Número de substrings de una string presente en otra

Supongamos que nos dan una string s1, necesitamos encontrar el número total de substrings (incluidas las múltiples apariciones de la misma substring) de s1 que están presentes en la string s2. 

Ejemplos: 

Input : s1 = aab
        s2 = aaaab
Output :6
Substrings of s1 are ["a", "a", "b", "aa", 
"ab", "aab"]. These all are present in s2. 
Hence, answer is 6.

Input :s1 = abcd
       s2 = swalencud
Output :3 

La idea es considerar todas las substrings de s1 y verificar si se presenta en s2.  

// CPP program to count number of substrings of s1
// present in s2.
#include<iostream>
#include<string>
using namespace std;
 
int countSubstrs(string s1, string s2)
{
    int ans = 0;
 
    for (int i = 0; i < s1.length(); i++) {
         
        // s3 stores all substrings of s1
        string s3;
        for (int j = i; j < s1.length(); j++) {
            s3 += s1[j];
 
            // check the presence of s3 in s2
            if (s2.find(s3) != string::npos)
                ans++;
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    string s1 = "aab", s2 = "aaaab";
    cout << countSubstrs(s1, s2);
    return 0;
}

// Java program to count number of
// substrings of s1 present in s2.
import java.util.*;
 
class GFG
{
 
static int countSubstrs(String s1,
                        String s2)
{
int ans = 0;
 
for (int i = 0; i < s1.length(); i++)
{
     
    // s3 stores all substrings of s1
    String s3 = "";
    char[] s4 = s1.toCharArray();
    for (int j = i; j < s1.length(); j++)
    {
        s3 += s4[j];
 
        // check the presence of s3 in s2
        if (s2.indexOf(s3) != -1)
            ans++;
    }
}
return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String s1 = "aab", s2 = "aaaab";
    System.out.println(countSubstrs(s1, s2));
}
}
 
// This code is contributed by ChitraNayal

# Python 3 program to count number of substrings of s1
# present in s2.
 
# Function for counting no. of substring
# of s1 present in s2
def countSubstrs(s1, s2) :
    ans = 0
    for i in range(len(s1)) :
        s3 = ""
 
        # s3 stores all substrings of s1
        for j in range(i, len(s1)) :
            s3 += s1[j]
 
            # check the presence of s3 in s2
            if s2.find(s3) != -1 :
                ans += 1
    return ans
 
# Driver code
if __name__ == "__main__" :
    s1 = "aab"
    s2 = "aaaab"
 
    # function calling
    print(countSubstrs(s1, s2))
     
# This code is contributed by ANKITRAI1

// C# program to count number of
// substrings of s1 present in s2.
using System;
 
class GFG
{
static int countSubstrs(String s1,
                        String s2)
{
int ans = 0;
 
for (int i = 0; i < s1.Length; i++)
{
     
    // s3 stores all substrings of s1
    String s3 = "";
    char[] s4 = s1.ToCharArray();
    for (int j = i; j < s1.Length; j++)
    {
        s3 += s4[j];
 
        // check the presence of s3 in s2
        if (s2.IndexOf(s3) != -1)
            ans++;
    }
}
return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    String s1 = "aab", s2 = "aaaab";
    Console.WriteLine(countSubstrs(s1, s2));
}
}
 
// This code is contributed
// by Kirti_Mangal

<?php
// PHP program to count number of
// substrings of s1 present in s2.
 
function countSubstrs($s1, $s2)
{
    $ans = 0;
 
    for ($i = 0; $i < strlen($s1); $i++)
    {
         
        // s3 stores all substrings of s1
        $s3 = "";
        for ($j = $i;
             $j < strlen($s1); $j++)
        {
            $s3 += $s1[$j];
 
            // check the presence of s3 in s2
            if (stripos($s2, $s3, 0) != -1)
                $ans++;
        }
    }
    return $ans;
}
 
// Driver code
$s1 = "aab";
$s2 = "aaaab";
echo countSubstrs($s1, $s2);
 
// This code is contributed
// by ChitraNayal
?>

<script>
 
// javascript program to count number of
// substrings of s1 present in s2.
 
function countSubstrs( s1, s2)
{
var ans = 0;
 
for (var i = 0; i < s1.length; i++)
{
     
    // s3 stores all substrings of s1
    var s3 = "";
    var s4 = s1 ;
    for (var j = i; j < s1.length; j++)
    {
        s3 += s4[j];
 
        // check the presence of s3 in s2
        if (s2.indexOf(s3) != -1)
            ans++;
    }
}
return ans;
}
 
// Driver code
 
    var s1 = "aab", s2 = "aaaab";
    document.write(countSubstrs(s1, s2));
 
</script>

6

Complejidad de tiempo: O(n*n*n), ya que se usan bucles anidados donde n es el tamaño de la string s1
Espacio auxiliar: O(1), ya que no se usa espacio adicional