Contains Duplicate II - Problem
You are given an integer array nums and a distance threshold k. Your task is to determine if there are any duplicate elements within a specific range.
More specifically, return true if there exist two distinct indices i and j in the array such that:
nums[i] == nums[j](the elements are equal)abs(i - j) <= k(the indices are at mostkpositions apart)
Otherwise, return false.
Example: In array [1,2,3,1] with k=3, we find that nums[0] = nums[3] = 1 and abs(0-3) = 3 <= 3, so we return true.
Input & Output
example_1.py — Basic Case
$
Input:
nums = [1,2,3,1], k = 3
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Output:
true
💡 Note:
The value 1 appears at indices 0 and 3. The distance between these indices is |3-0| = 3, which is exactly equal to k=3, so we return true.
example_2.py — No Valid Duplicates
$
Input:
nums = [1,0,1,1], k = 1
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Output:
true
💡 Note:
The value 1 appears at indices 0, 2, and 3. The distance between indices 2 and 3 is |3-2| = 1, which equals k=1, so we return true.
example_3.py — Distance Too Large
$
Input:
nums = [1,2,3,1,2,3], k = 2
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Output:
false
💡 Note:
Although there are duplicates (1 appears at indices 0,3 and 2 appears at indices 1,4), all duplicate pairs have distances greater than k=2, so we return false.
Constraints
- 1 ≤ nums.length ≤ 105
- -109 ≤ nums[i] ≤ 109
- 0 ≤ k ≤ 105
- Note: Two distinct indices means i ≠ j
Visualization
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Understanding the Visualization
1
Book Registry
Keep a record of when each book was last checked out (hash map)
2
New Checkout
When someone wants to check out a book, look up the last checkout date
3
Time Check
If the same book was checked out within k days, flag it as a violation
4
Update Record
Update the registry with the new checkout date
Key Takeaway
🎯 Key Insight: We only need to remember the most recent occurrence of each element, not all historical positions. This reduces both time and space complexity significantly!
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Explanation
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