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219. Contains Duplicate II

Description

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

Solutions

Solution 1: Hash Table

We use a hash table $d$ to store the nearest index of the number it has visited.

We traverse the array nums. For the current element $nums[i]$, if it exists in the hash table, and the difference between its index and the current index is no larger than $k$, then return true. Otherwise, we add the current element into the hash table.

After the traversal, return false.

The time complexity is $O(n)$ and the space complexity is $O(n)$. Here $n$ is the length of array nums.

  • class Solution {
        public boolean containsNearbyDuplicate(int[] nums, int k) {
            Map<Integer, Integer> d = new HashMap<>();
            for (int i = 0; i < nums.length; ++i) {
                if (i - d.getOrDefault(nums[i], -1000000) <= k) {
                    return true;
                }
                d.put(nums[i], i);
            }
            return false;
        }
    }
    
  • class Solution {
    public:
        bool containsNearbyDuplicate(vector<int>& nums, int k) {
            unordered_map<int, int> d;
            for (int i = 0; i < nums.size(); ++i) {
                if (d.count(nums[i]) && i - d[nums[i]] <= k) {
                    return true;
                }
                d[nums[i]] = i;
            }
            return false;
        }
    };
    
  • class Solution:
        def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
            mp = {}
            for i, v in enumerate(nums):
                if v in mp and i - mp[v] <= k:
                    return True
                mp[v] = i
            return False
    
    
  • func containsNearbyDuplicate(nums []int, k int) bool {
    	d := map[int]int{}
    	for i, x := range nums {
    		if j, ok := d[x]; ok && i-j <= k {
    			return true
    		}
    		d[x] = i
    	}
    	return false
    }
    
  • function containsNearbyDuplicate(nums: number[], k: number): boolean {
        const d: Map<number, number> = new Map();
        for (let i = 0; i < nums.length; ++i) {
            if (d.has(nums[i]) && i - d.get(nums[i])! <= k) {
                return true;
            }
            d.set(nums[i], i);
        }
        return false;
    }
    
    
  • class Solution {
        /**
         * @param Integer[] $nums
         * @param Integer $k
         * @return Boolean
         */
        function containsNearbyDuplicate($nums, $k) {
            $hashtable = [];
            for ($i = 0; $i < count($nums); $i++) {
                $tmp = $nums[$i];
                if (array_key_exists($tmp, $hashtable) && $k >= $i - $hashtable[$tmp]) {
                    return true;
                }
                $hashtable[$tmp] = $i;
            }
            return false;
        }
    }
    
  • public class Solution {
        public bool ContainsNearbyDuplicate(int[] nums, int k) {
            var d = new Dictionary<int, int>();
            for (int i = 0; i < nums.Length; ++i) {
                if (d.ContainsKey(nums[i]) && i - d[nums[i]] <= k) {
                    return true;
                }
                d[nums[i]] = i;
            }
            return false;
        }
    }
    

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