Contains Duplicate LeetCode Solution


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Problem Statement:

Contains Duplicate LeetCode Solution says that- Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

Example 1:

Input:

 nums = [1,2,3,1]

Output:

 true

Example 2:

Input:

 nums = [1,2,3,4]

Output:

 false

Example 3:

Input:

 nums = [1,1,1,3,3,4,3,2,4,2]

Output:

 true

 

Constraints:

  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109

 

 Algorithm –

Idea –

  • In order to find the Contains Duplicate. What will we do in this question? At first, we will focus on the duplicate element, and after seeing the duplicate element. we will simply return True and if the array doesn’t contain duplicates then we will Return False.
  • At first, we will create one HashSet after that we will traverse the whole array and check for the condition that if the element is present in the set then we will return True else we will add that element into the Set.

Approach –

  • Create one Hash-set for an efficient solution.
  • Iterate the whole loop from 0 to len(nums).
  • Check for the Condition that if the element is present in the set then we will return True.
  • Else if not Present then we will add that element into the set.
  • After traversing the whole array if it doesn’t return True, then finally we will return False after iterating the whole array.
  • Hence we will check the Contain Duplicate.

Image of Contain Duplicate-

Contains Duplicate LeetCode Solution

 

class Solution {
    public boolean containsDuplicate(int[] nums) {
        
         Set<Integer> s = new HashSet<Integer>();
     for(int i : nums){
             
        
       if (s.contains(i)){
                           return true;
             }
            
             else{
                 s.add(i);
             }
         }         
        return false;
    }
}

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        
        s = set()
        for i in nums:
            if i in s:
                return True
                
            else:
                s.add(i)
        return False

Complexity Analysis for Contains Duplicate LeetCode Solution:

Time complexity: O(N), As we have traversed through the whole array only one time.

Space complexity: O(N), As we have created one Hash set.

SIMILAR QUESTION – https://www.tutorialcup.com/interview/hashing/contains-duplicate.htm

 

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