considering for making equal.
As the window slides, we calculate how many operations it would take to make all the numbers inside the window equal to the largest number within that window.
If the operations needed are within the allowed limit, we record the size of the window. This size represents a potential maximum frequency.
If the operations needed exceed the limit, we shrink the window from the beginning (left side) until the operations are back within the allowed limit.
We continue sliding and adjusting the window, always checking if the current window size is the largest we've seen so far.
The largest window size recorded during this process is the maximum frequency of an element that can be achieved within the given operation limit.

Code Implementation

def maximum_frequency(numbers, allowed_operations):
    numbers.sort()
    maximum_frequency_so_far = 0
    window_start = 0
    current_sum = 0

    for window_end in range(len(numbers)):
        current_sum += numbers[window_end]

        # Shrink the window if operations exceed limit
        while (window_end - window_start + 1) * numbers[window_end] - current_sum > allowed_operations:
            current_sum -= numbers[window_start]
            window_start += 1

        # Track the largest valid window size
        maximum_frequency_so_far = max(maximum_frequency_so_far, window_end - window_start + 1)

    return maximum_frequency_so_far

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty or null array	Return 0 if the input array is null or empty since no elements exist.
Array with a single element	Return 1 if the array has only one element as that is the maximum frequency.
All elements are identical	The sliding window will expand to the entire array, and the result will be the size of the array.
Large input array with large numbers and large k	Ensure integer overflow doesn't occur when calculating the sum within the sliding window by using long data type.
All elements are zero	The sliding window will expand to the entire array and the result will be the size of the array.
Sorted array with a single large value at the end	The sliding window will expand until the operation budget is exhausted, demonstrating algorithm efficiency.
Input array contains negative numbers	The solution should correctly handle negative numbers as the goal is to minimize the difference to a common number with the given operation cost.
k is zero	Return the frequency of the most frequent number in the array directly, as no operations can be performed.