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package g2601_2700.s2670_find_the_distinct_difference_array;

// #Easy #Array #Hash_Table #2023_09_10_Time_5_ms_(92.12%)_Space_44.5_MB_(66.07%)

import java.util.HashSet;

/**
 * 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`
**/
public class Solution {
    public int[] distinctDifferenceArray(int[] nums) {
        int n = nums.length;
        HashSet prefixSet = new HashSet<>();
        HashSet suffixSet = new HashSet<>();
        int[] preList = new int[n];
        int[] sufList = new int[n];
        int[] ans = new int[n];
        for (int i = 0; i < nums.length; i++) {
            prefixSet.add(nums[i]);
            suffixSet.add(nums[n - 1 - i]);
            preList[i] = prefixSet.size();
            sufList[n - 1 - i] = suffixSet.size();
        }
        for (int i = 0; i < nums.length; i++) {
            if (i == nums.length - 1) {
                ans[i] = preList[i];
            } else {
                ans[i] = preList[i] - sufList[i + 1];
            }
        }
        return ans;
    }
}