questions C
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backend/data/questions/contains-duplicate-ii.yaml
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backend/data/questions/contains-duplicate-ii.yaml
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title: Contains Duplicate II
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slug: contains-duplicate-ii
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difficulty: easy
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leetcode_id: 219
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leetcode_url: https://leetcode.com/problems/contains-duplicate-ii/
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categories:
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- arrays
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- hash-tables
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patterns:
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- sliding-window
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description: |
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Given an integer array `nums` and an integer `k`, return `true` *if there are two **distinct indices*** `i` *and* `j` *in the array such that* `nums[i] == nums[j]` *and* `abs(i - j) <= k`.
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constraints: |
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- `1 <= nums.length <= 10^5`
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- `-10^9 <= nums[i] <= 10^9`
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- `0 <= k <= 10^5`
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examples:
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- input: "nums = [1,2,3,1], k = 3"
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output: "true"
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explanation: "The element 1 appears at index 0 and index 3. Since abs(0 - 3) = 3 <= k, we return true."
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- input: "nums = [1,0,1,1], k = 1"
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output: "true"
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explanation: "The element 1 appears at index 2 and index 3. Since abs(2 - 3) = 1 <= k, we return true."
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- input: "nums = [1,2,3,1,2,3], k = 2"
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output: "false"
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explanation: "While there are duplicates, no pair of duplicate values are within k = 2 indices of each other. The closest duplicate pair (1 at index 0 and 3) has distance 3 > k."
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explanation:
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intuition: |
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Imagine you're walking through a hallway with numbered rooms, and you need to find if any room number repeats within the last `k` rooms you've passed.
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The core insight is that we don't need to remember *every* room we've ever seen — we only care about rooms within our **sliding window** of the last `k` positions. If we encounter a room number we've seen within this window, we've found our duplicate.
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Think of it like this: as you move forward, you maintain a "memory" of the last `k` rooms. When you see a new room number, you check if it's already in your memory. If yes, you found a nearby duplicate. If not, add it to your memory and forget the oldest room (the one that's now more than `k` steps behind).
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This naturally suggests using a **hash set** as our memory — it gives us O(1) lookups to check for duplicates and O(1) insertions/deletions to maintain our sliding window.
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approach: |
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We solve this using a **Sliding Window with Hash Set** approach:
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**Step 1: Initialise a hash set**
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- Create an empty set `window` to store elements within our current window of size `k`
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- The set will contain at most `k` elements at any time
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**Step 2: Iterate through the array**
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- For each element at index `i`, check if it already exists in our `window` set
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- If yes, we found a duplicate within distance `k` — return `true`
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- If no, add the current element to the window
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**Step 3: Maintain window size**
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- If the window size exceeds `k`, remove the oldest element (the one at index `i - k`)
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- This ensures we only track elements within the valid distance
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**Step 4: Return the result**
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- If we complete the loop without finding duplicates, return `false`
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This approach efficiently combines the sliding window pattern with a hash set for O(1) operations, giving us an optimal O(n) solution.
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common_pitfalls:
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- title: The Brute Force Trap
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description: |
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A naive approach checks every pair of elements to see if they're equal and within distance `k`:
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- Outer loop `i` from `0` to `n-1`
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- Inner loop `j` from `i+1` to `min(i+k+1, n)`
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While this limits the inner loop to `k` iterations, it's still **O(n × k)** in the worst case. When both `n` and `k` are at their maximum (`10^5`), this results in up to 10 billion operations — causing a **Time Limit Exceeded (TLE)** error.
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wrong_approach: "Nested loops checking pairs within distance k"
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correct_approach: "Sliding window with hash set for O(n) time"
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- title: Using a Hash Map Instead of a Set
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description: |
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While a hash map (storing value → index) works, it's more complex than necessary. You'd need to update indices as you go and compare distances.
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A hash set is simpler: by maintaining exactly the last `k` elements, we implicitly guarantee any match is within the valid distance. If it's in the set, it's within range.
