#2121

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Intervals Between Identical Elements

Solve the problem of calculating intervals between identical elements using array scanning and hash lookup.

Problem Statement

You are given a 0-indexed array of n integers arr. The task is to calculate the sum of intervals between each element and all other elements that have the same value. Specifically, for each element in arr, its interval sum should be the sum of the absolute differences of its index with every other index where the same value appears.

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]. The problem can be efficiently solved using array scanning in combination with hash lookups to track the indices of identical elements as you traverse the array.

Examples

Example 1

Input: arr = [2,1,3,1,2,3,3]

Output: [4,2,7,2,4,4,5]

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]

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 Approach

Array Scanning with Hash Table Lookup

The core of the solution involves scanning through the array while maintaining a hash table of lists to store indices of identical elements. For each element, you will calculate the sum of intervals by referencing this table, which stores sorted indices of the same element found earlier in the array.

Optimizing Interval Calculation

For each element, instead of iterating over all previous occurrences, use the pre-stored indices to directly calculate the sum of intervals. This ensures a much faster calculation compared to brute-force methods.

Prefix Sum-like Optimization

For larger arrays, you can take inspiration from the prefix sum technique by storing cumulative interval sums for each element. This reduces redundant recalculations when checking intervals for each element.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on how efficiently the hash table is used. In the best case, when each element’s indices are processed in constant time, the solution will run in O(n). The space complexity is O(n) for storing the hash table of indices and interval sums.

What Interviewers Usually Probe

Look for candidates who use a hash table to store indices for efficient interval calculation.
A good solution will show the candidate’s understanding of optimizing array scanning operations using hash tables.
Watch out for inefficient brute-force approaches that don’t scale with larger arrays.

Common Pitfalls or Variants

Common pitfalls

Not properly handling cases where there are no other identical elements, leading to incorrect sum calculations.
Using brute-force methods to compare each element with every other element, resulting in a time complexity of O(n^2).
Not leveraging the hash table to efficiently look up the indices of identical elements, which significantly impacts performance.

Follow-up variants

Handling large arrays with more than one million elements.
Handling arrays with a small range of values or many repeated values.
Improving memory usage by optimizing how the indices are stored and accessed.

FAQ

How do I approach the 'Intervals Between Identical Elements' problem?

Use array scanning combined with hash table lookups to efficiently calculate the sum of intervals between identical elements in the array.

What is the time complexity of the optimal solution?

The optimal solution has a time complexity of O(n), assuming efficient use of the hash table to track element indices.

Can I solve this problem without a hash table?

While it's possible, a hash table greatly optimizes the process by reducing redundant work when looking up indices of identical elements.

What are common mistakes when solving the 'Intervals Between Identical Elements' problem?

Common mistakes include using a brute-force O(n^2) approach or failing to handle cases with only one occurrence of an element.

What is the space complexity of the solution?

The space complexity is O(n) due to the storage required for the hash table of indices and cumulative interval sums.

terminal

Solution

Solution 1

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class Solution:
    def getDistances(self, arr: List[int]) -> List[int]:
        d = defaultdict(list)
        n = len(arr)
        for i, v in enumerate(arr):
            d[v].append(i)
        ans = [0] * n
        for v in d.values():
            m = len(v)
            val = sum(v) - v[0] * m
            for i, p in enumerate(v):
                delta = v[i] - v[i - 1] if i >= 1 else 0
                val += i * delta - (m - i) * delta
                ans[p] = val
        return ans

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