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# Longest string which is prefix string of at least two strings

• Difficulty Level : Medium
• Last Updated : 25 May, 2021

Given a set of strings of the same length, we need to find the length of the longest string, which is a prefix string of at least two strings.

Examples:

```Input:  ["abcde", "abcsd", "bcsdf", "abcda", "abced"]
Output: 4
Explanation:
Longest prefix string is "abcd".

Input:  ["pqrstq", "pwxyza", "abcdef", "pqrstu"]
Output: 5```

Approach:

• Starting from the 0th position, iterate over every character and check if that character occurs in at least two of the strings at the current position or not.
• If it occurs, then recursively call on the next position. Otherwise,
• Update the max value by taking the current maximum with Current_position – 1.
• Finally, return the max value.

## C++

 `// C++ program to find longest``// string which is prefix string``// of at least two strings``#include``using` `namespace` `std;``int` `max1=0;` `// Function to find Max length``// of the prefix``int` `MaxLength(vector v, ``int` `i,``                                ``int` `m)``{``    ``// Base case``    ``if``(i>=m)``    ``{``        ``return` `m-1;``    ``}``    ` `    ``// Iterating over all the alphabets``    ``for``(``int` `k = 0; k < 26; k++)``    ``{``        ``char` `c = ``'a'` `+ k;``        ``vector v1;``        ` `        ``// Checking if char exists in``        ``// current string or not``        ``for``(``int` `j = 0; j < v.size(); j++)``        ``{``            ``if``(v[j][i] == c)``            ``{``                ``v1.push_back(v[j]);``            ``}``        ``}``        ` `        ``// If atleast 2 string have``        ``// that character``        ``if``(v1.size()>=2)``        ``{``           ``// Recursive call to i+1``           ``max1=max(max1,``                    ``MaxLength(v1, i+1, m));``        ``}``        ``else``        ``{``            ``max1=max(max1, i - 1);``        ``}``    ``}``    ``return` `max1;``}` `// Driver code``int` `main()``{``  ``// Initialising strings ``  ``string s1, s2, s3, s4, s5;``  ` `  ``s1 = ``"abcde"``;``  ``s2 = ``"abcsd"``;``  ``s3 = ``"bcsdf"``;``  ``s4 = ``"abcda"``;``  ``s5 = ``"abced"``;``     ` `  ``vector v;``    ` `  ``// push strings into vectors.``  ``v.push_back(s1);``  ``v.push_back(s2);``  ``v.push_back(s3);``  ``v.push_back(s4);``  ``v.push_back(s5);``    ` `  ``int` `m = v[0].size();``    ` `  ``cout<

## Java

 `// Java program to find longest``// String which is prefix String``// of at least two Strings``import` `java.util.*;``class` `GFG{``static` `int` `max1 = ``0``;` `// Function to find Max length``// of the prefix``static` `int` `MaxLength(Vector v,``                     ``int` `i, ``int` `m)``{``    ``// Base case``    ``if``(i>=m)``    ``{``        ``return` `m-``1``;``    ``}``    ` `    ``// Iterating over all the alphabets``    ``for``(``int` `k = ``0``; k < ``26``; k++)``    ``{``        ``char` `c = (``char``)(``'a'` `+ k);``        ``Vector v1 = ``new` `Vector();``        ` `        ``// Checking if char exists in``        ``// current String or not``        ``for``(``int` `j = ``0``; j < v.size(); j++)``        ``{``            ``if``(v.get(j).charAt(i) == c)``            ``{``                ``v1.add(v.get(j));``            ``}``        ``}``        ` `        ``// If atleast 2 String have``        ``// that character``        ``if``(v1.size() >= ``2``)``        ``{``           ``// Recursive call to i+1``           ``max1=Math.max(max1,``                         ``MaxLength(v1, i + ``1``, m));``        ``}``        ``else``        ``{``            ``max1=Math.max(max1, i - ``1``);``        ``}``    ``}``    ``return` `max1;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``  ``// Initialising Strings ``  ``String s1, s2, s3, s4, s5; ``  ``s1 = ``"abcde"``;``  ``s2 = ``"abcsd"``;``  ``s3 = ``"bcsdf"``;``  ``s4 = ``"abcda"``;``  ``s5 = ``"abced"``;``     ` `  ``Vector v = ``new` `Vector();``    ` `  ``// push Strings into vectors.``  ``v.add(s1);``  ``v.add(s2);``  ``v.add(s3);``  ``v.add(s4);``  ``v.add(s5);``    ` `  ``int` `m = v.get(``0``).length();   ``  ``System.out.print(MaxLength(v, ``0``, m) + ``1``);``}``}` `// This code is contributed by shikhasingrajput`

