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Minimum Number of Operations to Make Elements in Array Distinct

Easy

Amazon

4 views

Topics:

Arrays

You are given an integer array nums. You need to ensure that the elements in the array are distinct. To achieve this, you can perform the following operation any number of times:

Remove 3 elements from the beginning of the array. If the array has fewer than 3 elements, remove all remaining elements.

Note that an empty array is considered to have distinct elements. Return the minimum number of operations needed to make the elements in the array distinct.

For example:

nums = [1,2,3,4,2,3,3,5,7] should return 2
nums = [4,5,6,4,4] should return 2
nums = [6,7,8,9] should return 0

Write a function to efficiently determine the minimum number of operations to achieve distinct elements.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

Minimum Number of Operations to Make Elements in Array Distinct

Easy

Amazon

4 views

Topics:

Arrays

You are given an integer array nums. You need to ensure that the elements in the array are distinct. To achieve this, you can perform the following operation any number of times:

Remove 3 elements from the beginning of the array. If the array has fewer than 3 elements, remove all remaining elements.

Note that an empty array is considered to have distinct elements. Return the minimum number of operations needed to make the elements in the array distinct.

For example:

nums = [1,2,3,4,2,3,3,5,7] should return 2
nums = [4,5,6,4,4] should return 2
nums = [6,7,8,9] should return 0

Write a function to efficiently determine the minimum number of operations to achieve distinct elements.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

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 range of values within the `nums` array? Can I assume they are all non-negative integers?
What is the maximum size of the `nums` array?
If the input array is already distinct, should I return 0?
Can I modify the input array directly, or do I need to create a copy?
Are there any specific error conditions or edge cases I should handle, such as an empty input array or a very large input array that could lead to integer overflow issues during calculations?

Brute Force Solution

Approach

The brute force method for making array elements distinct involves exploring all possible ways to increase the array elements. We systematically try increasing each element, one at a time, to see if we can make all the elements unique.

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

Look at the first number in the list.
Check if this number is already present later in the list.
If it is, try increasing the first number by one.
Check again if this new number is present later in the list.
Keep increasing the first number until it becomes unique compared to the rest of the list.
Count how many times you had to increase the number.
Repeat this process for every number in the list, making sure it's different from all the numbers that come after it.
Add up all the increase counts from each number to get the total number of operations.
This total represents the minimum operations to make all numbers distinct using this specific order of changes.

Code Implementation

def min_operations_distinct_brute_force(numbers):
    operations_count = 0
    array_length = len(numbers)

    for current_index in range(array_length):
        current_number = numbers[current_index]

        #Check for duplicates later in the array.
        while True:
            is_duplicate = False
            for subsequent_index in range(current_index + 1, array_length):
                if current_number == numbers[subsequent_index]:
                    is_duplicate = True
                    break

            if not is_duplicate:
                break

            #Increase the current number until unique.
            current_number += 1
            operations_count += 1

        #Update array with the new distinct value.
        numbers[current_index] = current_number

    return operations_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the array. For each element, it checks for duplicates in the remaining portion of the array. In the worst case, for the first element, it compares with up to n-1 other elements, for the second element, it compares with up to n-2 other elements, and so on. The total number of comparisons can be approximated as n * (n-1) / 2, which simplifies to O(n²).

Space Complexity

O(1) – The described brute force algorithm iterates through the input array and modifies elements in place. While it checks for duplicates and increments values, it doesn't employ any auxiliary data structures like hash maps, lists, or sets to store visited numbers or intermediate results. It uses only a constant number of variables for comparisons and tracking the number of increments. Therefore, the auxiliary space complexity remains constant, regardless of the input array's size (N).

Optimal Solution

Approach

The goal is to make all the numbers in a list different from each other with the fewest possible increases. The best strategy is to consider the numbers in order and, whenever you find duplicates, increase the duplicate just enough to make it unique, and keep track of the total increases.

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

First, we need to arrange the numbers in increasing order, so it's easier to spot the duplicates.
Now, start from the beginning of the sorted list.
If a number is the same as or smaller than the number before it, we need to increase it.
Increase the current number to be one greater than the previous number.
Keep track of how much we increase each number so we can determine the total amount of operations at the end.
Move to the next number and repeat the process. If a number is already greater than the number before it, we do not change it.
Once we've gone through all the numbers, the total increase count is the minimum number of operations needed to make all the numbers unique.

Code Implementation

def min_operations_to_make_distinct(numbers):
    numbers.sort()
    operation_count = 0

    # To track the previous number for comparison.
    previous_number = -1

    for i in range(len(numbers)):

        # If the current number is less than or equal to the previous.
        if numbers[i] <= previous_number:

            # Increase the current number to be just greater than previous.
            operation_count += previous_number - numbers[i] + 1
            numbers[i] = previous_number + 1

            # Update previous_number after incrementing.
            previous_number = numbers[i]

        else:
            # No change needed, update previous number.
            previous_number = numbers[i]

    return operation_count

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 subsequent iteration through the sorted array to identify and resolve duplicates involves a single loop that runs in O(n) time. Since O(n log n) grows faster than O(n) as n increases, the overall time complexity is determined by the sorting step, making the entire process O(n log n).

