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Count of substrings of a string containing another given string as a substring | Set 2

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  • Last Updated : 16 Jun, 2022

Given two strings S and T of length N and M respectively, the task is to count the number of substrings of S that contains the string T in it as a substring.

Examples:

Input: S = “dabc”, T = “ab”
Output: 4
Explanation:
Substrings of S containing T as a substring are:

  1. S[0, 2] = “dab”
  2. S[1, 2] = “ab”
  3. S[1, 3] = “abc”
  4. S[0, 3] = “dabc”

Input: S = “hshshshs” T = “hs”
Output: 25

 

Naive Approach: For the simplest approach to solve the problem, refer to the previous post of this article.

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

Efficient Approach: To optimize the above approach, the idea is to find out all the occurrences of T in S. Whenever T is found in S, add all the substrings which contain this occurrence of T excluding the substrings which were already calculated in the previous occurrences. Follow the steps below to solve the problem:

  • Initialize a variable, say answer, to store the count of substrings.
  • Initialize a variable, say last, to store the starting index of the last occurrence of T in S.
  • Iterate over the range [0, N – M] using a variable, say i.
    • Check if the substring S[i, i + M] is equal to T or not. If found to be true, then add (i + 1 – last) * (N – (i + M – 1)) to answer and update last to (i + 1).
    • Otherwise, continue for the next iteration.
  • After completing the above steps, print the value of the answer as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the substrings of
// string containing another given
// string as a substring
void findOccurrences(string S, string T)
{
    // Store length of string S
    int n1 = S.size();
 
    // Store length of string T
    int n2 = T.size();
 
    // Store the required count of
    // substrings
    int ans = 0;
 
    // Store the starting index of
    // last occurrence of T in S
    int last = 0;
 
    // Iterate in range [0, n1-n2]
    for (int i = 0; i <= n1 - n2; i++) {
 
        // Check if substring from i
        // to i + n2 is equal to T
        bool chk = true;
 
        // Check if substring from i
        // to i + n2 is equal to T
        for (int j = 0; j < n2; j++) {
 
            // Mark chk as false and
            // break the loop
            if (T[j] != S[i + j]) {
                chk = false;
                break;
            }
        }
 
        // If chk is true
        if (chk) {
 
            // Add (i + 1 - last) *
            // (n1 - (i + n2 - 1))
            // to answer
            ans += (i + 1 - last)
                   * (n1 - (i + n2 - 1));
 
            // Update the last to i + 1
            last = i + 1;
        }
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver code
int main()
{
    string S = "dabc", T = "ab";
 
    // Function Call
    findOccurrences(S, T);
}

Java



// Java program for the above approach
class GFG{
   
// Function to count the substrings of
// string containing another given
// string as a substring
static void findOccurrences(String S, String T)
{
   
    // Store length of string S
    int n1 = S.length();
 
    // Store length of string T
    int n2 = T.length();
 
    // Store the required count of
    // substrings
    int ans = 0;
 
    // Store the starting index of
    // last occurrence of T in S
    int last = 0;
 
    // Iterate in range [0, n1-n2]
    for (int i = 0; i <= n1 - n2; i++)
    {
 
        // Check if substring from i
        // to i + n2 is equal to T
        boolean chk = true;
 
        // Check if substring from i
        // to i + n2 is equal to T
        for (int j = 0; j < n2; j++)
        {
 
            // Mark chk as false and
            // break the loop
            if (T.charAt(j) != S.charAt(i + j))
            {
                chk = false;
                break;
            }
        }
 
        // If chk is true
        if (chk)
        {
 
            // Add (i + 1 - last) *
            // (n1 - (i + n2 - 1))
            // to answer
            ans += (i + 1 - last)
                   * (n1 - (i + n2 - 1));
 
            // Update the last to i + 1
            last = i + 1;
        }
    }
 
    // Print the answer
    System.out.println(ans);
}
 
  // Driver code
  public static void main (String[] args)
  {
    String S = "dabc", T = "ab";
 
