Sub-strings of a string that are prefix of the same string

Last Updated : 20 Feb, 2023

Given a string str, the task is to count all possible sub-strings of the given string that are prefix of the same string.

Examples:

Input: str = “ababc”
Output: 7
All possible sub-string are “a”, “ab”, “aba”, “abab”, “ababc”, “a” and “ab”

Input: str = “abdabc”
Output: 8

Approach: Traverse the string character by character, if the current character is equal to the first character of the string then count all possible sub-strings starting from here that are also the prefixes of str and add it to count. After the complete string has been traversed, print the count.

Below is the implementation of the above approach:

C++14

   // C++ implementation of the approach
#include <iostream>
#include <string>
using namespace std;
 
// Function to return the 
// count of sub-strings starting
// from startIndex that are 
// also the prefixes of str
int subStringsStartingHere(string str, int n,
                            int startIndex)
{
    int count = 0, i = 1;
    while (i <= n)
    {
        if (str.substr(0,i) == 
                str.substr(startIndex, i))
        {
            count++;
        }
        else
            break;
        i++;
    }
 
    return count;
}
 
 
// Function to return the 
// count of all possible sub-strings
// of str that are also the prefixes of str
int countSubStrings(string str, int n)
{
    int count = 0;
    for (int i = 0; i < n; i++)
    {
 
        // If current character is equal to 
        // the starting character of str
        if (str[i] == str[0])
            count += subStringsStartingHere(str, 
                                           n, i);
    }
    return count;
}
 
// Driver code
int main() 
{
    string str = "abcda";
    int n = str.length();
   
    // Function Call
    cout << (countSubStrings(str, n));
}
 
// This code is contributed by harshvijeta0
 
  

Java

   // Java implementation of the approach
public class GFG 
{
 
  // Function to return 
  // the count of sub-strings starting
  // from startIndex that 
  // are also the prefixes of str
  public static int subStringsStartingHere(
                                String str, int n,
                                    int startIndex)
  {
    int count = 0, i = startIndex + 1;
    while (i <= n) 
    {
      if (str.startsWith(str.substring(
                                 startIndex, i))) 
      {
        count++;
      }
      else
        break;
      i++;
    }
    return count;
  }
 
  // Function to return the 
  // count of all possible sub-strings
  // of str that are also the prefixes of str
  public static int countSubStrings(String str, 
                                         int n)
  {
    int count = 0;
 
    for (int i = 0; i < n; i++) 
    {
 
      // If current character is equal to 
      // the starting character of str
      if (str.charAt(i) == str.charAt(0))
        count += subStringsStartingHere(str, n, i);
    }
 
    return count;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    String str = "ababc";
    int n = str.length();
    System.out.println(countSubStrings(str, n));
  }
}
 
  

Python3

   # Python3 implementation of the approach 
 
# Function to return the 
# count of sub-strings starting 
# from startIndex that are 
# also the prefixes of string 
def subStringsStartingHere(string, n, 
                           startIndex):
    count = 0
    i = startIndex + 1
     
    while(i <= n) :
        if string.startswith(
                 string[startIndex : i]):
            count += 1
        else :
            break
         
        i += 1
     
    return count
 
# Function to return the 
# count of all possible sub-strings 
# of string that are also 
# the prefixes of string 
def countSubStrings(string, n) :
    count = 0
     
    for i in range(n) :
         
        # If current character is equal to  
        # the starting character of str 
        if string[i] == string[0] :
            count += subStringsStartingHere(
                              string, n, i)
     
    return count
 
 
# Driver Code
if __name__ == "__main__" :
     
    string = "ababc"
    n = len(string)
    print(countSubStrings(string, n))
 
# this code is contributed by Ryuga
 
  

