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docs/DSA-Problem-Solution/Contains_duplicate.md

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+---
+id: contains-duplicate-leetcode-217
+title: Contains Duplicate
+sidebar_label: Leetcode 217
+tags: [Leetcode, Array, DSA, 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.
+---
+
+## 217. Contains Duplicate
+**Description**:  
+You are given an array of integers, `nums`, which may contain both positive and negative numbers. Your task is to determine whether any value appears more than once in the array. If at least one duplicate exists, return `true`. Otherwise, return `false`.
+
+## Example 1:
+
+**Input:**  
+`nums = [1, 2, 3, 1]`
+
+**Output:**  
+`true` (because 1 appears twice)
+
+**Explanation:**  
+- The frequency of 1 is 2  
+- The frequency of 2 is 1  
+- The frequency of 3 is 1  
+- Since 1 appears twice, there is a duplicate, so the output is `true`.
+
+## Example 2:
+
+**Input:**  
+`nums = [1, 2, 3, 4]`
+
+**Output:**  
+`false` (because all elements are distinct)
+
+**Explanation:**  
+- All elements are unique and appear only once.
+
+## C++ Code:
+```cpp
+#include <iostream>
+#include <unordered_set>
+#include <vector>
+
+using namespace std;
+
+bool containsDuplicate(const vector<int>& nums) {
+    unordered_set<int> uniques;
+
+    for (int num : nums) {
+        // If the number is already in the set, it means it's a duplicate
+        if (uniques.find(num) != uniques.end()) {
+            return true;
+        }
+        // Add the number to the set
+        uniques.insert(num);
+    }
+    return false;
+}
+
+int main() {
+    vector<int> nums = {1, 2, 3, 1}; // Example input
+    if (containsDuplicate(nums)) {
+        cout << "Array contains duplicates." << endl;
+    } else {
+        cout << "Array does not contain duplicates." << endl;
+    }
+
+    return 0;
+}
+
+```
+## Approach and code explanation
+**Using a Set for Uniqueness:**
+- A set data structure inherently stores unique values. In C++, we use an unordered_set for this purpose, which allows us to:
+Insert elements in O(1) average time.
+Check if an element exists (lookup) in O(1) average time.
+
+**Iterating Through the Array:**
+- The algorithm iterates through each element of the array.
+For each element:
+It checks if the element already exists in the set (using the find() method). If it does, this means the element is a duplicate, so we return true.
+If the element is not found in the set, it is added to the set to track that we’ve seen it before.
+
+**Returning the Result:**
+- If the loop completes without finding any duplicates, the function returns false, meaning all elements in the array are distinct.
+Code Flow:
+Initialization: We declare an `unordered_set<int>` uniques to store the unique numbers encountered during the iteration.
+
+**Loop through the array:**
+- For each number in the array:
+Check for duplication: If the number is already in the set, return true (duplicate found).
+Insert into the set: If not found, add the number to the set.
+Final Result: If the loop completes without finding any duplicates, return false.
+
+## complexcity
+## Time Complexity:
+Time complexity- O(n), where n is the size of the input array.
+
+## Space complexity:
+-O(n), as we are using an unordered_set to store up to n unique elements.