# Check if Duplicate Element Exists in an Array

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

Example 1

Input: [3, 5, 7, 1, 7]
Output: true
Explanation: 7 is repeated twice.

Example 2

Input: [3, 5, 7, 1, 2]
Output: false
Explanation: No element is repeated.

Example 3

Input: [3, 5, 7, 1, 2, 7, 1, 3, 5]
Output: true

## Approach 1: Brute Force

### Intuition

For every element in the array, check all the subsequent elements and if any of the subsequent element matches with the current element, return true.

If we do not find any match after checking element, the array contains distinct integers and false is returned.

### Implementation

bool containsDuplicate(vector<int>& nums) {
int n = nums.size();
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (nums[i] == nums[j]) return true;
}
}
return false;
}

### Complexity Analysis

• Time Complexity: At the worst case when there are no duplicates, number of comparisons = $\frac{N * (N - 1)}{2}$ making time complexity $O(N^2)$, where $N$ = number of elements in the array.
• Space Complexity: $O(1)$, only constant space is used.

## Approach 2: Sorting

### Intuition

If the array is sorted, same elements appear consecutively. For example, sorting [3, 5, 7, 1, 2, 7, 1, 3, 5] will yield [1, 1, 2, 3, 3, 5, 7, 7]. We can use this idea to check for duplicate elements.

### Implementation

bool containsDuplicate(vector<int>& nums) {
int n = nums.size();
sort(nums.begin(), nums.end());
for (int i = 0; i < n - 1; ++i) {
if (nums[i] == nums[i + 1]) return true;
}
return false;
}

### Complexity Analysis

• Time Complexity: $O(NlogN)$ for sorting.
• Space Complexity: $O(1)$, only constant space is used.

## Approach 3: Hash Table

### Intuition

While traversing the array, the elements are inserted into a hash table. At each element, we check if the element is seen before. If it is seen before, there exists a duplicate element. Since hash table provides $O(1)$ lookup time, overall time complexity does not exceed $O(N)$.

### Implementation

bool containsDuplicate(vector<int>& nums) {
unordered_set<int> seen;
for (int num : nums) {
// if the current number is found previously, return true
if (seen.find(num) != seen.end()) return true;
seen.insert(num);
}
return false;
}

### Complexity Analysis

• Time Complexity: $O(N)$, at the worst case (no duplicate elements), we have to traverse the entire array.
• Space Complexity: $O(N)$, at the worst case (no duplicate elements), all the elements will be inserted into the hash table.