Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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 <bits/stdc++.h>
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




<?php
// PHP program to find if a
// given corner string is
// present at corners.
 
function isCornerPresent($str,
                         $corner)
{
    $n = strlen($str);
    $cl = strlen($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 (!strcmp(substr($str, 0,
                           $cl), $corner) &&
            !strcmp(substr($str, $n -
                           $cl, $cl), $corner));
}
 
// Driver Code
$str = "geeksforgeeks";
$corner = "geeks";
if (isCornerPresent($str, $corner))
    echo ("Yes");
else
    echo ("No");
 
// This code is contributed by
// Manish Shaw(manishshaw1)
?>

Javascript




<script>
      // JavaScript program to find if a given corner string
      // is present at corners.
      function isCornerPresent(str, corner) {
        var n = str.length;
        var 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).localeCompare(corner) === 0 &&
          str.substring(n - cl, n).localeCompare(corner) === 0
        );
      }
 
      // Driver code
      var str = "geeksforgeeks";
      var corner = "geeks";
      if (isCornerPresent(str, corner)) document.write("Yes");
      else document.write("No");
    </script>

Output : 
 

Yes

 

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!