## Python3

 `# Python3 program to find longest``# string which is prefix string``# of at least two strings` `max1 ``=` `0` `# Function to find max length``# of the prefix``def` `MaxLength(v, i, m):``    ` `    ``global` `max1``    ` `    ``# Base case``    ``if``(i >``=` `m):``        ``return` `m ``-` `1``        ` `    ``# Iterating over all the alphabets``    ``for` `k ``in` `range``(``26``):``        ``c ``=` `chr``(``ord``(``'a'``) ``+` `k)``        ``v1 ``=` `[]``        ` `        ``# Checking if char exists in``        ``# current string or not``        ``for` `j ``in` `range``(``len``(v)):``            ``if``(v[j][i] ``=``=` `c):``                ``v1.append(v[j])``        ` `        ``# If atleast 2 string have``        ``# that character``        ``if``(``len``(v1) >``=` `2``):``            ` `            ``# Recursive call to i+1``            ``max1 ``=` `max``(max1, MaxLength(v1, i ``+` `1``, m))``        ``else``:``            ``max1 ``=` `max``(max1, i ``-` `1``)``            ` `    ``return` `max1` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Initialising strings``    ``s1 ``=` `"abcde"``    ``s2 ``=` `"abcsd"``    ``s3 ``=` `"bcsdf"``    ``s4 ``=` `"abcda"``    ``s5 ``=` `"abced"``    ``v ``=` `[]` `    ``# Push strings into vectors.``    ``v.append(s1)``    ``v.append(s2)``    ``v.append(s3)``    ``v.append(s4)``    ``v.append(s5)``    ` `    ``m ``=` `len``(v[``0``])``    ` `    ``print``(MaxLength(v, ``0``, m) ``+` `1``)` `# This code is contributed by BhupendraSingh`

## C#

 `// C# program to find longest``// String which is prefix String``// of at least two Strings``using` `System;``using` `System.Collections.Generic;` `class` `GFG{``    ` `static` `int` `max1 = 0;` `// Function to find Max length``// of the prefix``static` `int` `MaxLength(List v,``                     ``int` `i, ``int` `m)``{``    ` `    ``// Base case``    ``if` `(i >= m)``    ``{``        ``return` `m - 1;``    ``}``    ` `    ``// Iterating over all the alphabets``    ``for``(``int` `k = 0; k < 26; k++)``    ``{``        ``char` `c = (``char``)(``'a'` `+ k);``        ``List v1 = ``new` `List();``        ` `        ``// Checking if char exists in``        ``// current String or not``        ``for``(``int` `j = 0; j < v.Count; j++)``        ``{``            ``if` `(v[j][i] == c)``            ``{``                ``v1.Add(v[j]);``            ``}``        ``}``        ` `        ``// If atleast 2 String have``        ``// that character``        ``if` `(v1.Count >= 2)``        ``{``            ` `            ``// Recursive call to i+1``            ``max1 = Math.Max(max1,``                            ``MaxLength(v1, i + 1, m));``        ``}``        ``else``        ``{``            ``max1 = Math.Max(max1, i - 1);``        ``}``    ``}``    ``return` `max1;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ` `    ``// Initialising Strings``    ``String s1, s2, s3, s4, s5;``    ``s1 = ``"abcde"``;``    ``s2 = ``"abcsd"``;``    ``s3 = ``"bcsdf"``;``    ``s4 = ``"abcda"``;``    ``s5 = ``"abced"``;``        ` `    ``List v = ``new` `List();``        ` `    ``// push Strings into vectors.``    ``v.Add(s1);``    ``v.Add(s2);``    ``v.Add(s3);``    ``v.Add(s4);``    ``v.Add(s5);``        ` `    ``int` `m = v[0].Length;``    ` `    ``Console.Write(MaxLength(v, 0, m) + 1);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output:
`4`

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