# Longest Common Prefix

## Problem Statement

Given two strings, you have to find the longest common prefix of the two strings.

For example, if `apple` and `application` are the two strings, then `appl` is the longest common prefix for the two strings.

## Input

The first line of input is the first string: `string1`.

The second line of input is the second string: `string2`.

## Output

Print the longest common prefix of the two strings: `string1` and `string2`.

## As is

The boilerplate code is already there to read the input from user, in the `main()` function, and to print the value returned by the `longestCommonPrefix()` function. Do not edit this code.

## To do ✍

`largestCommonPrefix()` has two parameters: `string1` and `string2`. Complete the function that finds the longest common prefix of the two strings, and returns the same.

## Example 1

Test input

``````apple
application``````

Test output

``appl``

Explanation

For the two strings `appl` is the common, and also the longest common prefix. `a`, `ap`, and `app` are also common prefixes, but are not the longest.

## Example 2

Test input

``````bank
banana``````

Test output

``ban``

## Python Program

``````# Complete the following function
def longestCommonPrefix(string1, string2):

# Boilerplate code - Do not edit the following
def main():
string1 = input()
string2 = input()
print(longestCommonPrefix(string1, string2))

if __name__ == "__main__":
main()``````

Input and Outputs 1

``````apple
application``````
``appl``

Input and Outputs 2

``````apple
banana``````

Input and Outputs 3

``````banana
banking``````
``ban``

Input and Outputs 4

``````manali
manas``````
``mana``
``````# Complete the following function
def longestCommonPrefix(string1, string2):
length = min(len(string1), len(string2))
for i in range(length):
if string1[i] != string2[i]:
return string1[0:i]

# Boilerplate code - Do not edit the following
def main():
string1 = input()
string2 = input()
print(longestCommonPrefix(string1, string2))

if __name__ == "__main__":
main()``````