How do you solve Contains Duplicate III on LeetCode?

Contains Duplicate III (LeetCode #220) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log k) time complexity and O(k) space complexity. Key insight: Using a sliding window allows us to limit the number of comparisons and efficiently check conditions.

What is the time complexity of Contains Duplicate III?

The optimal solution for Contains Duplicate III runs in O(n log k) time and uses O(k) space. This complexity is due to maintaining a sorted set of size k (indexDiff), allowing us to perform efficient insertions and lookups.

What is the brute force solution for Contains Duplicate III?

The brute force approach for Contains Duplicate III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of indices in the array. It verifies if the conditions of index difference and value difference are satisfied for each pair. The optimal Optimal Solution approach improves this to O(n log k).

What are the interview tips for Contains Duplicate III?

Always clarify the problem constraints and edge cases before diving into coding. Discuss your thought process and approach with the interviewer to get feedback. Practice explaining your code as you write it, as this simulates the interview environment.

What are common mistakes when solving Contains Duplicate III?

Not considering the edge cases where indexDiff or valueDiff is zero. Confusing the conditions and checking them in the wrong order.

#220

Contains Duplicate III

Hard

ArraySliding WindowSortingBucket SortOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log k)Space O(k)

The optimal solution uses a sliding window approach combined with a balanced tree structure to maintain a window of k elements. This allows us to efficiently check for the required conditions without re-sorting.

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Algorithm

5 steps

1Step 1: Create a set to store the elements in the current sliding window of size indexDiff.
2Step 2: Iterate through the array, adding each element to the set.
3Step 3: For each new element, check if there exists an element in the set that meets the valueDiff condition using binary search.
4Step 4: If the size of the set exceeds indexDiff, remove the oldest element from the set.
5Step 5: If any valid pair is found during the checks, return true. If the loop completes without finding a pair, return false.

solution.py13 lines

1from sortedcontainers import SortedList
2
3def containsNearbyAlmostDuplicate(nums, indexDiff, valueDiff):
4    sorted_list = SortedList()
5    for i in range(len(nums)):
6        if i > indexDiff:
7            sorted_list.remove(nums[i - indexDiff - 1])
8        pos1 = sorted_list.bisect_left(nums[i] - valueDiff)
9        pos2 = sorted_list.bisect_right(nums[i] + valueDiff)
10        if pos1 != pos2:
11            return True
12        sorted_list.add(nums[i])
13    return False

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Complexity note: This complexity is due to maintaining a sorted set of size k (indexDiff), allowing us to perform efficient insertions and lookups.

1Using a sliding window allows us to limit the number of comparisons and efficiently check conditions.
2Maintaining a sorted structure helps in quickly finding if any number within the required range exists.

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