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Reduce Array Size to The Half

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ArraysGreedy Algorithms

You are given an integer array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.

Return the minimum size of the set so that at least half of the integers of the array are removed.

Example 1:

Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has a size greater than half of the size of the old array.

Example 2:

Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.

Constraints:

2 <= arr.length <= 10⁵
arr.length is even.
1 <= arr[i] <= 10⁵

Try coding it in LeetCode

Reduce Array Size to The Half

Medium

Asked by:

4 views

Topics:

ArraysGreedy Algorithms

You are given an integer array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.

Return the minimum size of the set so that at least half of the integers of the array are removed.

Example 1:

Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has a size greater than half of the size of the old array.

Example 2:

Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.

Constraints:

2 <= arr.length <= 10⁵
arr.length is even.
1 <= arr[i] <= 10⁵

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 is the data type and range of values within the input array?
Can the input array be empty or null?
If multiple sets of numbers can reduce the array size to half or less, should I return the set with the smallest number of elements?
Are duplicate numbers considered distinct when reducing the array size?
What should be returned if it is impossible to reduce the array size to half or less?

Brute Force Solution

Approach

The brute force approach to reduce an array's size to half involves exhaustively checking all possible combinations of elements we could remove. We try every possible group of numbers to remove and see if the remaining numbers are less than half the original array size. This is like trying every single possibility, no matter how inefficient.

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

First, figure out what 'half' the array size means for your specific array. For example, if your array has 10 elements, you need to remove enough elements so that 5 or fewer remain.
Start by considering removing just one unique number from the array and see if the remaining numbers have a count less than or equal to 'half'.
Next, consider removing all possible pairs of unique numbers and check if the remaining numbers have a count less than or equal to 'half'.
Continue this process, each time considering removing a larger set of unique numbers, like groups of three, then four, and so on.
For each set of unique numbers you try removing, count how many numbers are left in the original array after removing all occurrences of your chosen set.
Keep track of the smallest size of the set of unique numbers that, when removed, leaves you with no more than 'half' the original number of elements.
The smallest set of unique numbers you found is your answer. If you reach a point where removing all unique numbers doesn't achieve the result, something may be wrong with the question.

Code Implementation

def reduce_array_size_to_half_brute_force(array):
    array_length = len(array)
    half_array_length = (array_length + 1) // 2

    unique_numbers = list(set(array))
    number_of_unique_numbers = len(unique_numbers)

    for number_of_elements_to_remove in range(1, number_of_unique_numbers + 1):
        # Iterate through all possible combinations of unique numbers to remove
        for combination_indices in combinations(range(number_of_unique_numbers), number_of_elements_to_remove):
            elements_to_remove = [unique_numbers[i] for i in combination_indices]
            remaining_elements_count = 0

            # Count the remaining elements after removing the selected unique numbers
            for element in array:
                if element not in elements_to_remove:
                    remaining_elements_count += 1

            if remaining_elements_count <= half_array_length:
                return number_of_elements_to_remove

    return number_of_unique_numbers

from itertools import combinations

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach examines all possible subsets of unique numbers in the array. In the worst-case scenario, where all numbers are unique, the algorithm effectively generates the power set of the unique numbers. A set with n elements has 2^n subsets. For each subset, the algorithm iterates through the original array of size n to determine the number of elements remaining after removing the subset's elements. Thus, the time complexity is dominated by generating and processing these subsets, leading to O(n * 2^n). This can be simplified to O(2^n) because the 2^n term dominates the complexity.

Space Complexity

O(N!) – The brute force approach described involves checking all possible combinations of unique numbers to remove. In the worst case, it may need to explore a very large number of combinations, potentially up to examining the power set of the set of unique elements in the array. To facilitate this, intermediate sets of unique numbers are created, which could grow exponentially with the number of unique elements. Therefore, the space required to store these combinations of unique numbers to remove dominates the auxiliary space, approaching O(N!) where N represents the number of unique elements which in the worst case is N, the size of the input array. This is due to the algorithm's exhaustive exploration of combinations.

Optimal Solution

Approach

To find the smallest set of numbers to remove that gets the array down to half size, focus on how frequently each number appears. Removing the most frequent numbers first will get you to the goal faster. This way you are picking the elements that will have the biggest impact.

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

First, count how many times each number appears in the array.
Then, create a list of these counts and sort the list from largest to smallest.
Start removing groups of numbers based on their counts, beginning with the group that appears most often.
Keep track of how many numbers you've removed so far.
Stop removing groups as soon as the total number of removed elements is greater than or equal to half the size of the original array.
The number of groups you removed is the answer.

