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Maximum Number of Distinct Elements After Operations

Medium

Asked by:

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ArraysGreedy Algorithms

You are given an integer array nums and an integer k.

You are allowed to perform the following operation on each element of the array at most once:

Add an integer in the range [-k, k] to the element.

Return the maximum possible number of distinct elements in nums after performing the operations.

Example 1:

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

Output: 6

Explanation:

nums changes to [-1, 0, 1, 2, 3, 4] after performing operations on the first four elements.

Example 2:

Input: nums = [4,4,4,4], k = 1

Output: 3

Explanation:

By adding -1 to nums[0] and 1 to nums[1], nums changes to [3, 5, 4, 4].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
0 <= k <= 10⁹

Try coding it in LeetCode

Maximum Number of Distinct Elements After Operations

Medium

Asked by:

15 views

Topics:

ArraysGreedy Algorithms

You are given an integer array nums and an integer k.

You are allowed to perform the following operation on each element of the array at most once:

Add an integer in the range [-k, k] to the element.

Return the maximum possible number of distinct elements in nums after performing the operations.

Example 1:

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

Output: 6

Explanation:

nums changes to [-1, 0, 1, 2, 3, 4] after performing operations on the first four elements.

Example 2:

Input: nums = [4,4,4,4], k = 1

Output: 3

Explanation:

By adding -1 to nums[0] and 1 to nums[1], nums changes to [3, 5, 4, 4].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
0 <= k <= 10⁹

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 values within the input array and the number of operations `k`? Can they be negative?
Is the input array guaranteed to be non-empty, or should I handle the case where the input array is null or empty?
If after performing the operations, multiple arrays achieve the maximum number of distinct elements, is any valid array acceptable?
Can you clarify what happens if I use all `k` operations and still can't make all elements distinct? What should be returned in this case?
If there are duplicate values in the array, are there any specific rules or constraints on how those duplicates should be handled during the operations?

Brute Force Solution

Approach

The brute force way to solve this is to try out every single possibility. We will reduce the number of each element one at a time, seeing how many distinct elements we can make at each stage until we run out of operations.

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

Start with the original set of numbers.
Consider each number in the set individually.
For each number, try reducing its count by 1 (if possible and if you still have operations remaining).
After reducing a number's count, recalculate the number of distinct elements in the new set.
Keep track of the highest number of distinct elements seen so far.
Repeat the reduction process for all possible combinations of numbers and operation counts.
The maximum distinct count you've tracked is the result.

Code Implementation

def find_maximum_distinct_elements_brute_force(numbers, operation_count):
    maximum_distinct_count = 0

    def calculate_distinct_count(current_numbers):
        counts = {}
        for number in current_numbers:
            counts[number] = counts.get(number, 0) + 1
        return len(counts)

    def solve(current_numbers, operations_remaining):
        nonlocal maximum_distinct_count

        # Update max count, as we may have found a better combo
        maximum_distinct_count = max(maximum_distinct_count,
                                     calculate_distinct_count(current_numbers))

        # Base case: No more operations to perform
        if operations_remaining == 0:
            return

        for index in range(len(current_numbers)):
            # Do not process a number if it is already zero
            if current_numbers[index] > 0:

                # Create a copy so we don't modify the original list
                new_numbers = current_numbers[:]
                new_numbers[index] -= 1

                # Recurse with one fewer operation
                solve(new_numbers, operations_remaining - 1)

    solve(numbers, operation_count)
    return maximum_distinct_count

Big(O) Analysis

Time Complexity

O(n^k) – The algorithm explores all possible combinations of reducing the counts of each number in the input array. Let n be the number of elements in the input array and k be the number of allowed operations. In the worst-case scenario, where we explore every possible combination of reducing element counts, the time complexity becomes exponential with respect to both n and k. Because of the recursive nature of exploring all combinations the time complexity grows very fast. Therefore, the time complexity is O(n^k).

Space Complexity

O(N) – The brute force solution explores all possible combinations of reducing element counts using recursion. In the worst-case, each number can be potentially reduced by 1 in each recursive call. The maximum depth of the recursion, and thus the number of stack frames, can be proportional to N, where N is the number of elements in the input array. Each stack frame stores the modified array and intermediate calculations, thus leading to O(N) space due to the call stack.

Optimal Solution

Approach

The goal is to maximize distinct elements by strategically reducing the counts of the most frequent elements first. We want to minimize the number of duplicates present after performing our operations. This involves understanding how many operations it takes to bring an element's count down to zero or one.

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

Count how many times each number appears in the input.
Look at the numbers that appear more than once. We want to reduce these first because they contribute to the duplicate count.
Sort these counts from largest to smallest. This lets us focus on the most frequent numbers first.
Go through each count one by one. If we have enough operations, reduce the count to 1. If we don't have enough operations, stop and move to the next step.
Subtract the number of operations used from the total operations available.
Count the number of elements that are now distinct. This will be the number of elements with a count of 1 or greater.
If we have operations left, we need to consider elements that only appeared once in the initial array. We want to make sure we aren't wasting operations deleting single instances of numbers, if possible.
If we have more operations than elements with count 1, determine how many of these elements can be entirely removed from the input.
Add this value to our distinct count from step 6. Since any duplicate removal now removes an element with a count of one, our new result is the maximum number of distinct elements after the given number of operations.

