Check if a string is a scrambled form of another string
Given two strings S1 and S2 of equal length, the task is to determine if S2 is a scrambled form of S1.
Scrambled string:
Given string str, we can represent it as a binary tree by partitioning it into two non-empty substrings recursively.
Note: Scrambled string is not same as an Anagram
Below is one possible representation of str = “coder”:
coder / \ co der / \ / \ c o d er / \ e r
To scramble the string, we may choose any non-leaf node and swap its two children.
Suppose, we choose the node “co” and swap its two children, it produces a scrambled string “ocder”.
ocder / \ oc der / \ / \ o c d er / \ e r
Thus, “ocder” is a scrambled string of “coder”.
Similarly, if we continue to swap the children of nodes “der” and “er”, it produces a scrambled string “ocred”.
ocred / \ oc red / \ / \ o c re d / \ r e
Thus, “ocred” is a scrambled string of “coder”.
Examples:
Input: S1=”coder”, S2=”ocder”
Output: Yes
Explanation:
“ocder” is a scrambled form of “coder”
Input: S1=”abcde”, S2=”caebd”
Output: No
Explanation:
“caebd” is not a scrambled form of “abcde”
Approach
In order to solve this problem, we are using Divide and Conquer approach.
Given two strings of equal length (say n+1), S1[0…n] and S2[0…n]. If S2 is a scrambled form of S1, then there must exist an index i such that at least one of the following conditions is true:
- S2[0…i] is a scrambled string of S1[0…i] and S2[i+1…n] is a scrambled string of S1[i+1…n].
- S2[0…i] is a scrambled string of S1[n-i…n] and S2[i+1…n] is a scrambled string of S1[0…n-i-1].
Note: An optimization step to consider here is to check beforehand if the two strings are anagrams of each other. If not, it indicates that the strings contain different characters and can’t be a scrambled form of each other.
Below is the implementation of the above approach:
C++
// C++ Program to check if a // given string is a scrambled // form of another string #include <bits/stdc++.h> using namespace std; bool isScramble(string S1, string S2) { // Strings of non-equal length // cant' be scramble strings if (S1.length() != S2.length()) { return false ; } int n = S1.length(); // Empty strings are scramble strings if (n == 0) { return true ; } // Equal strings are scramble strings if (S1 == S2) { return true ; } // Check for the condition of anagram string copy_S1 = S1, copy_S2 = S2; sort(copy_S1.begin(), copy_S1.end()); sort(copy_S2.begin(), copy_S2.end()); if (copy_S1 != copy_S2) { return false ; } for ( int i = 1; i < n; i++) { // Check if S2[0...i] is a scrambled // string of S1[0...i] and if S2[i+1...n] // is a scrambled string of S1[i+1...n] if (isScramble(S1.substr(0, i), S2.substr(0, i)) && isScramble(S1.substr(i, n - i), S2.substr(i, n - i))) { return true ; } // Check if S2[0...i] is a scrambled // string of S1[n-i...n] and S2[i+1...n] // is a scramble string of S1[0...n-i-1] if (isScramble(S1.substr(0, i), S2.substr(n - i, i)) && isScramble(S1.substr(i, n - i), S2.substr(0, n - i))) { return true ; } } // If none of the above // conditions are satisfied return false ; } // Driver Code int main() { string S1 = "coder" ; string S2 = "ocred" ; if (isScramble(S1, S2)) { cout << "Yes" ; } else { cout << "No" ; } return 0; } |
Java
// Java program to check if a // given string is a scrambled // form of another string import java.util.*; class GFG{ static boolean isScramble(String S1, String S2) { // Strings of non-equal length // cant' be scramble strings if (S1.length() != S2.length()) { return false ; } int n = S1.length(); // Empty strings are scramble strings if (n == 0 ) { return true ; } // Equal strings are scramble strings if (S1.equals(S2)) { return true ; } // Converting string to // character array char [] tempArray1 = S1.toCharArray(); char [] tempArray2 = S2.toCharArray(); // Checking condition for Anagram Arrays.sort(tempArray1); Arrays.sort(tempArray2); String copy_S1 = new String(tempArray1); String copy_S2 = new String(tempArray2); if (!copy_S1.equals(copy_S2)) { return false ; } for ( int i = 1 ; i < n; i++) { // Check if S2[0...i] is a scrambled // string of S1[0...i] and if S2[i+1...n] // is a scrambled string of S1[i+1...n] if (isScramble(S1.substring( 0 , i), S2.substring( 0 , i)) && isScramble(S1.substring(i, n), S2.substring(i, n))) { return true ; } // Check if S2[0...i] is a scrambled // string of S1[n-i...n] and S2[i+1...n] // is a scramble string of S1[0...n-i-1] if (isScramble(S1.substring(n - i, n), S2.substring( 0 , i)) && isScramble(S1.substring( 0 , n - i), S2.substring(i, n))) { return true ; } } // If none of the above // conditions are satisfied return false ; } // Driver Code public static void main(String[] args) { String S1 = "coder" ; String S2 = "ocred" ; if (isScramble(S1, S2)) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } } // This code is contributed by dadi madhav |
