How do you solve Find All K-Distant Indices in an Array on LeetCode?

Find All K-Distant Indices in an Array (LeetCode #2200) 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 set helps avoid duplicate indices efficiently.

What is the brute force solution for Find All K-Distant Indices in an Array?

The brute force approach for Find All K-Distant Indices in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each index in the array to see if it is k-distant from any occurrence of the key. This is straightforward but inefficient since it examines every possible pair of indices. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find All K-Distant Indices in an Array?

Find All K-Distant Indices in an Array uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find All K-Distant Indices in an Array?

Always clarify the problem and constraints before jumping into coding. Discuss your thought process and potential edge cases with the interviewer. Start with a brute-force solution to demonstrate understanding, then optimize.

What are common mistakes when solving Find All K-Distant Indices in an Array?

Not using a set to store indices, leading to duplicates. Overlooking the bounds when calculating k-distant indices.

#2200

Find All K-Distant Indices in an Array

Easy

ArrayTwo PointersHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses a single pass to identify all indices of the key and then calculates the k-distant indices based on those positions. This is efficient because we avoid redundant checks.

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Algorithm

3 steps

1Step 1: Create a list to store indices of the key in the array.
2Step 2: For each index of the key found, calculate the range of k-distant indices and add them to a set to avoid duplicates.
3Step 3: Convert the set to a sorted list and return it.

solution.py7 lines

1def findKDistantIndices(nums, key, k):
2    key_indices = [i for i, num in enumerate(nums) if num == key]
3    k_distant_indices = set()
4    for index in key_indices:
5        for i in range(max(0, index - k), min(len(nums), index + k + 1)):
6            k_distant_indices.add(i)
7    return sorted(k_distant_indices)

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Complexity note: The time complexity is O(n) because we make a single pass to find key indices and then another pass to add k-distant indices. The space complexity is O(n) due to the storage of indices in a set.

1Using a set helps avoid duplicate indices efficiently.
2Understanding the range of indices to check is crucial for optimizing the solution.

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