2121. Intervals Between Identical Elements

Problem

You are given a 0-indexed array of n integers arr.

The interval between two elements in arr is defined as the absolute difference between their indices. More formally, the interval between arr[i] and arr[j] is |i - j|.

Return an array intervals of length n where intervals[i] **is *the sum of intervals* between arr[i] and each element in arr with the same value as arr[i].**

Note: |x| is the absolute value of x.

  Example 1:

Input: arr = [2,1,3,1,2,3,3]
Output: [4,2,7,2,4,4,5]
Explanation:
- Index 0: Another 2 is found at index 4. |0 - 4| = 4
- Index 1: Another 1 is found at index 3. |1 - 3| = 2
- Index 2: Two more 3s are found at indices 5 and 6. |2 - 5| + |2 - 6| = 7
- Index 3: Another 1 is found at index 1. |3 - 1| = 2
- Index 4: Another 2 is found at index 0. |4 - 0| = 4
- Index 5: Two more 3s are found at indices 2 and 6. |5 - 2| + |5 - 6| = 4
- Index 6: Two more 3s are found at indices 2 and 5. |6 - 2| + |6 - 5| = 5

Example 2:

Input: arr = [10,5,10,10]
Output: [5,0,3,4]
Explanation:
- Index 0: Two more 10s are found at indices 2 and 3. |0 - 2| + |0 - 3| = 5
- Index 1: There is only one 5 in the array, so its sum of intervals to identical elements is 0.
- Index 2: Two more 10s are found at indices 0 and 3. |2 - 0| + |2 - 3| = 3
- Index 3: Two more 10s are found at indices 0 and 2. |3 - 0| + |3 - 2| = 4

  Constraints:

• n == arr.length

• 1 <= n <= 10^5

• 1 <= arr[i] <= 10^5

Solution (Java)

class Solution {
    public long[] getDistances(int[] arr) {
        int n = arr.length;
        Map<Integer, List<Integer>> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            List<Integer> list = map.get(arr[i]);
            if (list == null) {
                list = new ArrayList<>();
            }
            list.add(i);
            map.put(arr[i], list);
        }
        long[] ans = new long[n];
        Arrays.fill(ans, 0);
        for (List<Integer> list : map.values()) {
            long sum = 0;
            int first = list.get(0);
            for (int i = 1; i < list.size(); i++) {
                sum = sum + list.get(i) - first;
            }
            ans[first] = sum;
            int prevElements = 0;
            int nextElements = list.size() - 2;
            for (int i = 1; i < list.size(); i++) {
                int diff = list.get(i) - list.get(i - 1);
                sum = sum + (long) diff * (prevElements - nextElements);
                ans[list.get(i)] = sum;
                prevElements++;
                nextElements--;
            }
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).