Find the Distinct Difference Array

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You are given a 0-indexed array nums of length n.

The distinct difference array of nums is an array diff of length n such that diff[i] is equal to the number of distinct elements in the suffix nums[i + 1, ..., n - 1] subtracted from the number of distinct elements in the prefix nums[0, ..., i].

Return the distinct difference array of nums.

Note that nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j inclusive. Particularly, if i > j then nums[i, ..., j] denotes an empty subarray.

Example 1:

Input: nums = [1,2,3,4,5]
Output: [-3,-1,1,3,5]
Explanation: For index i = 0, there is 1 element in the prefix and 4 distinct elements in the suffix. Thus, diff[0] = 1 - 4 = -3.
For index i = 1, there are 2 distinct elements in the prefix and 3 distinct elements in the suffix. Thus, diff[1] = 2 - 3 = -1.
For index i = 2, there are 3 distinct elements in the prefix and 2 distinct elements in the suffix. Thus, diff[2] = 3 - 2 = 1.
For index i = 3, there are 4 distinct elements in the prefix and 1 distinct element in the suffix. Thus, diff[3] = 4 - 1 = 3.
For index i = 4, there are 5 distinct elements in the prefix and no elements in the suffix. Thus, diff[4] = 5 - 0 = 5.

Example 2:

Input: nums = [3,2,3,4,2]
Output: [-2,-1,0,2,3]
Explanation: For index i = 0, there is 1 element in the prefix and 3 distinct elements in the suffix. Thus, diff[0] = 1 - 3 = -2.
For index i = 1, there are 2 distinct elements in the prefix and 3 distinct elements in the suffix. Thus, diff[1] = 2 - 3 = -1.
For index i = 2, there are 2 distinct elements in the prefix and 2 distinct elements in the suffix. Thus, diff[2] = 2 - 2 = 0.
For index i = 3, there are 3 distinct elements in the prefix and 1 distinct element in the suffix. Thus, diff[3] = 3 - 1 = 2.
For index i = 4, there are 3 distinct elements in the prefix and no elements in the suffix. Thus, diff[4] = 3 - 0 = 3.

Constraints:

1 <= n == nums.length <= 50
1 <= nums[i] <= 50

Try coding it in LeetCode

Find the Distinct Difference Array

Easy

Asked by:

5 views

Topics:

Arrays

You are given a 0-indexed array nums of length n.

The distinct difference array of nums is an array diff of length n such that diff[i] is equal to the number of distinct elements in the suffix nums[i + 1, ..., n - 1] subtracted from the number of distinct elements in the prefix nums[0, ..., i].

Return the distinct difference array of nums.

Note that nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j inclusive. Particularly, if i > j then nums[i, ..., j] denotes an empty subarray.

Example 1:

Input: nums = [1,2,3,4,5]
Output: [-3,-1,1,3,5]
Explanation: For index i = 0, there is 1 element in the prefix and 4 distinct elements in the suffix. Thus, diff[0] = 1 - 4 = -3.
For index i = 1, there are 2 distinct elements in the prefix and 3 distinct elements in the suffix. Thus, diff[1] = 2 - 3 = -1.
For index i = 2, there are 3 distinct elements in the prefix and 2 distinct elements in the suffix. Thus, diff[2] = 3 - 2 = 1.
For index i = 3, there are 4 distinct elements in the prefix and 1 distinct element in the suffix. Thus, diff[3] = 4 - 1 = 3.
For index i = 4, there are 5 distinct elements in the prefix and no elements in the suffix. Thus, diff[4] = 5 - 0 = 5.

Example 2:

Input: nums = [3,2,3,4,2]
Output: [-2,-1,0,2,3]
Explanation: For index i = 0, there is 1 element in the prefix and 3 distinct elements in the suffix. Thus, diff[0] = 1 - 3 = -2.
For index i = 1, there are 2 distinct elements in the prefix and 3 distinct elements in the suffix. Thus, diff[1] = 2 - 3 = -1.
For index i = 2, there are 2 distinct elements in the prefix and 2 distinct elements in the suffix. Thus, diff[2] = 2 - 2 = 0.
For index i = 3, there are 3 distinct elements in the prefix and 1 distinct element in the suffix. Thus, diff[3] = 3 - 1 = 2.
For index i = 4, there are 3 distinct elements in the prefix and no elements in the suffix. Thus, diff[4] = 3 - 0 = 3.

Constraints:

1 <= n == nums.length <= 50
1 <= nums[i] <= 50

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:

Can the input array contain duplicate numbers?
What are the possible value ranges for the numbers in the input array (e.g., integers, floats, and what are their minimum and maximum values)?
If the input array is empty or contains only one element, what should the function return?
Could you define more precisely what you mean by 'distinct difference'? Specifically, should I remove duplicates before or after calculating the differences?
If multiple 'distinct difference arrays' are possible, is there a specific criteria for selecting the one to return?

Brute Force Solution

Approach

Imagine you have a list of numbers. For each number, we want to find the difference between how many unique numbers come before it and how many unique numbers come after it. The brute force method essentially checks every single number in the list and then meticulously counts the unique numbers before and after it each time.

