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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.

• Java
• C++
• Python
• Go
• TypeScript
• Php
• C#

• 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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