Check if the given string is shuffled substring of another string - GeeksforGeeks

20 Jan, 2022
Given strings str1 and str2. The task is to find if str1 is a substring in the shuffled form of str2 or not. Print “YES” if str1 is a substring in shuffled form of str2 else print “NO”. 
Example 
Input: str1 = “onetwofour”, str2 = “hellofourtwooneworld” Output: YES Explanation: str1 is substring in shuffled form of str2 as str2 = “hello” + “fourtwoone” + “world” str2 = “hello” + str1 + “world”, where str1 = “fourtwoone” (shuffled form) Hence, str1 is a substring of str2 in shuffled form.
Input: str1 = “roseyellow”, str2 = “yellow” Output: NO Explanation: As the length of str1 is greater than str2. Hence, str1 is not a substring of str2.
Approach: Let n = length of str1, m = length of str2. 
If n > m, then string str1 can never be the substring of str2.
Else sort the string str1.
Traverse string str2 Put all the characters of str2 of length n in another string str.Sort the string str and Compare str and str1.If str = str1, then string str1 is a shuffled substring of string str2.else repeat the above process till ith index of str2 such that (i +n – 1 > m)(as after this index the length of remaining string str2 will be less than str1.If str is not equals to str1 in above steps, then string str1 can never be substring of str2.
Put all the characters of str2 of length n in another string str.Sort the string str and Compare str and str1.If str = str1, then string str1 is a shuffled substring of string str2.else repeat the above process till ith index of str2 such that (i +n – 1 > m)(as after this index the length of remaining string str2 will be less than str1.If str is not equals to str1 in above steps, then string str1 can never be substring of str2.
Put all the characters of str2 of length n in another string str.
Sort the string str and Compare str and str1.
If str = str1, then string str1 is a shuffled substring of string str2.
else repeat the above process till ith index of str2 such that (i +n – 1 > m)(as after this index the length of remaining string str2 will be less than str1.
If str is not equals to str1 in above steps, then string str1 can never be substring of str2.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ program to check if string// str1 is substring of str2 or not.#include <bits/stdc++.h>using namespace std; // Function two check string A// is shuffled  substring of B// or notbool isShuffledSubstring(string A, string B){    int n = A.length();    int m = B.length();     // Return false if length of    // string A is greater than    // length of string B    if (n > m) {        return false;    }    else {         // Sort string A        sort(A.begin(), A.end());         // Traverse string B        for (int i = 0; i < m; i++) {             // Return false if (i+n-1 >= m)            // doesn't satisfy            if (i + n - 1 >= m)                return false;             // Initialise the new string            string str = "";             // Copy the characters of            // string B in str till            // length n            for (int j = 0; j < n; j++)                str.push_back(B[i + j]);             // Sort the string str            sort(str.begin(), str.end());             // Return true if sorted            // string of "str" & sorted            // string of "A" are equal            if (str == A)                return true;        }    }} // Driver Codeint main(){    // Input str1 and str2    string str1 = "geekforgeeks";    string str2 = "ekegorfkeegsgeek";     // Function return true if    // str1 is shuffled substring    // of str2    bool a = isShuffledSubstring(str1, str2);     // If str1 is substring of str2    // print "YES" else print "NO"    if (a)        cout << "YES";    else        cout << "NO";    cout << endl;    return 0;}
// Java program to check if String// str1 is subString of str2 or not.import java.util.*; class GFG{ // Function two check String A// is shuffled subString of B// or notstatic boolean isShuffledSubString(String A, String B){    int n = A.length();    int m = B.length();     // Return false if length of    // String A is greater than    // length of String B    if (n > m)    {        return false;    }    else    {         // Sort String A        A = sort(A);         // Traverse String B        for (int i = 0; i < m; i++)        {             // Return false if (i + n - 1 >= m)            // doesn't satisfy            if (i + n - 1 >= m)                return false;             // Initialise the new String            String str = "";             // Copy the characters of            // String B in str till            // length n            for (int j = 0; j < n; j++)                str += B.charAt(i + j);             // Sort the String str            str = sort(str);             // Return true if sorted            // String of "str" & sorted            // String of "A" are equal            if (str.equals(A))                return true;        }    }    return false;} // Method to sort a string alphabeticallystatic String sort(String inputString){    // convert input string to char array    char tempArray[] = inputString.toCharArray();         // sort tempArray    Arrays.sort(tempArray);         // return new sorted string    return String.valueOf(tempArray);} // Driver Codepublic static void main(String[] args){    // Input str1 and str2    String str1 = "geekforgeeks";    String str2 = "ekegorfkeegsgeek";     // Function return true if    // str1 is shuffled subString    // of str2    boolean a = isShuffledSubString(str1, str2);     // If str1 is subString of str2    // print "YES" else print "NO"    if (a)        System.out.print("YES");    else        System.out.print("NO");    System.out.println();}} // This code is contributed by PrinciRaj1992
# Python3 program to check if string# str1 is subof str2 or not. # Function two check A# is shuffled subof B# or notdef isShuffledSubstring(A, B):    n = len(A)    m = len(B)     # Return false if length of    # A is greater than    # length of B    if (n > m):        return False    else:         # Sort A        A = sorted(A)         # Traverse B        for i in range(m):             # Return false if (i+n-1 >= m)            # doesn't satisfy            if (i + n - 1 >= m):                return False             # Initialise the new string            Str = ""             # Copy the characters of            # B in str till            # length n            for j in range(n):                Str += (B[i + j])             # Sort the str            Str = sorted(Str)             # Return true if sorted            # of "str" & sorted            # of "A" are equal            if (Str == A):                return True # Driver Codeif __name__ == '__main__':         # Input str1 and str2    Str1 = "geekforgeeks"    Str2 = "ekegorfkeegsgeek"     # Function return true if    # str1 is shuffled substring    # of str2    a = isShuffledSubstring(Str1, Str2)     # If str1 is subof str2    # print "YES" else print "NO"    if (a):        print("YES")    else:        print("NO") # This code is contributed by mohit kumar 29
// C# program to check if String// str1 is subString of str2 or not.using System; public class GFG{  // Function two check String A// is shuffled subString of B// or notstatic bool isShuffledSubString(String A, String B){    int n = A.Length;    int m = B.Length;      // Return false if length of    // String A is greater than    // length of String B    if (n > m)    {        return false;    }    else    {          // Sort String A        A = sort(A);          // Traverse String B        for (int i = 0; i < m; i++)        {              // Return false if (i + n - 1 >= m)            // doesn't satisfy            if (i + n - 1 >= m)                return false;              // Initialise the new String            String str = "";              // Copy the characters of            // String B in str till            // length n            for (int j = 0; j < n; j++)                str += B[i + j];              // Sort the String str            str = sort(str);              // Return true if sorted            // String of "str" & sorted            // String of "A" are equal            if (str.Equals(A))                return true;        }    }    return false;}  // Method to sort a string alphabeticallystatic String sort(String inputString){    // convert input string to char array    char []tempArray = inputString.ToCharArray();          // sort tempArray    Array.Sort(tempArray);          // return new sorted string    return String.Join("",tempArray);}  // Driver Codepublic static void Main(String[] args){    // Input str1 and str2    String str1 = "geekforgeeks";    String str2 = "ekegorfkeegsgeek";      // Function return true if    // str1 is shuffled subString    // of str2    bool a = isShuffledSubString(str1, str2);      // If str1 is subString of str2    // print "YES" else print "NO"    if (a)        Console.Write("YES");    else        Console.Write("NO");    Console.WriteLine();}} // This code is contributed by PrinciRaj1992
