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Length of all prefixes that are also the suffixes of given string
  • Difficulty Level : Easy
  • Last Updated : 23 Apr, 2021

Given a string S consisting of N characters, the task is to find the length of all prefixes of the given string S that are also suffixes of the same string S.

Examples:

Input: S = “ababababab”
Output: 2 4 6 8
Explanation: 
The prefixes of S that are also its suffixes are:

  1. “ab” of length = 2
  2. “abab” of length = 4
  3. “ababab” of length = 6
  4. “abababab” of length = 8

Input: S = “geeksforgeeks”
Output: 5

 

Naive Approach: The simplest approach to solve the given problem is to traverse the given string, S from the start, and in each iteration add the current character to the prefix string, and check if the prefix string is the same as the suffix of the same length or not. If found to be true, then print the length of the prefix string. Otherwise, check for the next prefix.



Below is the implementation of the above approach:

C++14




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to  find the length of all
// prefixes of the given string that
// are also suffixes of the same string
void countSamePrefixSuffix(string s, int n)
{
    // Stores the prefix string
    string prefix = "";
 
    // Traverse the string S
    for (int i = 0; i < n - 1; i++) {
 
        // Add the current character
        // to the prefix string
        prefix += s[i];
 
        // Store the suffix string
        string suffix = s.substr(
            n - 1 - i, n - 1);
 
 
        // Check if both the strings
        // are equal or not
        if (prefix == suffix) {
           cout << prefix.size() << " ";
        }
    }
}
 
// Driver Code
int main()
{
    string S = "ababababab";
    int N = S.size();
    countSamePrefixSuffix(S, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to  find the length of all
// prefixes of the given string that
// are also suffixes of the same string
static void countSamePrefixSuffix(String s, int n)
{
     
    // Stores the prefix string
    String prefix = "";
 
    // Traverse the string S
    for(int i = 0; i < n - 1; i++)
    {
         
        // Add the current character
        // to the prefix string
        prefix += s.charAt(i);
 
        // Store the suffix string
        String suffix = s.substring(n - 1 - i, n);
 
        // Check if both the strings
        // are equal or not
        if (prefix.equals(suffix))
        {
            System.out.print(prefix.length() + " ");
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
    String S = "ababababab";
    int N = S.length();
     
    countSamePrefixSuffix(S, N);
}
}
 
// This code is contributed by Kingash

Python3




# Python3 program for the above approach
 
# Function to  find the length of all
# prefixes of the given that
# are also suffixes of the same string
def countSamePrefixSuffix(s, n):
     
    # Stores the prefix string
    prefix = ""
 
    # Traverse the S
    for i in range(n - 1):
 
        # Add the current character
        # to the prefix string
        prefix += s[i]
 
        # Store the suffix string
        suffix = s[n - 1 - i: 2 * n - 2 - i]
 
        # Check if both the strings
        # are equal or not
        if (prefix == suffix):
            print(len(prefix), end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    S = "ababababab"
    N = len(S)
     
    countSamePrefixSuffix(S, N)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
 // Function to  find the length of all
// prefixes of the given string that
// are also suffixes of the same string
static void countSamePrefixSuffix(string s, int n)
{
    // Stores the prefix string
    string prefix = "";
 
    // Traverse the string S
    for (int i = 0; i < n - 1; i++) {
 
        // Add the current character
        // to the prefix string
        prefix += s[i];
 
        // Store the suffix string
        string suffix = s.Substring(n - 1 - i, i+1);
 
 
        // Check if both the strings
        // are equal or not
        if (prefix == suffix) {
           Console.Write(prefix.Length + " ");
        }
    }
}
 
// Driver Code
public static void Main()
{
    string S = "ababababab";
    int N = S.Length;
    countSamePrefixSuffix(S, N);   
}
}
 
// This code is contributed by SURENDRA_GANGWAR.
Output: 
2 4 6 8

 

Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach: The above approach can also be optimized by using hashing to store the prefixes of the given string. Then, iterate through all the suffixes and check if they are present in the hash map or not. Follow the steps below to solve the problem:

  • Initialize two deques, say prefix and suffix to store the prefix string and suffix strings of S.
  • Initialize a HashMap, say M to store all the prefixes of S.
  • Traverse the given string S over the range [0, N – 2] using the variable i
    • Push the current character at the back of prefix and suffix deque.
    • Mark prefix as true in the HashMap M.
  • After the loop, add the last character of the string, say S[N – 1] to the suffix.
  • Iterate over the range [0, N – 2] and perform the following steps:
    • Remove the front character of the suffix.
    • Now, check if the current deque is present in the HashMap M or not. If found to be true, then print the size of the deque.

Below is the implementation of the above approach:

C++14




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to  find the length of all
// prefixes of the  given string that
// are also suffixes of the same string
void countSamePrefixSuffix(string s, int n)
{
    // Stores the prefixes of the string
    map<deque<char>, int> cnt;
 
    // Stores the prefix & suffix strings
    deque<char> prefix, suffix;
 
    // Iterate in the range [0, n - 2]
    for (int i = 0; i < n - 1; i++) {
 
        // Add the current character to
        // the prefix and suffix strings
        prefix.push_back(s[i]);
        suffix.push_back(s[i]);
 
        // Mark the prefix as 1 in
        // the HashMap
        cnt[prefix] = 1;
    }
 
    // Add the last character to
    // the suffix
    suffix.push_back(s[n - 1]);
    int index = n - 1;
 
    // Iterate in the range [0, n - 2]
    for (int i = 0; i < n - 1; i++) {
 
        // Remove the character from
        // the front of suffix deque
        // to get the suffix string
        suffix.pop_front();
 
        // Check if the suffix is
        // present in HashMap or not
        if (cnt[suffix] == 1) {
            cout << index << " ";
        }
 
        index--;
    }
}
 
// Driver Code
int main()
{
    string S = "ababababab";
    int N = S.size();
    countSamePrefixSuffix(S, N);
 
    return 0;
}
Output: 
8 6 4 2

 

Time Complexity: O(N * log N)
Auxiliary Space: O(N)

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