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Frequency of the Most Frequent Element

Medium

Uber

1 view

Topics:

ArraysSliding WindowsGreedy Algorithms

You are given an integer array nums and an integer k. The frequency of an element is the number of times it occurs in an array. In one operation, you can choose an index of nums and increment the element at that index by 1. Return the maximum possible frequency of an element after performing at most k operations.

For example:

If nums = [1,2,4] and k = 5, the output should be 3. We can increment the first element three times and the second element two times to make nums = [4,4,4]. 4 has a frequency of 3.
If nums = [1,4,8,13] and k = 5, the output should be 2. There are multiple optimal solutions, like incrementing the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.
If nums = [3,9,6] and k = 2, the output should be 1.

How would you approach this problem, keeping in mind time and space complexity? Consider edge cases such as an empty input array or when k is zero.

Frequency of the Most Frequent Element

Medium

Uber

1 view

Topics:

ArraysSliding WindowsGreedy Algorithms

For example:

If nums = [1,2,4] and k = 5, the output should be 3. We can increment the first element three times and the second element two times to make nums = [4,4,4]. 4 has a frequency of 3.
If nums = [1,4,8,13] and k = 5, the output should be 2. There are multiple optimal solutions, like incrementing the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.
If nums = [3,9,6] and k = 2, the output should be 1.

How would you approach this problem, keeping in mind time and space complexity? Consider edge cases such as an empty input array or when k is zero.

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 value of `k`?
Can the elements in the `nums` array be negative, zero, or only positive?
If the array is empty, what should I return?
If there are multiple elements with the same maximum frequency after performing the operations, should I return the frequency of any one of them?
Can `k` be zero? If so, what should the output be?

Brute Force Solution

Approach

The brute force method is the most straightforward way to solve this. It involves checking every possible combination of elements to find the largest possible frequency by incrementing them.

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

Consider each element in the collection one at a time as the potential most frequent element.
For each element, examine all other elements to see how many of them could be increased to match it, given the allowed increment value.
Count how many elements you could increment to match the 'potential most frequent element'.
Remember this count. If it's the highest count you've seen so far, save it.
Repeat this process for every element in the collection.
Once you've gone through all the elements, the highest count you saved is the answer.

Code Implementation

def max_frequency_brute_force(numbers, increment_value):
    maximum_frequency = 0

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

        # Check other elements to find the maximum frequency
        for j in range(len(numbers)):
            if numbers[j] <= potential_most_frequent_element:
                difference = potential_most_frequent_element - numbers[j]

                # Increment count if the difference is within the increment_value
                if difference <= increment_value:
                    current_frequency += 1

        # Update the maximum frequency
        if current_frequency > maximum_frequency:
            maximum_frequency = current_frequency

    return maximum_frequency

Big(O) Analysis

Time Complexity

O(n²) – The brute force approach iterates through each of the n elements in the input array. For each element, it then iterates through the remaining elements to determine how many can be incremented to match the current element. This nested iteration results in approximately n * n/2 comparisons. Therefore, the time complexity is O(n²).

Space Complexity

O(1) – The brute force solution described only uses a few variables to store the potential most frequent element, a count of elements that can be incremented to match it, and the highest count seen so far. The number of variables does not depend on the size of the input array, denoted as N. Therefore, the auxiliary space used by the algorithm is constant, resulting in O(1) space complexity.

Optimal Solution

Approach

The goal is to find the longest possible sequence of numbers that, with some additions, can all be made equal. We can achieve this efficiently by sorting the numbers and then using a sliding window to check how many numbers we can make equal to the largest number within that window by incrementing the smaller numbers.

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

First, arrange the numbers from smallest to largest.
Imagine a window that slides across these sorted numbers.
For each position of the window, consider the largest number inside that window. The goal is to figure out how many of the smaller numbers inside the window can be increased to match that largest number, given our addition limit.
Calculate the total additions needed to make all the numbers in the window equal to the window's largest number.
If the total additions needed is less than or equal to our limit, this window is a possible solution. Remember its size.
Slide the window to the right, repeating the calculation of needed additions and updating the largest possible window size found so far.
The final largest window size found will be the answer, indicating the maximum number of elements that can be made equal.

