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Formatted question description: https://leetcode.ca/all/217.html

217. Contains Duplicate

Given an array of integers, find if the array contains any duplicates.

Your function should return true if any value appears at least twice in the array,
and it should return false if every element is distinct.



Use a hash table, traverse the entire array, if the hash table exists, return false, if not, put it into the hash table.

Or, sort the array first, and then compare whether two adjacent numbers are equal, the time complexity depends on the sorting method.



  • import java.util.Arrays;
    import java.util.HashSet;
    import java.util.Set;
    public class Contains_Duplicate {
        public class Solution_extraSpace {
            public boolean containsDuplicate(int[] nums) {
                Set<Integer> set = new HashSet<>();
                for (int i : nums) {
                    if (!set.add(i))// if there is same
                        return true;
                return false;
        public class Solution_extraOps {
            public boolean containsDuplicate(int[] nums) {
                for (int i = 1; i < nums.length; ++i) {
                    if (nums[i] == nums[i - 1])
                        return true;
                return false;
  • // OJ: https://leetcode.com/problems/contains-duplicate/
    // Time: O(N)
    // Space: O(N)
    class Solution {
        bool containsDuplicate(vector<int>& A) {
            unordered_set<int> s(begin(A), end(A));
            return s.size() != A.size();
  • class Solution:
        def containsDuplicate(self, nums: List[int]) -> bool:
            return len(set(nums)) < len(nums)
    class Solution(object):
      def containsDuplicate(self, nums):
        :type nums: List[int]
        :rtype: bool
        for i in range(0, len(nums) - 1):
          if nums[i] == nums[i + 1]:
            return True
        return False

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