219. Contains Duplicate II LeetCode Solution
In this guide, you will get 219. Contains Duplicate II LeetCode Solution with the best time and space complexity. The solution to Contains Duplicate II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Table of Contents
- Problem Statement
- Complexity Analysis
- Contains Duplicate II solution in C++
- Contains Duplicate II solution in Java
- Contains Duplicate II solution in Python
- Additional Resources

Problem Statement of Contains Duplicate II
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.
Example 1:
Input: nums = [1,2,3,1], k = 3
Output: true
Example 2:
Input: nums = [1,0,1,1], k = 1
Output: true
Example 3:
Input: nums = [1,2,3,1,2,3], k = 2
Output: false
Constraints:
1 <= nums.length <= 105
-109 <= nums[i] <= 109
0 <= k <= 105
Complexity Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
219. Contains Duplicate II LeetCode Solution in C++
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
unordered_set<int> seen;
for (int i = 0; i < nums.size(); ++i) {
if (!seen.insert(nums[i]).second)
return true;
if (i >= k)
seen.erase(nums[i - k]);
}
return false;
}
};
/* code provided by PROGIEZ */
219. Contains Duplicate II LeetCode Solution in Java
class Solution {
public boolean containsNearbyDuplicate(int[] nums, int k) {
Set<Integer> seen = new HashSet<>();
for (int i = 0; i < nums.length; ++i) {
if (!seen.add(nums[i]))
return true;
if (i >= k)
seen.remove(nums[i - k]);
}
return false;
}
}
// code provided by PROGIEZ
219. Contains Duplicate II LeetCode Solution in Python
class Solution:
def containsNearbyDuplicate(self, nums: list[int], k: int) -> bool:
seen = set()
for i, num in enumerate(nums):
if i > k:
seen.remove(nums[i - k - 1])
if num in seen:
return True
seen.add(num)
return False
# code by PROGIEZ
Additional Resources
- Explore all LeetCode problem solutions at Progiez here
- Explore all problems on LeetCode website here
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