# Find if a string starts and ends with another given string

• Difficulty Level : Easy
• Last Updated : 07 May, 2021

Given a string str and a corner string cs, we need to find out whether the string str starts and ends with the corner string cs or not.
Examples:

```Input : str = "geeksmanishgeeks", cs = "geeks"
Output : Yes

Input : str = "shreya dhatwalia", cs = "abc"
Output : No```

Hey! Looking for some great resources suitable for young ones? You've come to the right place. Check out our self-paced courses designed for students of grades I-XII

Start with topics like Python, HTML, ML, and learn to make some games and apps all with the help of our expertly designed content! So students worry no more, because GeeksforGeeks School is now here!

Algorithm

• Find length of given string str as well as corner string cs. Let this length be n and cl respectively.
• If cl>n, return false as cs can’t be greater than str.
• Otherwise, find the prefix and suffix of length cl from str. If both prefix and suffix match with corner string cs, return true otherwise return false.

## C++

 `// CPP program to find if a given corner string``// is present at corners.``#include ``using` `namespace` `std;` `bool` `isCornerPresent(string str, string corner)``{``    ``int` `n = str.length();``    ``int` `cl = corner.length();` `    ``// If length of corner string is more, it``    ``// cannot be present at corners.``    ``if` `(n < cl)``       ``return` `false``;` `    ``// Return true if corner string is present at``    ``// both corners of given string.``    ``return` `(str.substr(0, cl).compare(corner) == 0 &&``            ``str.substr(n-cl, cl).compare(corner) == 0);``}` `// Driver code``int` `main()``{``   ``string str = ``"geeksforgeeks"``;``   ``string corner = ``"geeks"``;``   ``if` `(isCornerPresent(str, corner))``      ``cout << ``"Yes"``;``   ``else``      ``cout << ``"No"``;``   ``return` `0;``}`

## Java

 `// Java program to find if a given corner``// string is present at corners.``import` `java.io.*;``class` `GFG {``    ` `    ``static` `boolean` `isCornerPresent(String str,``                                   ``String corner)``    ``{``        ``int` `n = str.length();``        ``int` `cl = corner.length();` `        ``// If length of corner string``        ``// is more, it cannot be present``        ``// at corners.``        ``if` `(n < cl)``        ``return` `false``;` `        ``// Return true if corner string``        ``// is present at both corners``        ``// of given string.``        ``return` `(str.substring(``0``, cl).equals(corner) &&``                ``str.substring(n - cl, n).equals(corner));``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``String str = ``"geeksforgeeks"``;``        ``String corner = ``"geeks"``;``        ``if` `(isCornerPresent(str, corner))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);``    ``}``}` `// This code is contributed by Manish_100`

## Python3

 `# Python program to find``# if a given corner string``# is present at corners.` `def` `isCornerPresent(``str``, corner) :` `    ``n ``=` `len``(``str``)``    ``cl ``=` `len``(corner)` `    ``# If length of corner``    ``# string is more, it``    ``# cannot be present``    ``# at corners.``    ``if` `(n < cl) :``        ``return` `False` `    ``# Return true if corner``    ``# string is present at``    ``# both corners of given``    ``# string.``    ``return` `((``str``[: cl] ``=``=` `corner) ``and``            ``(``str``[n ``-` `cl :] ``=``=` `corner))` `# Driver Code``str` `=` `"geeksforgeeks"``corner ``=` `"geeks"``if` `(isCornerPresent(``str``, corner)) :``    ``print` `(``"Yes"``)``else` `:``    ``print` `(``"No"``)` `# This code is contributed by``# Manish Shaw(manishshaw1)`

## C#

 `// C# program to find if a``// given corner string is``// present at corners.``using` `System;` `class` `GFG``{``static` `bool` `isCornerPresent(``string` `str,``                            ``string` `corner)``{``    ``int` `n = str.Length;``    ``int` `cl = corner.Length;` `    ``// If length of corner``    ``// string is more, it``    ``// cannot be present``    ``// at corners.``    ``if` `(n < cl)``        ``return` `false``;` `    ``// Return true if corner``    ``// string is present at``    ``// both corners of given``    ``// string.``    ``return` `(str.Substring(0,``            ``cl).Equals(corner) &&``            ``str.Substring(n - cl,``            ``cl).Equals(corner));``}` `// Driver Code``static` `void` `Main ()``{``    ``string` `str = ``"geeksforgeeks"``;``    ``string` `corner = ``"geeks"``;``    ``if` `(isCornerPresent(str, corner))``        ``Console.WriteLine(``"Yes"``);``    ``else``        ``Console.WriteLine(``"No"``);``}``}` `// This code is contributed by``// Manish Shaw(manishshaw1)`

## PHP

 ``

## Javascript

 ``

Output :

`Yes`

My Personal Notes arrow_drop_up