Python Program to Find the Second Largest Number in a List using Bubble Sort

This is a Python Program to find the second largest number in a list using bubble sort.

Problem Description

The program takes a list and finds the second largest number in the list using bubble sort.

Problem Solution

1. Take in the number of elements for the list and store it in a variable.
2. Take in the elements of the list one by one.
3. Then sort the list using bubble sort.
4. Print the second last element of the sorted list which is the second-largest number.
5. Exit.

Program/Source Code

Here is the source code of the Python Program to find the second largest number in a list using bubble sort. The program output is also shown below.

a=[]
n=int(input("Enter number of elements:"))
for i in range(1,n+1):
    b=int(input("Enter element:"))
    a.append(b)
for i in range(0,len(a)):
    for j in range(0,len(a)-i-1):
        if(a[j]>a[j+1]):
            temp=a[j]
            a[j]=a[j+1]
            a[j+1]=temp 
print('Second largest number is:',a[n-2])
Program Explanation

1. User must enter the number of elements for the list and store it in a variable.
2. User must then enter the elements of the list one by one using a for loop and store it in a list.
3. Then bubble sort algorithm may be used to sort the elements of the list.
4. The second last element of the sorted list is then printed which is basically the second largest element in the list.

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Runtime Test Cases
 
Case 1:
Enter number of elements:4
Enter element:56
Enter element:43
Enter element:78
Enter element:12
('Second largest number is:', 56)
 
Case 2:
Enter number of elements:6
Enter element:23
Enter element:45
Enter element:94
Enter element:10
Enter element:34
Enter element:95
('Second largest number is:', 94)

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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