How do you solve Intervals Between Identical Elements on LeetCode?

Intervals Between Identical Elements (LeetCode #2121) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap to store indices allows for efficient retrieval and calculation.

What is the brute force solution for Intervals Between Identical Elements?

The brute force approach for Intervals Between Identical Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the interval for each element by comparing it with every other element in the array. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Intervals Between Identical Elements?

Intervals Between Identical Elements uses the following concepts: Array, Hash Table, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Intervals Between Identical Elements?

Think about how to reduce time complexity early on in the problem. Explain your thought process clearly while coding. Practice dry runs of your solution to ensure correctness.

What are common mistakes when solving Intervals Between Identical Elements?

Not using a HashMap for index storage, leading to unnecessary repeated calculations. Failing to account for both directions when calculating intervals.

#2121

Intervals Between Identical Elements

Medium

ArrayHash TablePrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages a HashMap to store indices of each unique value, allowing us to calculate the intervals in a single pass. This significantly reduces the number of operations needed.

⚙️

Algorithm

3 steps

1Step 1: Create a HashMap to store each unique value and its corresponding list of indices.
2Step 2: Iterate through the array and populate the HashMap with indices for each value.
3Step 3: For each unique value, calculate the total intervals using the stored indices by leveraging prefix sums to efficiently compute the distances.

solution.py28 lines

1# Full working Python code
2
3def sum_of_intervals(arr):
4    from collections import defaultdict
5    n = len(arr)
6    intervals = [0] * n
7    index_map = defaultdict(list)
8    for i in range(n):
9        index_map[arr[i]].append(i)
10    for indices in index_map.values():
11        total = 0
12        prefix_sum = 0
13        for i in range(len(indices)):
14            idx = indices[i]
15            total += idx * i - prefix_sum
16            intervals[idx] += total
17            prefix_sum += idx
18        total = 0
19        prefix_sum = 0
20        for i in range(len(indices) - 1, -1, -1):
21            idx = indices[i]
22            total += prefix_sum - idx * (len(indices) - 1 - i)
23            intervals[idx] += total
24            prefix_sum += idx
25    return intervals
26
27# Example usage
28print(sum_of_intervals([2,1,3,1,2,3,3]))

ℹ

Complexity note: This complexity is achieved because we only traverse the array a few times, and the use of HashMap allows for efficient index storage and retrieval.

1Using a HashMap to store indices allows for efficient retrieval and calculation.
2Prefix sums help to compute intervals without redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.