2670. Find the Distinct Difference Array
Description
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
Solutions
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.
After preprocessing the suffix array $suf$, we clear the hash table $s$, and then traverse the array $nums$ again, using the hash table $s$ to maintain the elements that have appeared in the prefix. The answer at position $i$ is the number of distinct elements in $s$ minus $suf[i + 1]$, that is, $s.size() - suf[i + 1]$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
|