Here, We see Contains Duplicate II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

List of all LeetCode Solution

Contains Duplicate II LeetCode Solution

Problem Statement

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

Contains Duplicate II Leetcode Solution C++

class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k)
    {
       unordered_set<int> s;
       
       if (k <= 0) return false;
       if (k >= nums.size()) k = nums.size() - 1;
       
       for (int i = 0; i < nums.size(); i++)
       {
           if (i > k) s.erase(nums[i - k - 1]);
           if (s.find(nums[i]) != s.end()) return true;
           s.insert(nums[i]);
       }
       
       return false;
    }
};
Code language: C++ (cpp)

Contains Duplicate II Leetcode Solution Java

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Set<Integer> set = new HashSet<Integer>();
        for(int i = 0; i < nums.length; i++){
            if(i > k) set.remove(nums[i-k-1]);
            if(!set.add(nums[i])) return true;
        }
        return false;
 }
}
Code language: Java (java)

Contains Duplicate II Leetcode Solution JavaScript

var containsNearbyDuplicate = function(nums, k) {
    const hasmap = new Map();
    for (let idx = 0; idx < nums.length; idx++) {
        if (idx - hasmap.get(nums[idx]) <= k) {
            return true;
        }
        hasmap.set(nums[idx], idx);
    }
    return false;
};
Code language: JavaScript (javascript)

Contains Duplicate II Leetcode Solution Python

class Solution:
    def containsNearbyDuplicate(self, nums, k):
        dic = {}
        for i, v in enumerate(nums):
            if v in dic and i - dic[v] <= k:
                return True
            dic[v] = i
        return False
Code language: Python (python)