Largest Element in an array

1. Sort and return last index time complexity: nlog(n), space: O(1) for quick sort,
2. O(n), traverse through an array and keep max pointer

Second Largest Element in an array

1. Sort and return last index
   1. time complexity: nlog(n) for sorting + O(N) traverse for 0 to n-2
   2. space: O(1) for quick sort
2. O(n), traverse through an array and keep max pointer
   1. then traverse again and find the second largest. i.e largest
      1. if(a[i] < largest) and a[i]>sLargest
   2. Total time is 2 O(N)

3.  		if(a.length<2)
    			return -1;
    		largest = a[0], sLArgest = Math.INT_MIN;
    		for(int i=1;i<n;i++){
    				if(a[i]>largest){
    					sLArgest = largest;
    					largest = a[i];
    				}
    				if(a[i] < largest && a[i]>sLargest){
    					sLArgest = a[i];
    				}
    				
    		}
    		return sLargest;

Space: O(N)