<script>  // Javascript program to check if string// str1 is substring of str2 or not. // Function two check string A// is shuffled  substring of B// or notfunction isShuffledSubstring(A, B){    var n = A.length;    var m = B.length;     // Return false if length of    // string A is greater than    // length of string B    if (n > m) {        return false;    }    else {         // Sort string A        A = A.split('').sort().join('');         // Traverse string B        for (var i = 0; i < m; i++) {             // Return false if (i+n-1 >= m)            // doesn't satisfy            if (i + n - 1 >= m)                return false;             // Initialise the new string            var str = [];             // Copy the characters of            // string B in str till            // length n            for (var j = 0; j < n; j++)                str.push(B[i + j]);             // Sort the string str            str = str.sort()             // Return true if sorted            // string of "str" & sorted            // string of "A" are equal            if (str.join('') == A)                return true;        }    }} // Driver Code// Input str1 and str2var str1 = "geekforgeeks";var str2 = "ekegorfkeegsgeek";// Function return true if// str1 is shuffled substring// of str2var a = isShuffledSubstring(str1, str2);// If str1 is substring of str2// print "YES" else print "NO"if (a)    document.write( "YES");else    document.write( "NO");document.write("<br>"); </script>
YES
 
Time Complexity: O(m*n*log(n)), where n = length of string str1 and m = length of string str2 Auxiliary Space: O(n) 
Efficient Solution: This problem is a simpler version of Anagram Search. It can be solved in linear time using character frequency counting.We can achieve O(n) time complexity under the assumption that alphabet size is fixed which is typically true as we have maximum of 256 possible characters in ASCII. The idea is to use two count arrays:
1) The first count array stores frequencies of characters in a pattern. 2) The second count array stores frequencies of characters in the current window of text.The important thing to note is, time complexity to compare two counted arrays is O(1) as the number of elements in them is fixed (independent of pattern and text sizes). The following are steps of this algorithm. 1) Store counts of frequencies of pattern in first count array countP[]. Also, store counts of frequencies of characters in the first window of text in array countTW[].2) Now run a loop from i = M to N-1. Do following in loop. .....a) If the two count arrays are identical, we found an occurrence. .....b) Increment count of current character of text in countTW[] .....c) Decrement count of the first character in the previous window in countWT[]3) The last window is not checked by the above loop, so explicitly check it.
The following is the implementation of the above algorithm.
C++
Java
Python3
C#
Javascript
#include<iostream>#include<cstring>#define MAX 256using namespace std; // 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<MAX; i++)        if (arr1[i] != arr2[i])            return false;    return true;} // This function search for all permutations of pat[] in txt[]bool search(char *pat, char *txt){    int M = strlen(pat), N = strlen(txt);     // countP[]: Store count of all characters of pattern    // countTW[]: Store count of current window of text    int countP[MAX] = {0}, countTW[MAX] = {0};    for (int i = 0; i < M; i++)    {        countP[pat[i]]++;        countTW[txt[i]]++;    }     // Traverse through remaining characters of pattern    for (int i = M; i < N; i++)    {        // Compare counts of current window of text with        // counts of pattern[]        if (compare(countP, countTW))           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;} /* Driver program to test above function */int main(){    char txt[] = "BACDGABCDA";    char pat[] = "ABCD";    if (search(pat, txt))       cout << "Yes";    else       cout << "No";    return 0;}
import java.util.*; class GFG{ // This function returns true if// contents of arr1[] and arr2[]// are same, otherwise false.static boolean compare(int []arr1, int []arr2){    for(int i = 0; i < 256; i++)        if (arr1[i] != arr2[i])            return false;                 return true;} // This function search for all// permutations of pat[] in txt[]static boolean search(String pat, String txt){    int M = pat.length();    int N = txt.length();         // countP[]: Store count of all    // characters of pattern    // countTW[]: Store count of    // current window of text    int []countP = new int [256];    int []countTW = new int [256];    for(int i = 0; i < 256; i++)    {        countP[i] = 0;        countTW[i] = 0;    }             for(int i = 0; i < M; i++)    {        (countP[pat.charAt(i)])++;        (countTW[txt.charAt(i)])++;    }     // Traverse through remaining    // characters of pattern    for(int i = M; i < N; i++)    {                 // Compare counts of current        // window of text with        // counts of pattern[]        if (compare(countP, countTW))            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;} // Driver codepublic static void main(String[] args){    String txt = "BACDGABCDA";    String pat = "ABCD";         if (search(pat, txt))        System.out.println("Yes");    else        System.out.println("NO");}} // This code is contributed by Stream_Cipher