    // Function Call
    findOccurrences(S, T);
  }
}
 
// This code is contributed by AnkThon
Python3



# Python3 program for the above approach
 
# Function to count the substrings of
# containing another given
# as a sub
def findOccurrences(S, T):
   
    # Store length of S
    n1 = len(S)
 
    # Store length of T
    n2 = len(T)
 
    # Store the required count of
    # substrings
    ans = 0
 
    # Store the starting index of
    # last occurrence of T in S
    last = 0
 
    # Iterate in range [0, n1-n2]
    for i in range(n1 - n2 + 1):
 
        # Check if subfrom i
        # to i + n2 is equal to T
        chk = True
 
        # Check if subfrom i
        # to i + n2 is equal to T
        for j in range(n2):
 
            # Mark chk as false and
            # break the loop
            if (T[j] != S[i + j]):
                chk = False
                break
 
        # If chk is true
        if (chk):
 
            # Add (i + 1 - last) *
            # (n1 - (i + n2 - 1))
            # to answer
            ans += (i + 1 - last) * (n1 - (i + n2 - 1))
 
            # Update the last to i + 1
            last = i + 1
 
    # Print the answer
    print(ans)
 
# Driver code
if __name__ == '__main__':
    S,T = "dabc","ab"
 
    # Function Call
    findOccurrences(S, T)
 
# This code is contributed by mohit kumar 29
C#



// C# program for the above approach
using System;
 
class GFG
{
   
// Function to count the substrings of
// string containing another given
// string as a substring
static void findOccurrences(String S, String T)
{
   
    // Store length of string S
    int n1 = S.Length;
 
    // Store length of string T
    int n2 = T.Length;
 
    // Store the required count of
    // substrings
    int ans = 0;
 
    // Store the starting index of
    // last occurrence of T in S
    int last = 0;
 
    // Iterate in range [0, n1-n2]
    for (int i = 0; i <= n1 - n2; i++)
    {
 
        // Check if substring from i
        // to i + n2 is equal to T
        bool chk = true;
 
        // Check if substring from i
        // to i + n2 is equal to T
        for (int j = 0; j < n2; j++)
        {
 
            // Mark chk as false and
            // break the loop
            if (T[j] != S[i + j])
            {
                chk = false;
                break;
            }
        }
 
        // If chk is true
        if (chk)
        {
 
            // Add (i + 1 - last) *
            // (n1 - (i + n2 - 1))
            // to answer
            ans += (i + 1 - last)
                   * (n1 - (i + n2 - 1));
 
            // Update the last to i + 1
            last = i + 1;
        }
    }
 
    // Print the answer
    Console.WriteLine(ans);
}
 
  // Driver code
  public static void Main(String[] args)
  {
    String S = "dabc", T = "ab";
 
    // Function Call
    findOccurrences(S, T);
  }
}
 
 
// This code is contributed by 29AjayKumar
Javascript



<script>
 
// JavaScript program for above approach
 
// Function to count the substrings of
// string containing another given
// string as a substring
function findOccurrences(S, T)
{
    
    // Store length of string S
    let n1 = S.length;
  
    // Store length of string T
    let n2 = T.length;
  
    // Store the required count of
    // substrings
    let ans = 0;
  
    // Store the starting index of
    // last occurrence of T in S
    let last = 0;
  
    // Iterate in range [0, n1-n2]
    for (let i = 0; i <= n1 - n2; i++)
    {
  
        // Check if substring from i
        // to i + n2 is equal to T
        let chk = true;
  
        // Check if substring from i
        // to i + n2 is equal to T
        for (let j = 0; j < n2; j++)
        {
  
            // Mark chk as false and
            // break the loop
            if (T[j] != S[i + j])
            {
                chk = false;
                break;
            }
        }
  
        // If chk is true
        if (chk)
        {
  
            // Add (i + 1 - last) *
            // (n1 - (i + n2 - 1))
            // to answer
            ans += (i + 1 - last)
                   * (n1 - (i + n2 - 1));
  
            // Update the last to i + 1
            last = i + 1;
        }
    }
  
    // Print the answer
    document.write(ans);
}
 
// Driver Code
 
     let S = "dabc", T = "ab";
  
    // Function Call
    findOccurrences(S, T);
     
</script>
Output: 
4
 
Time Complexity: O(N*M)
Auxiliary Space: O(1)

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Article Contributed By :
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AmanGupta65
@AmanGupta65
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