Number of sub-strings which are anagram of any sub-string of another string

Given two strings S1 and S2, the task is to count the number of sub-strings of S1 that are anagrams of any sub-string of S2.

Examples:

Input: S1 = “ABB”, S2 = “BAB” 
Output: 5 
There are 6 sub-strings of S1 : “A”, “B”, “B”, “AB”, “BB” and “ABB” 
Out of which only “BB” is the one which is not an anagram of any sub-string of S2.

Input: S1 = “PLEASEHELPIMTRAPPED”, S2 = “INAKICKSTARTFACTORY” 
Output: 9 

Naive approach: A simple approach is to check all the sub-strings of S1 against all the sub-strings of S2 whether they are anagrams or not.

Efficient approach: Take all the sub-strings of S1 one by one say temp and check whether temp is an anagram of any sub-string of S2 by calculating the frequencies of all the characters of temp and comparing it with the character frequencies of sub-strings of S2 having length = length(temp). 

This can be done with a single traversal by taking the first length(temp) characters of S2 and then for every iteration, add the frequency of the next character of the string and remove the frequency of the first character of the previously chosen sub-string until the complete string is traversed.

Below is the implementation of the above approach: 

C++

// C++ program to find the number of sub-strings
// of s1 which are anagram of any sub-string of s2

#include <bits/stdc++.h>
using namespace std;

#define ALL_CHARS 256

// This function returns true if
// contents of arr1[] and arr2[]
// are same, otherwise false.
bool compare(char* arr1, char* arr2)
{
    for (int i = 0; i < ALL_CHARS; i++)
        if (arr1[i] != arr2[i])
            return false;

    return true;
}

// This function search for all permutations
// of string pat[] in string txt[]
bool search(string pat, string txt)
{
    int M = pat.length();
    int N = txt.length();

    int i;

    // countP[]: Store count of all characters
    // of pattern
    // countTW[]: Store count of current
    // window of text
    char countP[ALL_CHARS] = { 0 };
    char countTW[ALL_CHARS] = { 0 };
    for (i = 0; i < M; i++) {
        (countP[pat[i]])++;
        (countTW[txt[i]])++;
    }

    // Traverse through remaining
    // characters of pattern
    for (i = M; i < N; i++) {

        // Compare counts of current
        // window of text with
        // counts of pattern[]
        if (compare(countP, countTW)) {
            // cout<<pat<<" "<<txt<<" ";
            return true;
        }

        // Add current character to current window
        (countTW[txt[i]])++;

        // Remove the first character
        // of previous window
        countTW[txt[i - M]]--;
    }

    // Check for the last window in text
    if (compare(countP, countTW))
        return true;

    return false;
}

// Function to return the number of sub-strings of s1
// that are anagrams of any sub-string of s2
int calculatesubString(string s1, string s2, int n)
{
    // initializing variables
    int count = 0, j = 0, x = 0;

    // outer loop for picking starting point
    for (int i = 0; i < n; i++) {

        // loop for different length of substrings
        for (int len = 1; len <= n - i; len++) {

            // If s2 has any substring which is
            // anagram of s1.substr(i, len)
            if (search(s1.substr(i, len), s2)) {

                // increment the count
                count = count + 1;
            }
        }
    }
    return count;
}

// Driver code
int main()
{
    string str1 = "PLEASEHELPIMTRAPPED";
    string str2 = "INAKICKSTARTFACTORY";

    int len = str1.length();

    cout << calculatesubString(str1, str2, len);

    return 0;
}

Java

// Java program to find the number of sub-Strings
// of s1 which are anagram of any sub-String of s2
class GFG {

    static int MAX_LEN = 1005;
    static int MAX_CHAR = 26;

    static int ALL_CHARS = 256;

    // This function returns true if
    // contents of arr1[] and arr2[]
    // are same, otherwise false.
    static boolean compare(char[] arr1, char[] arr2)
    {
        for (int i = 0; i < ALL_CHARS; i++)
            if (arr1[i] != arr2[i])
                return false;

        return true;
    }

    // This function search for all permutations
    // of String pat[] in String txt[]
    static boolean search(String pat, String txt)
    {
        int M = pat.length();
        int N = txt.length();

        int i;