C#

   // C# implementation of the approach
using System;
class GFG 
{
  
    // Function to return the 
    // count of sub-strings starting
    // from startIndex that 
    // are also the prefixes of str
    static int subStringsStartingHere(
                               String str, int n,
                                   int startIndex)
    {
        int count = 0, i = startIndex + 1;
        while (i <= n) {
            if (str.StartsWith(str.Substring(
                  startIndex, i-startIndex))) 
            {
                count++;
            }
            else
                break;
            i++;
        }
  
        return count;
    }
  
    // Function to return the 
    // count of all possible sub-strings
    // of str that are also the prefixes of str
    static int countSubStrings(String str, int n)
    {
        int count = 0;
  
        for (int i = 0; i < n; i++) {
  
            // If current character is equal to 
            // the starting character of str
            if (str[i] == str[0])
                count += subStringsStartingHere(
                                        str, n, i);
        }
  
        return count;
    }
  
    // Driver code
    static public void Main(String []args)
    {
        String str = "ababc";
        int n = str.Length;
        Console.WriteLine(countSubStrings(str, n));
    }
}
//contributed by Arnab Kundu
 
  

Javascript

   <script>
 
// Javascript implementation of the approach
 
// Function to return the 
// count of sub-strings starting
// from startIndex that are 
// also the prefixes of str
function subStringsStartingHere(str, n, 
                                startIndex)
{
    var count = 0, i = startIndex + 1;
     
    while (i <= n) 
    {
        if (str.startsWith(
            str.substring(startIndex, i))) 
        {
            count++;
        }
        else
            break;
             
        i++;
    }
    return count;
}
 
// Function to return the count of all
// possible sub-strings of str that are
// also the prefixes of str
function countSubStrings(str, n)
{
    var count = 0;
    for(var i = 0; i < n; i++)
    {
         
        // If current character is equal to 
        // the starting character of str
        if (str[i] == str[0])
            count += subStringsStartingHere(str, 
                                            n, i);
    }
    return count;
}
 
// Driver code
var str = "abcda";
var n = str.length;
 
// Function Call
document.write(countSubStrings(str, n));
 
// This code is contributed by rutvik_56
 
</script>
 
  

Output

Complexity Analysis:

Time Complexity: O(N^2)
Auxiliary Space: O(1)

Efficient Approach:

Prerequisite: Z-Algorithm

Approach: Calculate the z-array of the string such that z[i] stores the length of the longest substring starting from i which is also a prefix of string s. Then to count all possible sub-strings of the string that are prefixes of the same string, we just need to add all the values of the z-array since the total number of substrings matching would be equal to the length of the longest substring.

Implementation:

C++

   #include <bits/stdc++.h>
using namespace std;
 
// returns an array z such that  z[i]
// stores length of the longest substring starting
// from i which is also a prefix of string s
vector<int> z_function(string s)
{
    int n = (int)s.length();
    vector<int> z(n);
    // consider a window [l,r]
    // which matches with prefix of s
    int l = 0, r = 0;
    z[0] = n;
    for (int i = 1; i < n; ++i) {
        // when i<=r, we make use of already computed z
        // value for some smaller index
        if (i <= r)
            z[i] = min(r - i + 1, z[i - l]);
 
        // if i>r nothing matches so we will calculate
        // z[i] using naive way.
        while (i + z[i] < n && s[z[i]] == s[i + z[i]])
            ++z[i];
        // update window size
        if (i + z[i] - 1 > r)
            l = i, r = i + z[i] - 1;
    }
    return z;
}
 
int main()
{
    string s = "abcda";
 
    int n = s.length();
 
    vector<int> z = z_function(s);
 
    // stores the count of
    // Sub-strings of a string that
    // are prefix of the same string
    int count = 0;
 
    for (auto x : z)
        count += x;
 
    cout << count << '\n';
 
    return 0;
}
 
  

Python3

   # returns an array z such that  z[i]
# stores length of the longest substring starting
# from i which is also a prefix of s
def z_function(s):
    n = len(s)
    z=[0]*n
    # consider a window [l,r]
    # which matches with prefix of s
    l = 0; r = 0
    z[0] = n
    for i in range(1, n) :
        # when i<=r, we make use of already computed z
        # value for some smaller index
        if (i <= r):
            z[i] = min(r - i + 1, z[i - l])
 
        # if i>r nothing matches so we will calculate
        # z[i] using naive way.
        while (i + z[i] < n and s[z[i]] == s[i + z[i]]):
            z[i]+=1
        # update window size
        if (i + z[i] - 1 > r):
            l = i; r = i + z[i] - 1
     
    return z
 
 
if __name__ == '__main__':
    s = "abcda"
 
    n = len(s)
 
    z = z_function(s)
 
