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# 10. Find second largest element

**Given an array of elements. Your task is to find the second maximum element in the array. If there does not exist any second largest element, then return -1.**\
**`Array contains duplicates`**

```
Example 1:
Input:
N=5
arr[] = { 2, 4, 5, 6, 7 }
Output:  6
Explanation:
The largest element is 7 and 
the second largest element is 6.

Example 2:
Input:
N=6
arr[] = { 7, 8, 2, 1, 4, 3 }
Output:  7
```

## Solution: (Two Traversal)

**Approach:** \
**Find the maximum element**\
**Find the second maximum which is not equal to maximum**

```cpp
class Solution
{
public:
    int print2largest(int arr[], int arr_size)
    {

        int mx = INT_MIN;
        for (int i = 0; i < arr_size; i++)
        {
            if (arr[i] > mx)
            {
                mx = arr[i];
            }
        }

        int mx2 = INT_MIN;
        for (int i = 0; i < arr_size; i++)
        {
            if (arr[i] < mx && arr[i] > mx2)
            {
                mx2 = arr[i];
            }
        }

        if (mx2 == INT_MIN)
        {
            return -1;
        }
        return mx2;
    }
};
```

**Time Complexity: O(n)**

## Solution: (Using one traversal)

**Use two variables max and second max**\
**If value is greater than max:**\
&#x20;        **second max = max**\
&#x20;        **max = value**\
**If value lies in between:**\
&#x20;        **second max = value**&#x20;

```cpp
class Solution
{
public:
    int print2largest(int arr[], int arr_size)
    {
        int max = arr[0];
        int sec_max = INT_MIN;

        for (int i = 1; i < arr_size; i++)
        {
            if (arr[i] > max)
            {
                sec_max = max;
                max = arr[i];
            }
            else if (arr[i] > sec_max && arr[i] < max)
            {
                sec_max = arr[i];
            }
        }

        if (sec_max == INT_MIN)
        {
            return -1;
        }
        return sec_max;
    }
};
```

**Time Complexity: O(n)**