Finding leaders in an array

Here I have written the Two approach one is naive approach which will take O(n^2) time complexity and another is the efficient approach which will take O(n) time complexity.

Naive solution

#include <iostream>

// naive approach;

int leader(int arr[], int n)

{

    for (int i = 0; i < n; i++)

    {

        bool flag = false;

        for (int j = i + 1; j < n; j++)

        {

            if (arr[i] <= arr[j])

            {

                flag = true;

                break;

            }

        }

        if (flag == false)

        {

            std::cout << arr[i] << ' ';

        }

    }

}

int main()

{

    int arr[] = {2, 5, 10, 7, 4};

    int n = 5;

    std::cout << "The size of the array = " << n << std::endl;

    leader(arr, n);

    return 0;

}

Efficient solution

#include <iostream>

//efficient solution

void leader(int arr[], int n)

{

    int curr_lad = arr[n - 1];

    std::cout << curr_lad << ' ';

    for (int i = n - 2; i >= 0; i--)

    {

        if (curr_lad < arr[i])

        {

            curr_lad = arr[i];

            std::cout << curr_lad << ' ';

        }

    }

}

int main()

{

    int arr[] = {1, 3, 4, 10, 2};

    int size = sizeof(arr) / sizeof(arr[0]);

    std::cout << "The size of the array = " << size << std::endl;

    leader(arr, size);

    return 0;

}