    # stores the count of
    # Sub-strings of a that
    # are prefix of the same string
    count = 0
 
    for x in z:
        count += x
 
    print(count)
 
  

C#

   using System;
class GFG {
 
  // returns an array z such that  z[i]
  // stores length of the longest substring starting
  // from i which is also a prefix of string s
  static int[] z_function(string s)
  {
    int n = s.Length;
    int[] z = new int[n];
 
    // consider a window [l,r]
    // which matches with prefix of s
    int l = 0, r = 0;
    z[0] = n;
    for (int i = 1; i < n; ++i)
    {
 
      // when i<=r, we make use of already computed z
      // value for some smaller index
      if (i <= r)
        z[i] = Math.Min(r - i + 1, z[i - l]);
 
      // if i>r nothing matches so we will calculate
      // z[i] using naive way.
      while (i + z[i] < n && s[z[i]] == s[i + z[i]])
        ++z[i];
      // update window size
      if (i + z[i] - 1 > r)
        l = i;
      r = i + z[i] - 1;
    }
    return z;
  }
 
  public static void Main()
  {
    string s = "abcda";
 
    int n = s.Length;
 
    int[] z = z_function(s);
 
    // stores the count of
    // Sub-strings of a string that
    // are prefix of the same string
    int count = 0;
 
    for (int i = 0; i < z.Length; i++)
      count += z[i];
 
    Console.WriteLine(count);
  }
}
 
// This code is contributed by Samim Hossain Mondal.
 
  

Javascript

   <script>
 
// JavaScript code for the approach
 
// returns an array z such that  z[i]
// stores length of the longest substring starting
// from i which is also a prefix of string s
function z_function(s)
{
    let n = s.length;
    let z = new Array(n).fill(0);
     
    // consider a window [l,r]
    // which matches with prefix of s
    let l = 0, r = 0;
    z[0] = n;
    for (let i = 1; i < n; i++)
    {
     
        // when i<=r, we make use of already computed z
        // value for some smaller index
        if (i <= r)
            z[i] = Math.min(r - i + 1, z[i - l]);
 
        // if i>r nothing matches so we will calculate
        // z[i] using naive way.
        while (i + z[i] < n && s[z[i]] == s[i + z[i]])
            z[i]++;
        // update window size
        if (i + z[i] - 1 > r)
            l = i, r = i + z[i] - 1;
    }
    return z;
}
 
// driver code
let s = "abcda";
let n = s.length;
let z = z_function(s);
 
// stores the count of
// Sub-strings of a string that
// are prefix of the same string
let count = 0;
for (let x of z)
    count += x;
 
document.write(count)
 
// This code is contributed by shinjanpatra
 
</script>
 
  

Java

   import java.util.*;
 
public class Main {
     
    public static ArrayList<Integer> z_function(String s) {
        int n = s.length();
        ArrayList<Integer> z = new ArrayList<Integer>(Collections.nCopies(n, 0));
        int l = 0, r = 0;
        z.set(0, n);
        for (int i = 1; i < n; ++i) {
            if (i <= r) {
                z.set(i, Math.min(r - i + 1, z.get(i - l)));
            }
            while (i + z.get(i) < n && s.charAt(z.get(i)) == s.charAt(i + z.get(i))) {
                int value = z.get(i) + 1;
z.set(i, value);
 
            }
            if (i + z.get(i) - 1 > r) {
                l = i;
                r = i + z.get(i) - 1;
            }
        }
        return z;
    }
 
    public static void main(String[] args) {
        String s = "abcda";
        int n = s.length();
        ArrayList<Integer> z = z_function(s);
        int count = 0;
        for (int x : z) {
            count += x;
        }
        System.out.println(count);
    }
}
 
  

Output

Complexity Analysis:

Time Complexity: O(n)
Auxiliary Space: O(n)

Largest substring of str2 which is a prefix of str1

rituraj_jain

Improve

Article Tags :

Practice Tags :

Sub-strings of a string that are prefix of the same string

C++14

Java

Python3

C#

Javascript

C++

Python3

C#

Javascript

Java

Similar Reads