Space Complexity

O(1) – The provided algorithm sorts the input array in place, which might require O(log N) space depending on the sorting algorithm used. However, the core logic for finding and adjusting duplicate numbers operates directly on the sorted array using a single variable to keep track of the previous element. Therefore, the dominant auxiliary space used is constant, independent of the input size N, where N is the number of elements in the input array, since no extra data structures with size dependent on N are created.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 if the input is null or empty, as no operations are needed to make an empty array distinct.
Array with one element	Return 0 if the array has only one element, as it is already distinct.
Array with all identical elements	The algorithm should increment elements until all are distinct, accumulating the total increments.
Array with already distinct elements	The algorithm should correctly return 0 operations needed when elements are already distinct.
Array with negative numbers	The algorithm must correctly handle negative numbers, allowing incrementing them to achieve distinctness.
Array with large positive numbers that might cause overflow during increment	Use a data type that can accommodate large numbers (e.g., long in Java) to prevent integer overflow.
Array with a very large number of elements (scaling issue)	An efficient sorting algorithm (e.g., merge sort or quicksort with O(n log n) time complexity) should be used for scalability.
Input contains zero	Zero should be treated like any other number during comparisons and incrementations.

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 range of values within the `nums` array? Can I assume they are all non-negative integers?
What is the maximum size of the `nums` array?
If the input array is already distinct, should I return 0?
Can I modify the input array directly, or do I need to create a copy?
Are there any specific error conditions or edge cases I should handle, such as an empty input array or a very large input array that could lead to integer overflow issues during calculations?

Brute Force Solution

Approach

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

Look at the first number in the list.
Check if this number is already present later in the list.
If it is, try increasing the first number by one.
Check again if this new number is present later in the list.
Keep increasing the first number until it becomes unique compared to the rest of the list.
Count how many times you had to increase the number.
Repeat this process for every number in the list, making sure it's different from all the numbers that come after it.
Add up all the increase counts from each number to get the total number of operations.
This total represents the minimum operations to make all numbers distinct using this specific order of changes.

Code Implementation

def min_operations_distinct_brute_force(numbers):
    operations_count = 0
    array_length = len(numbers)

    for current_index in range(array_length):
        current_number = numbers[current_index]

        #Check for duplicates later in the array.
        while True:
            is_duplicate = False
            for subsequent_index in range(current_index + 1, array_length):
                if current_number == numbers[subsequent_index]:
                    is_duplicate = True
                    break

            if not is_duplicate:
                break

            #Increase the current number until unique.
            current_number += 1
            operations_count += 1

        #Update array with the new distinct value.
        numbers[current_index] = current_number

    return operations_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, we need to arrange the numbers in increasing order, so it's easier to spot the duplicates.
Now, start from the beginning of the sorted list.
If a number is the same as or smaller than the number before it, we need to increase it.
Increase the current number to be one greater than the previous number.
Keep track of how much we increase each number so we can determine the total amount of operations at the end.
Move to the next number and repeat the process. If a number is already greater than the number before it, we do not change it.
Once we've gone through all the numbers, the total increase count is the minimum number of operations needed to make all the numbers unique.

Code Implementation

def min_operations_to_make_distinct(numbers):
    numbers.sort()
    operation_count = 0

    # To track the previous number for comparison.
    previous_number = -1

    for i in range(len(numbers)):

        # If the current number is less than or equal to the previous.
        if numbers[i] <= previous_number:

            # Increase the current number to be just greater than previous.
            operation_count += previous_number - numbers[i] + 1
            numbers[i] = previous_number + 1

            # Update previous_number after incrementing.
            previous_number = numbers[i]

        else:
            # No change needed, update previous number.
            previous_number = numbers[i]

    return operation_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 if the input is null or empty, as no operations are needed to make an empty array distinct.
Array with one element	Return 0 if the array has only one element, as it is already distinct.
Array with all identical elements	The algorithm should increment elements until all are distinct, accumulating the total increments.
Array with already distinct elements	The algorithm should correctly return 0 operations needed when elements are already distinct.
Array with negative numbers	The algorithm must correctly handle negative numbers, allowing incrementing them to achieve distinctness.
Array with large positive numbers that might cause overflow during increment	Use a data type that can accommodate large numbers (e.g., long in Java) to prevent integer overflow.
Array with a very large number of elements (scaling issue)	An efficient sorting algorithm (e.g., merge sort or quicksort with O(n log n) time complexity) should be used for scalability.
Input contains zero	Zero should be treated like any other number during comparisons and incrementations.