Q : Finding second largest element in the array(PYTHON):

Approach 1:

First we define two variables large and sec_large and assign values 0 to them.

Next we iterate through the entire list and make sec_large to large and large to a number greater than large in the list .

def Second_largest(list):
    largest = 0
    sec_largest = 0
    for item in list:
        if largest < item:
            sec_largest = largest
            largest = item
    return sec_largest
print(Second_largest([10,20,30,40,60]))

Approach 2:

First we sort the array in decsending order ,then return the second last element which is not equal to last.

def sec_large(arr):
    if len(arr)<2: 
        return "No Second Element Exist."
    else:
        arr.sort() #sorting the array in descending order 
        for index in range(len(arr)-2,-1,-1): #iterating from second last to first element
            if arr[index] != arr[len(arr)-1]: #checking which element is not equal to last
                return arr[index]
        else:
            return "No Second Element Exist." 
print(sec_large([2,5,3,6,6]))