Code Implementation

def reduce_array_size(array_to_reduce):
    element_counts = {}
    for element in array_to_reduce:
        element_counts[element] = element_counts.get(element, 0) + 1

    counts = list(element_counts.values())
    counts.sort(reverse=True)

    removed_elements_count = 0
    removed_groups_count = 0
    array_length = len(array_to_reduce)

    # Iterate through the sorted counts
    for count in counts:
        removed_elements_count += count
        removed_groups_count += 1

        # Stop when at least half the array is removed
        if removed_elements_count >= array_length / 2:

            return removed_groups_count

    return removed_groups_count

Big(O) Analysis

Time Complexity

O(n log n) – The first step involves counting the frequency of each number in the input array of size n, which takes O(n) time. Next, we create a list of these frequencies, which can be at most n in size. Sorting this list of frequencies, in the worst case, takes O(n log n) time. The final step iterates through the sorted frequencies, summing them until we reach half the original array size. This iteration, in the worst case, processes all n frequencies and thus runs in O(n) time. Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity

O(N) – The solution utilizes a hash map to count the frequency of each number, which in the worst case (where all numbers are distinct) requires storing N key-value pairs, where N is the size of the input array. Additionally, it creates a list to store these frequencies. In the worst case, this list can also have N elements. Therefore, the auxiliary space used is proportional to the input size N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0, as no elements need to be removed when the array is already empty.
Array size is already less than or equal to half the original size.	Return 0, as the array already meets the condition.
All elements in the array are distinct.	Return n/2 rounded up, as removing one occurrence of each element will result in the minimum number of removals.
All elements in the array are the same.	Return 1, as removing all but one unique number will reduce the array size to 1.
Array contains negative numbers.	Frequency counting method will still work with negative numbers as keys.
Array with a large number of elements (performance considerations).	Choose frequency counting with hash map which offers O(n) time complexity rather than sorting based approaches.
Integer overflow in frequency counts	Use a data type that can handle large frequency counts, such as long, or check for potential overflow when incrementing counters.
Multiple sets of numbers to remove that reduce the size to half.	The algorithm should return the minimum number of unique numbers to remove for reducing the array size to at least half, so choose the most frequent ones.

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 data type and range of values within the input array?
Can the input array be empty or null?
If multiple sets of numbers can reduce the array size to half or less, should I return the set with the smallest number of elements?
Are duplicate numbers considered distinct when reducing the array size?
What should be returned if it is impossible to reduce the array size to half or less?

Brute Force Solution

Approach

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

First, figure out what 'half' the array size means for your specific array. For example, if your array has 10 elements, you need to remove enough elements so that 5 or fewer remain.
Start by considering removing just one unique number from the array and see if the remaining numbers have a count less than or equal to 'half'.
Next, consider removing all possible pairs of unique numbers and check if the remaining numbers have a count less than or equal to 'half'.
Continue this process, each time considering removing a larger set of unique numbers, like groups of three, then four, and so on.
For each set of unique numbers you try removing, count how many numbers are left in the original array after removing all occurrences of your chosen set.
Keep track of the smallest size of the set of unique numbers that, when removed, leaves you with no more than 'half' the original number of elements.
The smallest set of unique numbers you found is your answer. If you reach a point where removing all unique numbers doesn't achieve the result, something may be wrong with the question.

Code Implementation

def reduce_array_size_to_half_brute_force(array):
    array_length = len(array)
    half_array_length = (array_length + 1) // 2

    unique_numbers = list(set(array))
    number_of_unique_numbers = len(unique_numbers)

    for number_of_elements_to_remove in range(1, number_of_unique_numbers + 1):
        # Iterate through all possible combinations of unique numbers to remove
        for combination_indices in combinations(range(number_of_unique_numbers), number_of_elements_to_remove):
            elements_to_remove = [unique_numbers[i] for i in combination_indices]
            remaining_elements_count = 0

            # Count the remaining elements after removing the selected unique numbers
            for element in array:
                if element not in elements_to_remove:
                    remaining_elements_count += 1

            if remaining_elements_count <= half_array_length:
                return number_of_elements_to_remove

    return number_of_unique_numbers

from itertools import combinations

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, count how many times each number appears in the array.
Then, create a list of these counts and sort the list from largest to smallest.
Start removing groups of numbers based on their counts, beginning with the group that appears most often.
Keep track of how many numbers you've removed so far.
Stop removing groups as soon as the total number of removed elements is greater than or equal to half the size of the original array.
The number of groups you removed is the answer.

Code Implementation

def reduce_array_size(array_to_reduce):
    element_counts = {}
    for element in array_to_reduce:
        element_counts[element] = element_counts.get(element, 0) + 1

    counts = list(element_counts.values())
    counts.sort(reverse=True)

    removed_elements_count = 0
    removed_groups_count = 0
    array_length = len(array_to_reduce)

    # Iterate through the sorted counts
    for count in counts:
        removed_elements_count += count
        removed_groups_count += 1

        # Stop when at least half the array is removed
        if removed_elements_count >= array_length / 2:

            return removed_groups_count

    return removed_groups_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0, as no elements need to be removed when the array is already empty.
Array size is already less than or equal to half the original size.	Return 0, as the array already meets the condition.
All elements in the array are distinct.	Return n/2 rounded up, as removing one occurrence of each element will result in the minimum number of removals.
All elements in the array are the same.	Return 1, as removing all but one unique number will reduce the array size to 1.
Array contains negative numbers.	Frequency counting method will still work with negative numbers as keys.
Array with a large number of elements (performance considerations).	Choose frequency counting with hash map which offers O(n) time complexity rather than sorting based approaches.
Integer overflow in frequency counts	Use a data type that can handle large frequency counts, such as long, or check for potential overflow when incrementing counters.
Multiple sets of numbers to remove that reduce the size to half.	The algorithm should return the minimum number of unique numbers to remove for reducing the array size to at least half, so choose the most frequent ones.