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g2601_2700.s2670_find_the_distinct_difference_array.readme.md Maven / Gradle / Ivy

2670\. Find the Distinct Difference Array

Easy

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`