Code Implementation

def max_distinct_elements(numbers, operations_available):
    element_counts = {}
    for number in numbers:
        element_counts[number] = element_counts.get(number, 0) + 1

    counts_greater_than_one = []
    single_count_elements = 0

    for count in element_counts.values():
        if count > 1:
            counts_greater_than_one.append(count)
        else:
            single_count_elements += 1

    # Sort counts to efficiently reduce most frequent first
    counts_greater_than_one.sort(reverse=True)

    for i in range(len(counts_greater_than_one)):
        count = counts_greater_than_one[i]

        reduction_needed = count - 1
        if operations_available >= reduction_needed:
            operations_available -= reduction_needed
            counts_greater_than_one[i] = 1
        else:
            counts_greater_than_one[i] -= operations_available
            operations_available = 0
            break

    distinct_elements = 0
    for count in counts_greater_than_one:
        if count >= 1:
            distinct_elements += 1

    distinct_elements += single_count_elements

    # Remove single count elements only if operations remain
    if operations_available > 0:

        # Remove as many single elements as possible
        elements_to_remove = min(single_count_elements, operations_available)

        # Update the distinct count accordingly
        distinct_elements -= elements_to_remove

        # Account for excess operations that can't remove more single elements
        operations_available -= elements_to_remove

    return distinct_elements

Big(O) Analysis

Time Complexity

O(n log n) – Counting the frequency of each number takes O(n) time. Sorting the frequencies takes O(m log m) time, where m is the number of distinct elements, and m <= n, thus O(n log n). Iterating through the sorted frequencies and reducing counts takes O(m) which is O(n) time. Therefore, the dominant operation is sorting the frequencies, leading to a time complexity of O(n log n).

Space Complexity

O(N) – The algorithm uses a hash map to count the frequency of each number in the input array, which takes O(N) space in the worst case, where N is the number of elements in the input. Additionally, a list of counts is created and sorted which could be of size N in the worst case (all distinct elements). Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as no distinct elements can be achieved with an empty input.
k = 0 (no operations allowed)	Return the number of distinct elements in the original array.
k >= n (operations exceed array size)	If k is greater or equal to n, all elements can be potentially made equal, thus the answer is 1.
Array with all identical elements	If all elements are identical and k > 0, the answer is 1; otherwise, the answer is 1.
Array with a large number of duplicates for a single element	The algorithm should correctly handle cases where many elements are the same and focus on minimizing their impact by increasing them to be equal.
Large input array size that could lead to memory issues	Use an efficient data structure like a hash map or counter to store element frequencies to minimize memory usage, and potentially use the `collections.Counter` library to improve this
Elements that, when incremented by one, cause integer overflow	The algorithm should avoid arithmetic overflows with very large array element values, and modulo arithmetic should be considered if the prompt specifies upper boundaries to avoid the integer overflow.
Input array already contains only distinct elements	Return n - k if k < n else 1 where n is the array's size to account for the given operation.

Null or empty input array

How to Handle:

Return 0, as no distinct elements can be achieved with an empty input.

k = 0 (no operations allowed)

How to Handle:

Return the number of distinct elements in the original array.

k >= n (operations exceed array size)

How to Handle:

If k is greater or equal to n, all elements can be potentially made equal, thus the answer is 1.

Array with all identical elements

How to Handle:

If all elements are identical and k > 0, the answer is 1; otherwise, the answer is 1.

Array with a large number of duplicates for a single element

How to Handle:

The algorithm should correctly handle cases where many elements are the same and focus on minimizing their impact by increasing them to be equal.

Large input array size that could lead to memory issues

How to Handle:

Use an efficient data structure like a hash map or counter to store element frequencies to minimize memory usage, and potentially use the `collections.Counter` library to improve this

Elements that, when incremented by one, cause integer overflow

How to Handle:

The algorithm should avoid arithmetic overflows with very large array element values, and modulo arithmetic should be considered if the prompt specifies upper boundaries to avoid the integer overflow.

Input array already contains only distinct elements

How to Handle:

Return n - k if k < n else 1 where n is the array's size to account for the given operation.

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 values within the input array and the number of operations `k`? Can they be negative?
Is the input array guaranteed to be non-empty, or should I handle the case where the input array is null or empty?
If after performing the operations, multiple arrays achieve the maximum number of distinct elements, is any valid array acceptable?
Can you clarify what happens if I use all `k` operations and still can't make all elements distinct? What should be returned in this case?
If there are duplicate values in the array, are there any specific rules or constraints on how those duplicates should be handled during the operations?

Brute Force Solution

Approach

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

Start with the original set of numbers.
Consider each number in the set individually.
For each number, try reducing its count by 1 (if possible and if you still have operations remaining).
After reducing a number's count, recalculate the number of distinct elements in the new set.
Keep track of the highest number of distinct elements seen so far.
Repeat the reduction process for all possible combinations of numbers and operation counts.
The maximum distinct count you've tracked is the result.