Python3
# Python3 program to check if a # given string is a scrambled # form of another string def isScramble(S1: str , S2: str ): # Strings of non-equal length # cant' be scramble strings if len (S1) ! = len (S2): return False n = len (S1) # Empty strings are scramble strings if not n: return True # Equal strings are scramble strings if S1 = = S2: return True # Check for the condition of anagram if sorted (S1) ! = sorted (S2): return False for i in range ( 1 , n): # Check if S2[0...i] is a scrambled # string of S1[0...i] and if S2[i+1...n] # is a scrambled string of S1[i+1...n] if (isScramble(S1[:i], S2[:i]) and isScramble(S1[i:], S2[i:])): return True # Check if S2[0...i] is a scrambled # string of S1[n-i...n] and S2[i+1...n] # is a scramble string of S1[0...n-i-1] if (isScramble(S1[ - i:], S2[:i]) and isScramble(S1[: - i], S2[i:])): return True # If none of the above # conditions are satisfied return False # Driver Code if __name__ = = "__main__" : S1 = "coder" S2 = "ocred" if (isScramble(S1, S2)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by sgshah2 |
C#
// C# program to check if a // given string is a scrambled // form of another string using System; using System.Collections.Generic; class GFG { static bool isScramble( string S1, string S2) { // Strings of non-equal length // cant' be scramble strings if (S1.Length != S2.Length) { return false ; } int n = S1.Length; // Empty strings are scramble strings if (n == 0) { return true ; } // Equal strings are scramble strings if (S1.Equals(S2)) { return true ; } // Converting string to // character array char [] tempArray1 = S1.ToCharArray(); char [] tempArray2 = S2.ToCharArray(); // Checking condition for Anagram Array.Sort(tempArray1); Array.Sort(tempArray2); string copy_S1 = new string (tempArray1); string copy_S2 = new string (tempArray2); if (!copy_S1.Equals(copy_S2)) { return false ; } for ( int i = 1; i < n; i++) { // Check if S2[0...i] is a scrambled // string of S1[0...i] and if S2[i+1...n] // is a scrambled string of S1[i+1...n] if (isScramble(S1.Substring(0, i), S2.Substring(0, i)) && isScramble(S1.Substring(i, n - i), S2.Substring(i, n - i))) { return true ; } // Check if S2[0...i] is a scrambled // string of S1[n-i...n] and S2[i+1...n] // is a scramble string of S1[0...n-i-1] if (isScramble(S1.Substring(0, i), S2.Substring(n - i, i)) && isScramble(S1.Substring(i, n - i), S2.Substring(0, n - i))) { return true ; } } // If none of the above // conditions are satisfied return false ; } // Driver code static void Main() { string S1 = "coder" ; string S2 = "ocred" ; if (isScramble(S1, S2)) { Console.WriteLine( "Yes" ); } else { Console.WriteLine( "No" ); } } } // This code is contributed by divyeshrabadiya07 |
Javascript
<script> // Javascript program to check if a // given string is a scrambled // form of another string function isScramble(S1, S2) { // Strings of non-equal length // can't be scramble strings if (S1.length != S2.length) { return false ; } let n = S1.length; // Empty strings are scramble strings if (n == 0) { return true ; } // Equal strings are scramble strings if (S1 == S2) { return true ; } // Converting string to // character array let tempArray1 = S1.split(' '); let tempArray2 = S2.split(' '); // Checking condition for Anagram tempArray1.sort(); tempArray2.sort(); let copy_S1 = tempArray1.join( "" ); let copy_S2 = tempArray2.join( "" ); if (copy_S1 != copy_S2) { return false ; } for (let i = 1; i < n; i++) { // Check if S2[0...i] is a scrambled // string of S1[0...i] and if S2[i+1...n] // is a scrambled string of S1[i+1...n] if (isScramble(S1.substring(0, i), S2.substring(0, i)) && isScramble(S1.substring(i, i + n), S2.substring(i, i + n))) { return true ; } // Check if S2[0...i] is a scrambled // string of S1[n-i...n] and S2[i+1...n] // is a scramble string of S1[0...n-i-1] if (isScramble(S1.substring(n - i, n - i + n), S2.substring(0, i)) && isScramble(S1.substring(0, n - i), S2.substring(i, i + n))) { return true ; } } // If none of the above // conditions are satisfied return false ; } let S1 = "coder" ; let S2 = "ocred" ; if (isScramble(S1, S2)) { document.write( "Yes" ); } else { document.write( "No" ); } // This code is contributed by decode2207. </script> |
Yes
Time Complexity: O(2^k + 2^(n-k)), where k and n-k are the length of the two substrings.
