Find the Distinct Difference Array Leetcode solution

2670.  Find the Distinct Difference Array

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

solution :

class Solution { public:     vector<int>pre(vector<int>&nums ,int size)     {          unordered_map<int, int>map;         vector<int>result;         int count =0;         for(int i =0; i < size; i++)         {             if(map[nums[i]] ==0)             {                 count++;                 map[nums[i]]++;             }             result.push_back(count);                   }         return result;     }     vector<int>suff(vector<int>nums , int size)     {          unordered_map<int , int>map;         vector<int>result(size , 0);         // map[nums[i]] =1;         int count =0;         for(int i = size -2 ; i >=0 ; i--)         {              if(map[nums[i+1]]==0)             {                 count++;                 map[nums[i+1]] = 1;             }             result[i] = count;         }   return result;       }     vector<int> distinctDifferenceArray(vector<int>& nums) {         int size = nums.size();         vector<int>pre1 = pre(nums , size);         vector<int>suf1 = suff(nums , size);         vector<int>result;         for(int i =0; i < size ; i++)         {             result.push_back(pre1[i] - suf1[i]);         }         return result;     } };