Code Implementation

def find_maximum_distinct_elements_brute_force(numbers, operation_count):
    maximum_distinct_count = 0

    def calculate_distinct_count(current_numbers):
        counts = {}
        for number in current_numbers:
            counts[number] = counts.get(number, 0) + 1
        return len(counts)

    def solve(current_numbers, operations_remaining):
        nonlocal maximum_distinct_count

        # Update max count, as we may have found a better combo
        maximum_distinct_count = max(maximum_distinct_count,
                                     calculate_distinct_count(current_numbers))

        # Base case: No more operations to perform
        if operations_remaining == 0:
            return

        for index in range(len(current_numbers)):
            # Do not process a number if it is already zero
            if current_numbers[index] > 0:

                # Create a copy so we don't modify the original list
                new_numbers = current_numbers[:]
                new_numbers[index] -= 1

                # Recurse with one fewer operation
                solve(new_numbers, operations_remaining - 1)

    solve(numbers, operation_count)
    return maximum_distinct_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Count how many times each number appears in the input.
Look at the numbers that appear more than once. We want to reduce these first because they contribute to the duplicate count.
Sort these counts from largest to smallest. This lets us focus on the most frequent numbers first.
Go through each count one by one. If we have enough operations, reduce the count to 1. If we don't have enough operations, stop and move to the next step.
Subtract the number of operations used from the total operations available.
Count the number of elements that are now distinct. This will be the number of elements with a count of 1 or greater.
If we have operations left, we need to consider elements that only appeared once in the initial array. We want to make sure we aren't wasting operations deleting single instances of numbers, if possible.
If we have more operations than elements with count 1, determine how many of these elements can be entirely removed from the input.
Add this value to our distinct count from step 6. Since any duplicate removal now removes an element with a count of one, our new result is the maximum number of distinct elements after the given number of operations.

Code Implementation

def max_distinct_elements(numbers, operations_available):
    element_counts = {}
    for number in numbers:
        element_counts[number] = element_counts.get(number, 0) + 1

    counts_greater_than_one = []
    single_count_elements = 0

    for count in element_counts.values():
        if count > 1:
            counts_greater_than_one.append(count)
        else:
            single_count_elements += 1

    # Sort counts to efficiently reduce most frequent first
    counts_greater_than_one.sort(reverse=True)

    for i in range(len(counts_greater_than_one)):
        count = counts_greater_than_one[i]

        reduction_needed = count - 1
        if operations_available >= reduction_needed:
            operations_available -= reduction_needed
            counts_greater_than_one[i] = 1
        else:
            counts_greater_than_one[i] -= operations_available
            operations_available = 0
            break

    distinct_elements = 0
    for count in counts_greater_than_one:
        if count >= 1:
            distinct_elements += 1

    distinct_elements += single_count_elements

    # Remove single count elements only if operations remain
    if operations_available > 0:

        # Remove as many single elements as possible
        elements_to_remove = min(single_count_elements, operations_available)

        # Update the distinct count accordingly
        distinct_elements -= elements_to_remove

        # Account for excess operations that can't remove more single elements
        operations_available -= elements_to_remove

    return distinct_elements

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as no distinct elements can be achieved with an empty input.
k = 0 (no operations allowed)	Return the number of distinct elements in the original array.
k >= n (operations exceed array size)	If k is greater or equal to n, all elements can be potentially made equal, thus the answer is 1.
Array with all identical elements	If all elements are identical and k > 0, the answer is 1; otherwise, the answer is 1.
Array with a large number of duplicates for a single element	The algorithm should correctly handle cases where many elements are the same and focus on minimizing their impact by increasing them to be equal.
Large input array size that could lead to memory issues	Use an efficient data structure like a hash map or counter to store element frequencies to minimize memory usage, and potentially use the `collections.Counter` library to improve this
Elements that, when incremented by one, cause integer overflow	The algorithm should avoid arithmetic overflows with very large array element values, and modulo arithmetic should be considered if the prompt specifies upper boundaries to avoid the integer overflow.
Input array already contains only distinct elements	Return n - k if k < n else 1 where n is the array's size to account for the given operation.

Null or empty input array

How to Handle:

Return 0, as no distinct elements can be achieved with an empty input.

k = 0 (no operations allowed)

How to Handle:

Return the number of distinct elements in the original array.

k >= n (operations exceed array size)

How to Handle:

If k is greater or equal to n, all elements can be potentially made equal, thus the answer is 1.

Array with all identical elements

How to Handle:

If all elements are identical and k > 0, the answer is 1; otherwise, the answer is 1.

Array with a large number of duplicates for a single element

How to Handle:

The algorithm should correctly handle cases where many elements are the same and focus on minimizing their impact by increasing them to be equal.

Large input array size that could lead to memory issues

How to Handle:

Use an efficient data structure like a hash map or counter to store element frequencies to minimize memory usage, and potentially use the `collections.Counter` library to improve this

Elements that, when incremented by one, cause integer overflow

How to Handle:

Input array already contains only distinct elements

How to Handle:

Return n - k if k < n else 1 where n is the array's size to account for the given operation.