#220

Hard

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Contains Duplicate III

The problem involves finding a pair of indices in an array where the index and value differences are within given limits.

array sliding window sorting bucket sort ordered set

Problem Statement

You are given an integer array nums and two integers, indexDiff and valueDiff. Your task is to find a pair of indices (i, j) such that the following conditions hold:

abs(i - j) <= indexDiff, 2. abs(nums[i] - nums[j]) <= valueDiff, and 3. i != j. Return true if such a pair exists, otherwise return false.

Examples

Example 1

Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0

Output: true

We can choose (i, j) = (0, 3). We satisfy the three conditions: i != j --> 0 != 3 abs(i - j) abs(0 - 3) abs(1 - 1) <= 0

Example 2

Input: nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3

Output: false

After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.

Constraints

2 <= nums.length <= 105
-109 <= nums[i] <= 109
1 <= indexDiff <= nums.length
0 <= valueDiff <= 109

Solution Approach

Sliding Window with Running State Updates

The problem can be solved using a sliding window approach combined with running state updates. As we iterate through the array, we maintain a window of size indexDiff, ensuring that elements within this window meet the valueDiff condition. This ensures that we only need to check a limited number of elements at any given time, optimizing performance.

Efficient Data Structure Usage

To maintain the window and check the conditions efficiently, an ordered set or a bucket sort can be used. These structures allow fast lookups and updates, helping us quickly determine whether a valid pair exists within the window. The choice of data structure impacts the overall time complexity.

Handling Edge Cases

Edge cases, such as arrays with small lengths or when the difference in values is zero, must be handled carefully. Special conditions where no valid pair can be found should also be considered, and we must ensure that the window slides correctly when the indexDiff constraint is near the array’s bounds.

Complexity Analysis

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

The time complexity of this problem depends on the approach and the data structure used. Using a sliding window with efficient data structures like an ordered set or bucket sort can result in an O(n log k) time complexity, where n is the length of the array and k is the window size. Space complexity is O(k) due to the space required for storing the elements in the window.

What Interviewers Usually Probe

Look for understanding of sliding window techniques combined with efficient state management.
Assess the candidate's ability to handle edge cases where no valid pair exists.
Evaluate the candidate's knowledge of data structures like ordered sets or bucket sorts for optimization.

Common Pitfalls or Variants

Common pitfalls

Overlooking the need for an efficient data structure to track the sliding window.
Failing to properly manage the constraints, especially when dealing with large inputs.
Incorrectly handling edge cases, such as when the valueDiff is zero or when no valid pair exists.

Follow-up variants

Allow a larger range for indexDiff and valueDiff, making the sliding window approach more complex.
Limit the array length, requiring optimization of both space and time complexity.
Implement the solution without relying on specific data structures like ordered sets, which may require a brute-force approach.

FAQ

What is the sliding window technique in Contains Duplicate III?

The sliding window technique helps to efficiently track elements within a range of indices. It limits the number of elements checked at once, which is critical for optimizing performance in problems like Contains Duplicate III.

How do I handle edge cases in Contains Duplicate III?

Edge cases include arrays of small length or situations where no valid pair satisfies the constraints. Properly managing the window size and handling these cases is essential to solving the problem correctly.

What is the time complexity of the sliding window approach for Contains Duplicate III?

Using a sliding window with an efficient data structure results in a time complexity of O(n log k), where n is the length of the array and k is the window size.

How do ordered sets help in solving Contains Duplicate III?

Ordered sets allow for fast lookups and updates, ensuring that we can efficiently check if the current window meets the valueDiff condition. They help maintain an optimized solution for larger arrays.

What are the potential pitfalls when solving Contains Duplicate III?

Common pitfalls include failing to use the correct data structure to manage the sliding window, neglecting edge cases, and not optimizing for large input sizes, leading to suboptimal performance.

terminal

Solution

Solution 1: Sliding Window + Ordered Set

We maintain a sliding window of size $k$, and the elements in the window are kept in order.

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class Solution:
    def containsNearbyAlmostDuplicate(
        self, nums: List[int], indexDiff: int, valueDiff: int
    ) -> bool:
        s = SortedSet()
        for i, v in enumerate(nums):
            j = s.bisect_left(v - valueDiff)
            if j < len(s) and s[j] <= v + valueDiff:
                return True
            s.add(v)
            if i >= indexDiff:
                s.remove(nums[i - indexDiff])
        return False

Continue Practicing

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