#2670

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Find the Distinct Difference Array

Calculate the distinct difference array of a given list of integers by counting distinct elements in the prefix and suffix of the array.

array hash table

Problem Statement

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.

Examples

Example 1

Input: nums = [1,2,3,4,5]

Output: [-3,-1,1,3,5]

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]

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

Solution Approach

Use Hash Set to Count Distinct Elements

The most efficient approach for counting distinct elements in the prefix and suffix is to use a hash set. You can keep track of the distinct elements in the prefix and suffix by iterating over the array while maintaining two hash sets, one for the prefix and one for the suffix.

Iterate Through the Array Efficiently

Scan through the array while updating the counts of distinct elements in the prefix and suffix. For each index, calculate the difference between the distinct count in the prefix and the suffix. The prefix count is updated as the index moves forward, and the suffix count decreases accordingly.

Optimized Time Complexity with Hashing

The time complexity can be optimized to O(n) by iterating over the array once while utilizing hash sets for constant time insertion and lookup. This eliminates the need for nested loops, significantly improving performance.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the length of the array, as the solution iterates over the array once and performs constant time operations with the hash sets. The space complexity is also O(n) because we use hash sets to track distinct elements in the prefix and suffix.

What Interviewers Usually Probe

Focus on how efficiently you can update distinct counts in both prefix and suffix.
Watch for candidates who may attempt brute force methods like nested loops that result in O(n^2) time complexity.
Evaluate the candidate's approach to optimizing space usage, especially when dealing with hash sets.

Common Pitfalls or Variants

Common pitfalls

Using nested loops for counting distinct elements leads to inefficient O(n^2) time complexity.
Not updating the suffix count correctly while iterating through the array.
Forgetting to handle the case where there are no elements in the suffix for the last element.

Follow-up variants

Modify the problem to find the distinct difference array for subarrays of different lengths.
Allow negative numbers in the array and handle distinct counting for negative values.
Increase the array size limit and evaluate performance under larger inputs.

FAQ

What is the time complexity of the solution for the distinct difference array?

The time complexity of the optimal solution is O(n), where n is the length of the input array. This is achieved by using hash sets to count distinct elements in the prefix and suffix while iterating through the array once.

How can I optimize space complexity in the distinct difference array problem?

Space complexity can be optimized by using hash sets for the prefix and suffix, ensuring that the distinct element counts are tracked in constant space relative to the array's size.

What pattern should I use to solve the Find the Distinct Difference Array problem?

The primary pattern is array scanning combined with hash lookups, where hash sets efficiently track distinct elements in both the prefix and suffix of the array.

What is the distinct difference array in this problem?

The distinct difference array is an array where each element diff[i] represents the difference between the distinct elements in the prefix nums[0...i] and the distinct elements in the suffix nums[i+1...n-1].

What should I focus on when solving the Find the Distinct Difference Array problem?

Focus on efficiently counting distinct elements using hash sets, iterating over the array in linear time while ensuring the correct update of distinct counts for both prefix and suffix.

terminal

Solution

Solution 1: Hash Table + Preprocessed Suffix

We can preprocess a suffix array $suf$, where $suf[i]$ represents the number of distinct elements in the suffix $nums[i, ..., n - 1]$. During the preprocessing, we use a hash table $s$ to maintain the elements that have appeared in the suffix, so we can query the number of distinct elements in the suffix in $O(1)$ time.

class Solution:
    def distinctDifferenceArray(self, nums: List[int]) -> List[int]:
        n = len(nums)
        suf = [0] * (n + 1)
        s = set()
        for i in range(n - 1, -1, -1):
            s.add(nums[i])
            suf[i] = len(s)
        s.clear()
        ans = [0] * n
        for i, x in enumerate(nums):
            s.add(x)
            ans[i] = len(s) - suf[i + 1]
        return ans

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