algoadvance

We are given an integer array nums. We need to construct a difference array diff such that for each i (0 <= i < n):

• diff[i] is equal to the difference between the number of distinct elements in the subarray nums[0:i+1] and the number of distinct elements in the subarray nums[i+1:n].

We need to return the diff array.

---

Clarifying Questions

1. Input Size: Is there any constraint on the size of the input array nums?
2. Elements Range: What kinds of integers can the array nums contain (positive, negative, zero)?
3. Edge Cases: What should be returned if the input array is empty?

Assume for our purposes:

1. The input size n can be up to (10^5).
2. The integers in nums can be positive, negative, and zero.
3. For an empty input, since there are no elements to compare, it should return an empty array.

Given this, let’s proceed with the solution.

---

Strategy

1. Preprocessing: Use a prefix approach. Compute:
   • The number of distinct elements in the prefix subarray nums[0:i+1].
   • The number of distinct elements in the suffix subarray nums[i+1:n].
2. Prefix Array: Construct an array prefix_distinct such that prefix_distinct[i] stores the count of distinct elements from the start to index i.

3. Suffix Array: Construct an array suffix_distinct such that suffix_distinct[i] stores the count of distinct elements from index i to the end.

4. Difference Calculation: Compute the diff array by subtracting the suffix_distinct values from prefix_distinct values.

5. Edge Cases: Handle cases where the array might be empty or very short.

---

Code

def find_distinct_difference_array(nums):
    n = len(nums)
    if n == 0:
        return []
    
    # Initialize the prefix and suffix distinct count arrays
    prefix_distinct = [0] * n
    suffix_distinct = [0] * n
    
    seen_prefix = set()
    seen_suffix = set()
    
    # Constructing prefix_distinct array
    for i in range(n):
        seen_prefix.add(nums[i])
        prefix_distinct[i] = len(seen_prefix)
    
    # Constructing suffix_distinct array
    for i in range(n-1, -1, -1):
        seen_suffix.add(nums[i])
        suffix_distinct[i] = len(seen_suffix)
    
    # Calculating the difference array
    diff = [0] * n
    for i in range(n):
        if i + 1 < n:
            diff[i] = prefix_distinct[i] - suffix_distinct[i+1]
        else:
            diff[i] = prefix_distinct[i]
    
    return diff

# Example usage
nums = [1, 2, 3, 4, 5]
print(find_distinct_difference_array(nums))  # Output should be [1, 1, 1, 1, 1]

---

Time Complexity

• Prefix Processing: Constructing the prefix_distinct array involves iterating over the array and maintaining a set. This is (O(n)).
• Suffix Processing: Constructing the suffix_distinct array also involves a single pass with a set, making it (O(n)).
• Difference Calculation: Computing the final diff array comprises a simple iteration which is (O(n)).

Therefore, the overall time complexity of the solution is (O(n)).

Space Complexity

• Prefix and Suffix Arrays: We store two arrays of size n.
• Sets: We use two sets that in the worst case can grow up to size n.

Thus, the space complexity is (O(n)).

Try our interview co-pilot at AlgoAdvance.com

• Got blindsided by a question you didn’t expect?

• Spend too much time studying?

• Or simply don’t have the time to go over all 3000 questions?