Problem 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.
Examples
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
Python Solution
def containsDuplicate(nums: List[int]) -> bool:
return len(nums) != len(set(nums))
Alternative Solution:
def containsDuplicate(nums: List[int]) -> bool:
seen = set()
for num in nums:
if num in seen:
return True
seen.add(num)
return False
Java Solution
class Solution {
public boolean containsDuplicate(int[] nums) {
Set seen = new HashSet<>();
for (int num : nums) {
if (!seen.add(num)) {
return true;
}
}
return false;
}
}
C++ Solution
class Solution {
public:
bool containsDuplicate(vector& nums) {
unordered_set seen;
for (int num : nums) {
if (seen.count(num)) {
return true;
}
seen.insert(num);
}
return false;
}
};
JavaScript Solution
/**
* @param {number[]} nums
* @return {boolean}
*/
var containsDuplicate = function(nums) {
return new Set(nums).size !== nums.length;
Alternative Solution:
var containsDuplicate = function(nums) {
const seen = new Set();
for (const num of nums) {
if (seen.has(num)) {
return true;
}
seen.add(num);
}
return false;
};
C# Solution
public class Solution {
public bool ContainsDuplicate(int[] nums) {
HashSet seen = new HashSet();
foreach (int num in nums) {
if (!seen.Add(num)) {
return true;
}
}
return false;
}
}
Complexity Analysis
- Time Complexity: O(n) where n is the length of the array
- Space Complexity: O(n) to store the hash set
Solution Explanation
There are two main approaches to solve this problem:
- Set Comparison:
- Convert array to set
- Compare lengths
- More concise solution
- Always processes all elements
- Hash Set Tracking:
- Track seen elements
- Early termination
- More efficient for large arrays
- Better memory usage
Key Points
- Hash set usage
- Early termination
- Memory efficiency
- Language specifics
Alternative Approaches
- Sorting (O(n log n))
- Brute force (O(n²))
- Bit manipulation
- Stream operations
Edge Cases
- Empty array
- Single element
- All duplicates
- No duplicates