Longest string which is prefix string of at least two strings
Last Updated :
23 Feb, 2023
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<bits/stdc++.h>
using namespace std;
int max1=0;
// Function to find Max length
// of the prefix
int MaxLength(vector<string> 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<string> 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<string> 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<<MaxLength(v, 0, m) + 1<<endl;
return 0;
}
// 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<String> 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<String> v1 = new Vector<String>();
// 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<String> v = new Vector<String>();
// 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 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# 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<String> 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<String> v1 = new List<String>();
// 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<String> v = new List<String>();
// 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
<script>
// Javascript program to find longest
// string which is prefix string
// of at least two strings
let max1=0;
// Function to find Max length
// of the prefix
function MaxLength(v, i, m)
{
// Base case
if(i>=m)
{
return m-1;
}
// Iterating over all the alphabets
for(let k = 0; k < 26; k++)
{
let c = String.fromCharCode('a'.charCodeAt(0) + k);
let v1 = [];
// Checking if char exists in
// current string or not
for(let j = 0; j < v.length; j++)
{
if(v[j][i] == c)
{
v1.push(v[j]);
}
}
// If atleast 2 string have
// that character
if(v1.length >= 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
// Initialising strings
let s1, s2, s3, s4, s5;
s1 = "abcde";
s2 = "abcsd";
s3 = "bcsdf";
s4 = "abcda";
s5 = "abced";
let v = [];
// push strings into vectors.
v.push(s1);
v.push(s2);
v.push(s3);
v.push(s4);
v.push(s5);
let m = v[0].length;
document.write(MaxLength(v, 0, m) + 1 + "<br>");
</script>
Time Complexity: O(N*26)
Auxiliary Space: O(1)
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