Code Implementation

def max_frequency(numbers, allowed_addition):
    numbers.sort()
    window_start = 0
    max_frequency_length = 0
    window_sum = 0

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

        # Shrink the window until additions are <= allowed_addition.
        while (window_end - window_start + 1) * numbers[window_end] - window_sum > allowed_addition:

            window_sum -= numbers[window_start]
            window_start += 1

        # Update max length.
        max_frequency_length = max(max_frequency_length, window_end - window_start + 1)

    return max_frequency_length

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 then iterates through the sorted array once (O(n)). Inside the sliding window, we calculate the sum of differences, but this can be done in O(1) by maintaining a running sum, so the sliding window part is O(n). Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity

O(1) – The dominant space complexity factor stems from the sorting operation. While the problem description doesn't specify the sorting algorithm, in-place sorting algorithms like heapsort or quicksort (with optimizations to minimize stack usage) are commonly employed. Assuming an in-place sorting algorithm is used, it modifies the input array directly and requires minimal auxiliary space, typically for a few index variables or temporary storage during swaps. This auxiliary space remains constant regardless of the input array's size (N). Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array (nums is null or empty).	Return 0, as no frequency can be calculated with an empty array.
k is zero and nums has distinct elements.	Return 1, as the maximum frequency achievable without operations is 1.
nums has only one element.	Return 1, as the frequency of that single element is 1.
All elements in nums are identical.	Return the length of nums, since no operations are needed.
k is very large (can increment all numbers to the maximum value).	Return the length of the array, since we can make all numbers equal to the maximum value through operations.
nums contains large positive integers and k is also a large integer that could cause an overflow.	Use long data types to prevent integer overflow during calculations involving sums.
nums contains a wide range of values (very small to very large) and k is small.	The sliding window approach correctly handles this because it adjusts its window based on the cost of incrementing the smaller numbers to become the largest number within the window.
nums is very large and k is also very large; algorithm complexity should be considered.	Using a sliding window approach with sorting achieves O(n log n) complexity, avoiding time limit exceeded errors.

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 value of `k`?
Can the elements in the `nums` array be negative, zero, or only positive?
If the array is empty, what should I return?
If there are multiple elements with the same maximum frequency after performing the operations, should I return the frequency of any one of them?
Can `k` be zero? If so, what should the output be?

Brute Force Solution

Approach

The brute force method is the most straightforward way to solve this. It involves checking every possible combination of elements to find the largest possible frequency by incrementing them.

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

Consider each element in the collection one at a time as the potential most frequent element.
For each element, examine all other elements to see how many of them could be increased to match it, given the allowed increment value.
Count how many elements you could increment to match the 'potential most frequent element'.
Remember this count. If it's the highest count you've seen so far, save it.
Repeat this process for every element in the collection.
Once you've gone through all the elements, the highest count you saved is the answer.

Code Implementation

def max_frequency_brute_force(numbers, increment_value):
    maximum_frequency = 0

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

        # Check other elements to find the maximum frequency
        for j in range(len(numbers)):
            if numbers[j] <= potential_most_frequent_element:
                difference = potential_most_frequent_element - numbers[j]

                # Increment count if the difference is within the increment_value
                if difference <= increment_value:
                    current_frequency += 1

        # Update the maximum frequency
        if current_frequency > maximum_frequency:
            maximum_frequency = current_frequency

    return maximum_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 from smallest to largest.
Imagine a window that slides across these sorted numbers.
For each position of the window, consider the largest number inside that window. The goal is to figure out how many of the smaller numbers inside the window can be increased to match that largest number, given our addition limit.
Calculate the total additions needed to make all the numbers in the window equal to the window's largest number.
If the total additions needed is less than or equal to our limit, this window is a possible solution. Remember its size.
Slide the window to the right, repeating the calculation of needed additions and updating the largest possible window size found so far.
The final largest window size found will be the answer, indicating the maximum number of elements that can be made equal.

Code Implementation

def max_frequency(numbers, allowed_addition):
    numbers.sort()
    window_start = 0
    max_frequency_length = 0
    window_sum = 0

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

        # Shrink the window until additions are <= allowed_addition.
        while (window_end - window_start + 1) * numbers[window_end] - window_sum > allowed_addition:

            window_sum -= numbers[window_start]
            window_start += 1

        # Update max length.
        max_frequency_length = max(max_frequency_length, window_end - window_start + 1)

    return max_frequency_length

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array (nums is null or empty).	Return 0, as no frequency can be calculated with an empty array.
k is zero and nums has distinct elements.	Return 1, as the maximum frequency achievable without operations is 1.
nums has only one element.	Return 1, as the frequency of that single element is 1.
All elements in nums are identical.	Return the length of nums, since no operations are needed.
k is very large (can increment all numbers to the maximum value).	Return the length of the array, since we can make all numbers equal to the maximum value through operations.
nums contains large positive integers and k is also a large integer that could cause an overflow.	Use long data types to prevent integer overflow during calculations involving sums.
nums contains a wide range of values (very small to very large) and k is small.	The sliding window approach correctly handles this because it adjusts its window based on the cost of incrementing the smaller numbers to become the largest number within the window.
nums is very large and k is also very large; algorithm complexity should be considered.	Using a sliding window approach with sorting achieves O(n log n) complexity, avoiding time limit exceeded errors.