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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2601_2700.s2615_sum_of_distances;

// #Medium #Array #Hash_Table #Prefix_Sum #2023_08_30_Time_13_ms_(100.00%)_Space_70.4_MB_(43.45%)

import java.util.HashMap;

/**
 * 2615 - Sum of Distances\.
 *
 * Medium
 *
 * You are given a **0-indexed** integer array `nums`. There exists an array `arr` of length `nums.length`, where `arr[i]` is the sum of `|i - j|` over all `j` such that `nums[j] == nums[i]` and `j != i`. If there is no such `j`, set `arr[i]` to be `0`.
 *
 * Return _the array_ `arr`_._
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,3,1,1,2]
 *
 * **Output:** [5,0,3,4,0]
 *
 * **Explanation:**
 *
 * When i = 0, nums[0] == nums[2] and nums[0] == nums[3]. Therefore, arr[0] = |0 - 2| + |0 - 3| = 5.
 *
 * When i = 1, arr[1] = 0 because there is no other index with value 3.
 *
 * When i = 2, nums[2] == nums[0] and nums[2] == nums[3]. Therefore, arr[2] = |2 - 0| + |2 - 3| = 3.
 *
 * When i = 3, nums[3] == nums[0] and nums[3] == nums[2]. Therefore, arr[3] = |3 - 0| + |3 - 2| = 4.
 *
 * When i = 4, arr[4] = 0 because there is no other index with value 2.
 *
 * **Example 2:**
 *
 * **Input:** nums = [0,5,3]
 *
 * **Output:** [0,0,0]
 *
 * **Explanation:** Since each element in nums is distinct, arr[i] = 0 for all i.
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁵
 * *   0 <= nums[i] <= 10⁹
**/
public class Solution {
    public long[] distance(int[] nums) {
        HashMap map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            long[] temp = map.computeIfAbsent(nums[i], k -> new long[4]);
            temp[0] += i;
            temp[2]++;
        }
        long[] ans = new long[nums.length];
        for (int i = 0; i < nums.length; i++) {
            long[] temp = map.get(nums[i]);
            ans[i] += i * temp[3] - temp[1];
            temp[1] += i;
            temp[3]++;
            ans[i] += temp[0] - temp[1] - i * (temp[2] - temp[3]);
        }
        return ans;
    }
}