class Main {
    public static void main(String[] args) {
        int arr[] = {1,2,3,4,5,6,7,7};
        int n = arr.length;//length of the array is n
        int largest = arr[n - 1];// the largest is always the last in sorted array 
        int second_largest = 0;// initaialze a variable for 2nd 

        for (int i = n - 2; i >= 0; i--) {
            if (arr[i] != largest) {
                second_largest = arr[i];
                break;
            }
        }

        System.out.println("the second largest element:-"+second_largest);
    }
}
        
//required logic in case of array size < 2 
//        if(arr.length<2){// u can use n variable (n<2)
//            System.out.println(" no element in array");
//            return ;
//        }


//for loop:-
// iteration from the first or second last element , as we are searching for second largest no need to search for n-1 , but we can start from n-1 also , no problem 


//This is a brute force approach 
//TC=O(nlogn) due to sorting taking more time
//SC=O(n)