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Java-based LeetCode algorithm problem solutions, regularly updated
2121\. Intervals Between Identical Elements
Medium
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 <= 105
* 1 <= arr[i] <= 105