> For the complete documentation index, see [llms.txt](https://gl01.gitbook.io/gfg-editorials/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://gl01.gitbook.io/gfg-editorials/2023/09-2023-sep-23/24-find-duplicates-in-an-array.md).

# 24. Find duplicates in an array

The problem can be found at the following link: [Question Link](https://practice.geeksforgeeks.org/problems/find-duplicates-in-an-array/1)

## My Approach

To find duplicates in the given array,

* I use an array `cnt` to keep track of the count of each element in the input array `arr`.
* Then, I iterate through the `cnt` array and push the indices of elements with a count greater than 1 to the `out` vector.
* Finally, if there are duplicates, I return the `out` vector; otherwise, I return {-1} to indicate no duplicates.

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(n)`, where `n` is the size of the input array. This is because we iterate through the input array once to count the occurrences of each element and then iterate through the `cnt` array once to find duplicates.
* **Auxiliary Space Complexity**: `O(n)`, as we use an additional array `cnt` to store the counts of elements.

## Code (C++)

```cpp
class Solution {
public:
    vector<int> duplicates(int arr[], int n) {
        int cnt[n] = {0};
        for(int i = 0; i < n; ++i)
            ++cnt[arr[i]];

        vector<int> out;
        for(int i = 0; i < n; ++i)
            if(cnt[i] > 1)
                out.push_back(i);

        if(out.size())
            return out;

        return {-1};
    }
};
```

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