        // countP[]: Store count of all characters
        // of pattern
        // countTW[]: Store count of current
        // window of text
        char countP[] = new char[ALL_CHARS];
        char countTW[] = new char[ALL_CHARS];
        for (i = 0; i < M; i++) {
            (countP[pat.charAt(i)])++;
            (countTW[txt.charAt(i)])++;
        }

        // Traverse through remaining
        // characters of pattern
        for (i = M; i < N; i++) {

            // Compare counts of current
            // window of text with
            // counts of pattern[]
            if (compare(countP, countTW)) {
                // cout<<pat<<" "<<txt<<" ";
                return true;
            }

            // Add current character to current window
            (countTW[txt.charAt(i)])++;

            // Remove the first character
            // of previous window
            countTW[txt.charAt(i - M)]--;
        }

        // Check for the last window in text
        if (compare(countP, countTW))
            return true;

        return false;
    }

    // Function to return the number of sub-Strings of s1
    // that are anagrams of any sub-String of s2
    static int calculatesubString(String s1, String s2, int n)
    {
        // initializing variables
        int count = 0, j = 0, x = 0;

        // outer loop for picking starting point
        for (int i = 0; i < n; i++) {

            // loop for different length of subStrings
            for (int len = 1; len <= n - i; len++) {

                // If s2 has any subString which is
                // anagram of s1.substr(i, len)
                if (search(s1.substring(i, i + len), s2)) {

                    // increment the count
                    count = count + 1;
                }
            }
        }
        return count;
    }

    // Driver code
    public static void main(String args[])
    {
        String str1 = "PLEASEHELPIMTRAPPED";
        String str2 = "INAKICKSTARTFACTORY";

        int len = str1.length();

        System.out.println(calculatesubString(str1, str2, len));
    }
}

// This code has been contributed by 29AjayKumar

Python3

# Python3 program to find the number of sub-strings
# of s1 which are anagram of any sub-string of s2
ALL_CHARS = 256

# This function returns true if
# contents of arr1[] and arr2[]
# are same, otherwise false.
def compare(arr1, arr2):
    for i in range(ALL_CHARS):
        if arr1[i] != arr2[i]:
            return False
    return True

# This function search for all permutations
# of string pat[] in string txt[]
def search(pat, txt):
    M = len(pat)
    N = len(txt)

    # countP[]: Store count of all characters
    # of pattern
    # countTW[]: Store count of current
    # window of text
    countP = [0] * ALL_CHARS
    countTW = [0] * ALL_CHARS
    for i in range(M):
        countP[ord(pat[i])] += 1
        countTW[ord(txt[i])] += 1

    # Traverse through remaining
    # characters of pattern
    for i in range(M, N):

        # Compare counts of current
        # window of text with
        # counts of pattern[]
        if compare(countP, countTW):
            return True

        # Add current character to current window
        countTW[ord(txt[i])] += 1

        # Remove the first character
        # of previous window
        countTW[ord(txt[i - M])] -= 1

    # Check for the last window in text
    if compare(countP, countTW):
        return True

    return False

# Function to return the number of sub-strings of s1
# that are anagrams of any sub-string of s2
def calculateSubString(s1, s2, n):

    # initializing variables
    count, j, x = 0, 0, 0

    # outer loop for picking starting point
    for i in range(n):

        # loop for different length of substrings
        for length in range(1, n - i + 1):

            # If s2 has any substring which is
            # anagram of s1.substr(i, len)
            if search(s1[i:i + length], s2):

                # increment the count
                count += 1

    return count

# Driver Code
if __name__ == "__main__":
    str1 = "PLEASEHELPIMTRAPPED"
    str2 = "INAKICKSTARTFACTORY"

    length = len(str1)

    print(calculateSubString(str1, str2, length))

# This code is contributed by
# sanjeev2552

C#

// C# program to find the number of sub-Strings
// of s1 which are anagram of any sub-String of s2
using System;
using System.Collections.Generic;

class GFG {

    static int MAX_LEN = 1005;
    static int MAX_CHAR = 26;

    static int ALL_CHARS = 256;

    // This function returns true if
    // contents of arr1[] and arr2[]
    // are same, otherwise false.
    static bool compare(char[] arr1, char[] arr2)
    {
        for (int i = 0; i < ALL_CHARS; i++)
            if (arr1[i] != arr2[i])
                return false;

        return true;
    }

    // This function search for all permutations
    // of String pat[] in String txt[]
    static bool search(String pat, String txt)
    {
        int M = pat.Length;
        int N = txt.Length;

        int i;

