Introduction

In this blog post, we will explore how to solve the LeetCode problem "220" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array nums and two integers indexDiff and valueDiff. Find a pair of indices (i, j) such that: i != j, abs(i - j) <= indexDiff. abs(nums[i] - nums[j]) <= valueDiff, and Return true if such pair exists or false otherwise. Example 1: Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0 Output: true Explanation: We can choose (i, j) = (0, 3). We satisfy the three conditions: i != j --> 0 != 3 abs(i - j) <= indexDiff --> abs(0 - 3) <= 3 abs(nums[i] - nums[j]) <= valueDiff --> abs(1 - 1) <= 0 Example 2: Input: nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3 Output: false Explanation: 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

Explanation

• Sliding Window with Sorted Set: Maintain a sorted set (using a binary search tree implementation like SortedList in Python) of the numbers within the current indexDiff window. For each new number, efficiently find if there exists a number within the window that satisfies the valueDiff condition.
  • Binary Search for Efficient Lookup: Use binary search within the sorted set to find the potential candidates that satisfy the valueDiff condition. This avoids a linear scan of the window.
  • Maintain Window: Slide the window by adding the new number to the sorted set and removing the number that falls outside the window.

• Runtime Complexity: O(N log K), where N is the length of the input array nums, and K is the window size (indexDiff). The log K factor comes from the insertion, deletion, and binary search operations on the sorted set. Storage Complexity: O(K), due to storing at most indexDiff elements in the sorted set.

Code

    from sortedcontainers import SortedList

def containsNearbyAlmostDuplicate(nums, indexDiff, valueDiff):
    """
    Checks if there are two distinct indices i and j in nums such that
    abs(i - j) <= indexDiff and abs(nums[i] - nums[j]) <= valueDiff.

    Args:
        nums: The input list of integers.
        indexDiff: The maximum allowed difference between the indices.
        valueDiff: The maximum allowed difference between the values.

    Returns:
        True if such indices exist, False otherwise.
    """
    if not nums or len(nums) < 2:
        return False

    window = SortedList()
    for i in range(len(nums)):
        # Find the smallest element in the window that is >= nums[i] - valueDiff
        left = nums[i] - valueDiff
        idx = window.bisect_left(left)

        if idx < len(window) and abs(window[idx] - nums[i]) <= valueDiff:
            return True

        window.add(nums[i])

        if i >= indexDiff:
            window.remove(nums[i - indexDiff])

    return False