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

For the first number in the list, carefully look at all the numbers that come before it. Since it's the first number, there are none before it, so the count of unique numbers before it is zero.
Now, for that same first number, carefully look at all the numbers that come after it in the list.
Remove any duplicate numbers from that list of numbers that come after it. Only keep one copy of each unique number.
Count how many unique numbers are left in that list. This is the number of unique numbers after the first number.
Subtract the number of unique numbers after from the number of unique numbers before (which was zero). This is the difference for the first number.
Write down this difference.
Now, repeat these steps for the second number in the list, then the third, and so on, until you have calculated and written down the difference for every number in the original list.

Code Implementation

def find_distinct_difference_array(numbers):
    distinct_difference_list = []

    for current_index in range(len(numbers)):
        # Initialize the count of unique numbers before the current number.
        unique_before_count = 0
        unique_numbers_before = set()

        # Count unique numbers before the current number.
        for index_before in range(current_index):
            unique_numbers_before.add(numbers[index_before])

        unique_before_count = len(unique_numbers_before)

        # Get the numbers after the current number and remove duplicates.
        numbers_after = numbers[current_index + 1:]
        unique_numbers_after = []
        seen = set()

        # Remove duplicate numbers from numbers_after.
        for number in numbers_after:
            if number not in seen:
                unique_numbers_after.append(number)
                seen.add(number)

        # Calculate the difference and add it to the result list.
        difference = unique_before_count - len(unique_numbers_after)
        distinct_difference_list.append(difference)

    return distinct_difference_list

Big(O) Analysis

Time Complexity

O(n³) – The algorithm iterates through each of the n elements in the input array. For each element, determining the unique elements before it takes O(i) time in the ith iteration where i goes from 0 to n-1 and determining the unique elements after it takes O(n-i) time, but requires removing duplicates first. The duplicate removal can be implemented using a set which takes O(n) time. Therefore, in the worst case, each iteration of the outer loop requires O(n) time complexity, leading to O(n*n) for the inner loops. However, since determining the unique elements after it takes O(n) time, and we perform this operation for n elements, the total runtime comes to O(n³).

Space Complexity

O(N) – The provided solution, when calculating the unique numbers after each element, creates a temporary list to store the elements after the current element. This list can, in the worst-case scenario where all elements after the current one are unique, be of size N-1, where N is the size of the input list. Furthermore, duplicate removal from this list likely uses a set or similar data structure, also potentially of size N-1 in the worst case. Therefore, the auxiliary space used is proportional to N, resulting in a space complexity of O(N).

Optimal Solution

Approach

The goal is to find the difference between the count of unique numbers to the left and to the right for each number in the sequence. We can cleverly use two sets to track unique numbers on either side as we go through the sequence, allowing us to efficiently compute the differences.

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

First, create two empty sets, one to track the unique numbers we've seen so far from the left, and another to track the unique numbers from the right.
Also, make a new list that will eventually hold the differences we calculate.
Initially, populate the right-side set with all the unique numbers from the original sequence.
Now, go through the original sequence one number at a time.
For each number, remove it from the right-side set, if its present.
Count how many unique numbers are currently in the left-side set (this represents the unique numbers to the left).
Count how many unique numbers are now in the right-side set (this represents the unique numbers to the right).
Subtract the count of unique numbers on the right from the count of unique numbers on the left, and add this difference to our difference list.
Add the current number to the left-side set.
After doing this for every number in the original sequence, the difference list will contain the desired distinct differences.

Code Implementation

def find_distinct_difference_array(sequence):
    left_unique_numbers = set()
    right_unique_numbers = set()
    distinct_differences = []

    # Populate right set initially
    for number in sequence:
        right_unique_numbers.add(number)

    for number in sequence:
        if number in right_unique_numbers:
            right_unique_numbers.remove(number)

        # Calculate difference and append
        left_count = len(left_unique_numbers)
        right_count = len(right_unique_numbers)
        difference = left_count - right_count
        distinct_differences.append(difference)

        # Add current number to left set.
        left_unique_numbers.add(number)

    return distinct_differences

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once. Inside the loop, set operations (insertion, deletion, and size retrieval) take constant time on average, assuming a good hash function is used. The initial population of the right set also takes O(n) time, but it doesn't dominate the single loop through the array. Therefore, the time complexity is dominated by the single pass through the input array.

Space Complexity

O(N) – The solution uses two sets, leftSet and rightSet, to store unique numbers. In the worst-case scenario, all numbers in the input sequence are distinct, meaning both sets could potentially hold up to N elements, where N is the length of the input sequence. Additionally, a new list named differenceList is created to store the calculated differences, which will also have a size of N. Therefore, the auxiliary space required is proportional to N, leading to a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or undefined input array	Return an empty list or throw an IllegalArgumentException if the input array is null or undefined.
Array with fewer than two elements	Return an empty list because calculating the distinct difference requires at least two elements.
Array with all identical elements	The left distinct count will always be 0, potentially leading to predictable behavior; ensure correct handling by checking map after element addition.
Array containing large numbers (potential integer overflow)	Use a data type that can handle large numbers or consider using modular arithmetic if the problem allows it.
Array with a large number of elements (performance)	Ensure the chosen algorithm has acceptable time complexity; using an O(n) or O(n log n) algorithm is preferable for scaling.
Array with negative numbers	The algorithm should correctly handle negative numbers without any modification if using hash maps or sets.
Array containing duplicate elements with varying frequencies	The frequency of elements should not affect the correctness as we're concerned with distinct counts.
Maximum possible integer value present in the array	Carefully handle edge cases where the value can be equal to the maximum possible integer to avoid integer overflow when calculating left and right distinct counts.