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2121. Intervals Between Identical Elements

Difficulty Topics

Description

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

Solution

intervals-between-identical-elements.py
class Solution:
    def getDistances(self, arr: List[int]) -> List[int]:
        res = []
        mp = collections.defaultdict(list)
        prefix = collections.defaultdict(list)

        for i, x in enumerate(arr):
            mp[x].append(i)
            prefix[x].append(i)

        for k, p in prefix.items():
            for i in range(1, len(p)):
                p[i] += p[i - 1]
            prefix[k] = [0] + prefix[k]

        for i, x in enumerate(arr):
            n = len(mp[x])

            index = bisect.bisect_left(mp[x], i)

            left = index * i - (prefix[x][index] - prefix[x][0])
            right = (prefix[x][-1] - prefix[x][index + 1]) - i * (n - index - 1)

            res.append(left + right)

        return res