        // countP[]: Store count of all characters
        // of pattern
        // countTW[]: Store count of current
        // window of text
        char[] countP = new char[ALL_CHARS];
        char[] countTW = new char[ALL_CHARS];
        for (i = 0; i < M; i++) {
            (countP[pat[i]])++;
            (countTW[txt[i]])++;
        }

        // Traverse through remaining
        // characters of pattern
        for (i = M; i < N; i++) {

            // Compare counts of current
            // window of text with
            // counts of pattern[]
            if (compare(countP, countTW)) {
                // cout<<pat<<" "<<txt<<" ";
                return true;
            }

            // Add current character to current window
            (countTW[txt[i]])++;

            // Remove the first character
            // of previous window
            countTW[txt[i - M]]--;
        }

        // Check for the last window in text
        if (compare(countP, countTW))
            return true;

        return false;
    }

    // Function to return the number of sub-Strings of s1
    // that are anagrams of any sub-String of s2
    static int calculatesubString(String s1, String s2, int n)
    {
        // initializing variables
        int count = 0, j = 0, x = 0;

        // outer loop for picking starting point
        for (int i = 0; i < n; i++) {

            // loop for different length of subStrings
            for (int len = 1; len <= n - i; len++) {

                // If s2 has any subString which is
                // anagram of s1.substr(i, len)
                if (search(s1.Substring(i, len), s2)) {

                    // increment the count
                    count = count + 1;
                }
            }
        }
        return count;
    }

    // Driver code
    public static void Main(String[] args)
    {
        String str1 = "PLEASEHELPIMTRAPPED";
        String str2 = "INAKICKSTARTFACTORY";

        int len = str1.Length;

        Console.WriteLine(calculatesubString(str1, str2, len));
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>

    // JavaScript program to find the number of sub-Strings
    // of s1 which are anagram of any sub-String of s2

    let MAX_LEN = 1005;
    let MAX_CHAR = 26;

    let ALL_CHARS = 256;

    // This function returns true if
    // contents of arr1[] and arr2[]
    // are same, otherwise false.
    function compare(arr1, arr2)
    {
        for (let i = 0; i < ALL_CHARS; i++)
            if (arr1[i] != arr2[i])
                return false;

        return true;
    }

    // This function search for all permutations
    // of String pat[] in String txt[]
    function search(pat, txt)
    {
        let M = pat.length;
        let N = txt.length;

        let i;

        // countP[]: Store count of all characters
        // of pattern
        // countTW[]: Store count of current
        // window of text
        let countP = new Array(ALL_CHARS);
        countP.fill(0);
        let countTW = new Array(ALL_CHARS);
        countTW.fill(0);
        for (i = 0; i < M; i++) {
            countP[pat[i].charCodeAt()]++;
            countTW[txt[i].charCodeAt()]++;
        }

        // Traverse through remaining
        // characters of pattern
        for (i = M; i < N; i++) {

            // Compare counts of current
            // window of text with
            // counts of pattern[]
            if (compare(countP, countTW)) {
                // cout<<pat<<" "<<txt<<" ";
                return true;
            }

            // Add current character to current window
            countTW[txt[i].charCodeAt()]++;

            // Remove the first character
            // of previous window
            countTW[txt[i - M].charCodeAt()]--;
        }

        // Check for the last window in text
        if (compare(countP, countTW))
            return true;

        return false;
    }

    // Function to return the number of sub-Strings of s1
    // that are anagrams of any sub-String of s2
    function calculatesubString(s1, s2, n)
    {
        // initializing variables
        let count = 0, j = 0, x = 0;

        // outer loop for picking starting point
        for (let i = 0; i < n; i++) {

            // loop for different length of subStrings
            for (let len = 1; len <= n - i; len++) {

                // If s2 has any subString which is
                // anagram of s1.substr(i, len)
                if (search(s1.substring(i, i + len), s2)) {

                    // increment the count
                    count = count + 1;
                }
            }
        }
        return count;
    }

    let str1 = "PLEASEHELPIMTRAPPED";
    let str2 = "INAKICKSTARTFACTORY";

    let len = str1.length;

    document.write(calculatesubString(str1, str2, len));

</script>

Output

9

Complexity Analysis:

• Time Complexity:  O(N³)
• Space Complexity:  O(M) where M is the maximum length of characters ALL_CHARS.