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Java-based LeetCode algorithm problem solutions, regularly updated
package g2101_2200.s2121_intervals_between_identical_elements;
// #Medium #Array #Hash_Table #Prefix_Sum #2022_06_02_Time_93_ms_(79.63%)_Space_188.7_MB_(66.05%)
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
/**
* 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
**/
public class Solution {
public long[] getDistances(int[] arr) {
int n = arr.length;
Map> map = new HashMap<>();
for (int i = 0; i < n; i++) {
List 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 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;
}
}
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