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 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 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# 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
<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>
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;
}
# 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)
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.
<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>
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);
}
}
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(n)
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