Maximum consecutive integers present in an array

Given:  An array of integers, we have to find the maximum length such that the elements are consecutive integers (the consecutive numbers can be in any order).

EXAMPLES:

Input : xyz[] = {5, 54, 3, 45, 6, 92, 4};

Output : 4 

 The maximum consecutive elements are 3, 4, 5, 6. 

Code:

#include <bits/stdc++.h>

int findlongsub(int array[], int no_of_elements)
{
    std::unordered_set<int> uno_set;  //taking a set of input
    for (int i = 0; i < no_of_elements; i++)
        uno_set.insert(array[i]);

    int answer = 0;
    for (int i = 0; i < no_of_elements; i++)
    {
        // to check for required elements in starting of set
        if (uno_set.find(array[i] - 1) == uno_set.end())
        {
            // Checking for all the consecutive elements in set
            int j = array[i];
            // increment the value of array element and again search for the consecutive elements
            while (uno_set.find(j) != uno_set.end())
                j++;

            answer = std::max(answer, j - array[i]);
        }
    }
    return answer;
}

// Main code
int main()
{
    int array[] = { 5, 54, 3, 45, 6, 92, 4 };
    //here 3,4,5,6 are the longest subsequence
    int no_of_elements = sizeof(array) / sizeof(int);
    std::cout << findlongsub(array, no_of_elements) << std::endl;
    return 0;
}

Output

4

Time complexity : O(n)