Code-Memo

Contains Duplicate II

Given an array of integers nums and an integer k, return True if there are two distinct indices i and j in the array such that nums[i] equals nums[j] and the absolute difference between i and j is at most k.

Example:

nums = [1, 2, 3, 1]
k = 3
# Output: True (because nums[0] and nums[3] have the same value 1, and the difference between their indices is 3-0 = 3 <= k)

Solution: Using HashMap

class Solution(object):
    def containsNearbyDuplicate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        nums_map = {}  # Initialize an empty dictionary to store seen numbers and their indices
        for i in range(len(nums)):
            if nums[i] in nums_map and abs(nums_map[nums[i]] - i) <= k:
                return True  # If the number exists in nums_map and its index difference is <= k, return True
            nums_map[nums[i]] = i  # Store the current number and its index in nums_map
        return False  # If no duplicates within k distance are found, return False

Explanation:

• Time Complexity: O(n)
• Space Complexity: O(n)

Steps:

1. Initialization: Create an empty dictionary nums_map to store seen numbers (nums[i]) and their indices (i).

2. Iteration: Iterate through each number in nums using a for loop and index i.

3. Check for Duplicates within k Distance:

   • For each nums[i], check if nums[i] already exists in nums_map and if the absolute difference between nums_map[nums[i]] (the index where nums[i] was first seen) and i is less than or equal to k.
   • If both conditions are met, return True indicating that there exists a pair of indices i and j such that nums[i] == nums[j] and |i - j| <= k.

4. Store in HashMap: If the conditions in step 3 are not met, store nums[i] and its index i in nums_map for future reference.

5. Return Default: If the loop completes without finding any duplicates within k distance, return False.