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# String from prefix and suffix of given two strings

Given two strings a and b, form a new string of length l, from these strings by combining the prefix of string a and suffix of string b.

Examples :

```Input : string a = remuneration
string b = acquiesce
length of pre/suffix(l) = 5
Output :remuniesce

obstreperous
6

Approach :

1.  Get first l letters from string a, and last l letters from string b.
2. Combine both results, and this will be resultant string.

Implementation:

## C++

 `// CPP code to form new string from``// pre/suffix of given strings.``#include``using` `namespace` `std;` `// Returns a string which contains first l``// characters of 'a' and last l characters of 'b'.``string GetPrefixSuffix(string a, string b, ``int` `l)``{``    ``// Getting prefix of first``    ``// string of given length``    ``string prefix = a.substr(0, l);``    ` `    ``// length of string b``    ``int` `lb = b.length();``    ` `    ``// Calculating suffix of second string``    ``string suffix = b.substr(lb - l);``    ` `    ``// Concatenating both prefix and suffix``    ``return` `(prefix + suffix);``}` `// Driver code``int` `main()``{``    ``string a = ``"remuneration"` `,``           ``b = ``"acquiesce"``;``    ``int` `l = 5;``    ``cout << GetPrefixSuffix(a, b, l);``    ``return` `0;``}`

## Java

 `// Java Program to form new string``// from pre/suffix of given strings``import` `java.io.*;` `class` `GFG``{``    ``// Returns a string which contains first l``    ``// characters of 'a' and last l characters of 'b'.``    ``public` `static` `String prefixSuffix(String a,``                                      ``String b,``                                      ``int` `l)``    ``{``        ``// Calculating prefix of first``        ``// string of given length``        ``String prefix = a.substring(``0``, l);``        ``int` `lb = b.length();` `        ``// Calculating suffix of second``        ``// string of given length``        ``String suffix = b.substring(lb - l);``        ``return` `(prefix + suffix);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``                            ``throws` `IOException``    ``{``        ``String a = ``"remuneration"` `,``               ``b = ``"acquiesce"``;``        ``int` `l = ``5``;``        ``System.out.println(prefixSuffix(a, b, l));``    ``}``}`

## Python3

 `# Python code to form new from``# pre/suffix of given strings.` `# Returns a string which contains first l``# characters of 'a' and last l characters of 'b'.``def` `GetPrefixSuffix(a, b, l):``    ``# Getting prefix of first``    ``# of given length``    ``prefix ``=` `a[: l];``    ` `    ``# length of string b``    ``lb ``=` `len``(b);``    ` `    ``# Calculating suffix of second string``    ``suffix ``=` `b[lb ``-` `l:];``    ` `    ``# Concatenating both prefix and suffix``    ``return` `(prefix ``+` `suffix);`  `# Driver code``a ``=` `"remuneration"``;``b ``=` `"acquiesce"``;``l ``=` `5``;``print``(GetPrefixSuffix(a, b, l));`  `# This code contributed by Rajput-Ji`

## C#

 `// C# Program to form new string``// from pre/suffix of given strings.``using` `System;` `class` `GFG``{``    ``// Returns a string which contains first l``    ``// characters of 'a' and last l characters of 'b'.``    ``public` `static` `String prefixSuffix(String a,``                                      ``String b,``                                      ``int` `l)``    ``{``        ``// Calculating prefix of first``        ``// string of given length``        ``String prefix = a.Substring(0, l);``        ``int` `lb = b.Length;` `        ``// Calculating suffix of second``        ``// string of given length``        ``String suffix = b.Substring(lb - l);``        ``return` `(prefix + suffix);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``String a = ``"remuneration"` `,``               ``b = ``"acquiesce"``;``        ``int` `l = 5;``        ``Console.Write(prefixSuffix(a, b, l));``    ``}``}` `// This code is contributed by Nitin Mittal.`

## PHP

 ``

## Javascript

 ``

Output

`remuniesce`

Time Complexity: O(n + m), where n and m are the lengths of the given string.
Auxiliary Space: O(n + m), where n and m are the lengths of the given string.

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