How do you solve Find the Distinct Difference Array on LeetCode?

Find the Distinct Difference Array (LeetCode #2670) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a frequency map helps efficiently track distinct elements.

What is the brute force solution for Find the Distinct Difference Array?

The brute force approach for Find the Distinct Difference Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the distinct elements in both the prefix and suffix for each index. It uses nested loops to count distinct elements, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Distinct Difference Array?

Find the Distinct Difference Array uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Distinct Difference Array?

Always think about how to reduce the number of iterations needed. Consider using data structures like HashMaps for counting problems. Explain your thought process clearly while coding.

What are common mistakes when solving Find the Distinct Difference Array?

Not using a set or map to track distinct elements efficiently. Overlooking the need to update counts correctly while iterating.

#2670

Find the Distinct Difference Array

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses two passes through the array to maintain counts of distinct elements in a single pass for both prefix and suffix, leveraging a HashMap for efficiency.

⚙️

Algorithm

6 steps

1Step 1: Create a frequency map to count occurrences of each element in the array.
2Step 2: Initialize two variables: 'prefixCount' for distinct elements in the prefix and 'suffixCount' for distinct elements in the suffix.
3Step 3: Calculate the initial suffixCount based on the frequency map.
4Step 4: Iterate through the array, updating prefixCount and suffixCount as you go, and calculate diff[i] as prefixCount - suffixCount.
5Step 5: Update the frequency map as you include elements in the prefix.
6Step 6: Return the 'diff' list.

solution.py22 lines

1# Full working Python code
2
3def distinctDifferenceArray(nums):
4    from collections import Counter
5    n = len(nums)
6    diff = []
7    freq = Counter(nums)
8    suffixCount = len(freq)
9    prefixCount = 0
10
11    for i in range(n):
12        if freq[nums[i]] == 1:
13            suffixCount -= 1
14        freq[nums[i]] -= 1
15        if freq[nums[i]] == 0:
16            del freq[nums[i]]
17        prefixCount += 1
18        diff.append(prefixCount - suffixCount)
19    return diff
20
21# Example usage
22print(distinctDifferenceArray([1,2,3,4,5]))

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the array to build the frequency map and another pass to compute the distinct counts. The space complexity is O(n) due to the storage of the frequency map.

1Using a frequency map helps efficiently track distinct elements.
2Two passes through the array can optimize the counting process.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.