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wrong_approach: "Hash map with index tracking and distance calculation"
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correct_approach: "Hash set with sliding window of size k"
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- title: Off-by-One in Window Size
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description: |
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Be careful about when to remove elements from the window. The condition `abs(i - j) <= k` means indices can be up to `k` apart, so your window should contain `k` previous elements (not `k-1` or `k+1`).
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Remove the element at index `i - k` only when `i >= k`, ensuring the window never exceeds `k` elements from the past.
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wrong_approach: "Removing when i > k or keeping k+1 elements"
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correct_approach: "Remove element at index i - k when i >= k"
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key_takeaways:
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- "**Sliding window + hash set**: When you need to find duplicates within a range, combine a fixed-size window with a set for O(1) lookups"
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- "**Implicit distance guarantee**: By maintaining exactly `k` elements, any match is automatically within the valid distance — no need to track indices"
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- "**Set vs Map tradeoff**: Choose the simpler data structure when it suffices; a set is often cleaner than a map when you don't need the stored values"
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- "**Related problems**: This pattern extends to 'Contains Duplicate III' (within range *and* value difference) and other sliding window problems"
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time_complexity: "O(n). We traverse the array once, with O(1) hash set operations (add, remove, lookup) at each step."
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space_complexity: "O(min(n, k)). The hash set stores at most `min(n, k)` elements at any time."
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solutions:
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- approach_name: Sliding Window with Hash Set
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is_optimal: true
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code: |
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def contains_nearby_duplicate(nums: list[int], k: int) -> bool:
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# Set to track elements in our current window of size k
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window = set()
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for i, num in enumerate(nums):
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# If we've seen this number in our window, we found a duplicate
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if num in window:
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return True
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# Add current element to the window
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window.add(num)
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# Maintain window size: remove element that's now too far behind
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if i >= k:
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window.remove(nums[i - k])
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# No nearby duplicates found
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return False
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explanation: |
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**Time Complexity:** O(n) — Single pass through the array with O(1) set operations.
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**Space Complexity:** O(min(n, k)) — The set contains at most k elements.
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We maintain a sliding window of the last k elements using a hash set. For each new element, we check if it's already in the window (O(1) lookup). If found, we have a duplicate within distance k. Otherwise, we add it and remove the oldest element to maintain the window size.
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- approach_name: Hash Map with Index Tracking
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is_optimal: false
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code: |
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def contains_nearby_duplicate(nums: list[int], k: int) -> bool:
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# Map each value to its most recent index
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last_seen = {}
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for i, num in enumerate(nums):
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# Check if we've seen this number before
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if num in last_seen:
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# Check if the previous occurrence is within distance k
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if i - last_seen[num] <= k:
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return True
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# Update the most recent index for this number
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last_seen[num] = i
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return False
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explanation: |
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**Time Complexity:** O(n) — Single pass with O(1) hash map operations.
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**Space Complexity:** O(n) — In the worst case, all elements are unique and stored in the map.
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This approach stores the last seen index for each value. When we encounter a number we've seen before, we check if the distance is within k. While correct and efficient, it uses more space than the sliding window approach when k is small relative to n.
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- approach_name: Brute Force
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is_optimal: false
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code: |
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def contains_nearby_duplicate(nums: list[int], k: int) -> bool:
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n = len(nums)
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# Check each element against the next k elements
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for i in range(n):
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# Only check within the valid range
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for j in range(i + 1, min(i + k + 1, n)):
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if nums[i] == nums[j]:
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return True
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return False
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explanation: |
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**Time Complexity:** O(n × k) — For each element, we check up to k subsequent elements.
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**Space Complexity:** O(1) — No additional data structures used.
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This straightforward approach checks every valid pair. While it passes small test cases, it will TLE on large inputs where both n and k approach 10^5. Included to illustrate why the hash-based approaches are necessary.
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