MAX = 256 # This function returns true if contents# of arr1[] and arr2[] are same,# otherwise false.def compare(arr1, arr2):         global MAX     for i in range(MAX):        if (arr1[i] != arr2[i]):            return False                 return True # This function search for all permutations# of pat[] in 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 for i in range(MAX)]    countTW = [0 for i in range(MAX)]     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 # Driver codetxt = "BACDGABCDA"pat = "ABCD" if (search(pat, txt)):    print("Yes")else:    print("No") # This code is contributed by avanitrachhadiya2155
using System.Collections.Generic;using System; class GFG{ // This function returns true if// contents of arr1[] and arr2[]// are same, otherwise false.static bool compare(int []arr1, int []arr2){    for(int i = 0; i < 256; i++)        if (arr1[i] != arr2[i])            return false;                 return true;} // This function search for all// permutations of pat[] in txt[]static bool search(String pat, String txt){    int M = pat.Length;    int N = txt.Length;     // countP[]: Store count of all    // characters of pattern    // countTW[]: Store count of    // current window of text    int []countP = new int [256];    int []countTW = new int [256];         for(int i = 0; i < 256; i++)    {        countP[i] = 0;        countTW[i] = 0;    }             for(int i = 0; i < M; i++)    {        (countP[pat[i]])++;        (countTW[txt[i]])++;    }     // Traverse through remaining    // characters of pattern    for(int i = M; i < N; i++)    {                 // Compare counts of current        // window of text with        // counts of pattern[]        if (compare(countP, countTW))            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;} // Driver codepublic static void Main(){    string txt = "BACDGABCDA";    string pat = "ABCD";         if (search(pat, txt))        Console.WriteLine("Yes");    else        Console.WriteLine("NO");}} // This code is contributed by Stream_Cipher
<script> // This function returns true if// contents of arr1[] and arr2[]// are same, otherwise false.function compare(arr1,arr2){    for(let i = 0; i < 256; i++)        if (arr1[i] != arr2[i])            return false;                  return true;} // This function search for all// permutations of pat[] in txt[]function search(pat,txt){    let M = pat.length;    let N = txt.length;          // countP[]: Store count of all    // characters of pattern    // countTW[]: Store count of    // current window of text    let countP = new Array(256);    let countTW = new Array(256);    for(let i = 0; i < 256; i++)    {        countP[i] = 0;        countTW[i] = 0;    }    for(let i = 0; i < 256; i++)    {        countP[i] = 0;        countTW[i] = 0;    }              for(let i = 0; i < M; i++)    {        (countP[pat[i].charCodeAt(0)])++;        (countTW[txt[i].charCodeAt(0)])++;    }      // Traverse through remaining    // characters of pattern    for(let i = M; i < N; i++)    {                  // Compare counts of current        // window of text with        // counts of pattern[]        if (compare(countP, countTW))            return true;          // Add current character to        // current window        (countTW[txt[i].charCodeAt(0)])++;          // Remove the first character        // of previous window        countTW[txt[i - M].charCodeAt(0)]--;    }      // Check for the last window in text    if (compare(countP, countTW))        return true;        return false;} // Driver codelet txt = "BACDGABCDA";let pat = "ABCD"; if (search(pat, txt))    document.write("Yes");else    document.write("NO");  // This code is contributed by ab2127</script>
Yes
 
princiraj1992
mohit kumar 29
hawkanubhav
Stream_Cipher
avanitrachhadiya2155
noob2000
anikaseth98
ab2127
amansidd17861
abhishek2x
substring
Competitive Programming
Pattern Searching
Sorting
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