Contains-Duplicate

217. Contains Duplicate

Description

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

Solutions

Python3

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        return len(nums) != len(set(nums))

Java

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

TypeScript

function containsDuplicate(nums: number[]): boolean {
    let unique: Set<number> = new Set(nums);
    return unique.size != nums.length;
}

C++

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_set<int> s;
        for (int e : nums)
        {
            if (s.count(e)) return true;
            s.insert(e);
        }
        return false;
    }
};

Go

func containsDuplicate(nums []int) bool {
	s := make(map[int]bool)
	for _, e := range nums {
		if s[e] {
			return true
		}
		s[e] = true
	}
	return false
}

C#

public class Solution {
    public bool ContainsDuplicate(int[] nums) {
        return nums.Distinct().Count() < nums.Length;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var containsDuplicate = function (nums) {
    return new Set(nums).size !== nums.length;
};

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