Auxiliary Space: O(2^N), recursion stack.
Dynamic Programming Solution: The above recursive Code can be optimized by storing Boolean values of substrings in an unordered map, so if the same substrings have to be checked again we can easily just get value from the map instead of performing function calls.
Memoized Code:
C++
// C++ Program to check if a // given string is a scrambled // form of another string #include <bits/stdc++.h> using namespace std; // map declaration for storing key value pair // means for storing subproblem result unordered_map<string, bool > mp; bool isScramble(string S1, string S2) { // Strings of non-equal length // cant' be scramble strings if (S1.length() != S2.length()) { return false ; } int n = S1.length(); // Empty strings are scramble strings if (n == 0) { return true ; } // Equal strings are scramble strings if (S1 == S2) { return true ; } // Check for the condition of anagram string copy_S1 = S1, copy_S2 = S2; sort(copy_S1.begin(), copy_S1.end()); sort(copy_S2.begin(), copy_S2.end()); if (copy_S1 != copy_S2) { return false ; } // make key of type string for search in map string key = (S1 + " " + S2); // checking if both string are before calculated or not // if calculated means find in map then return it's // value if (mp.find(key) != mp.end()) { return mp[key]; } // declaring flag variable to store result bool flag = false ; for ( int i = 1; i < n; i++) { // Check if S2[0...i] is a scrambled // string of S1[0...i] and if S2[i+1...n] // is a scrambled string of S1[i+1...n] if (isScramble(S1.substr(0, i), S2.substr(0, i)) && isScramble(S1.substr(i, n - i), S2.substr(i, n - i))) { flag = true ; return true ; } // Check if S2[0...i] is a scrambled // string of S1[n-i...n] and S2[i+1...n] // is a scramble string of S1[0...n-i-1] if (isScramble(S1.substr(0, i), S2.substr(n - i, i)) && isScramble(S1.substr(i, n - i), S2.substr(0, n - i))) { flag = true ; return true ; } } // add key & flag value to map (store for future use) // so next time no required to calculate it again mp[key] = flag; // If none of the above conditions are satisfied return false ; } // Driver Code int main() { string S1 = "coder" ; string S2 = "ocred" ; if (isScramble(S1, S2)) { cout << "Yes" ; } else { cout << "No" ; } return 0; } |
Java
// Java Program to check if a // given string is a scrambled // form of another string import java.util.*; public class Main { // map declaration for storing key value pair // means for storing subproblem result static HashMap<String, Boolean> mp = new HashMap<String, Boolean>(); static boolean isScramble(String S1, String S2) { // Strings of non-equal length // cant' be scramble strings if (S1.length() != S2.length()) { return false ; } int n = S1.length(); // Empty strings are scramble strings if (n == 0 ) { return true ; } // Equal strings are scramble strings if (S1.equals(S2)) { return true ; } // Check for the condition of anagram String copy_S1 = S1, copy_S2 = S2; char [] t1 = copy_S1.toCharArray(); char [] t2 = copy_S2.toCharArray(); Arrays.sort(t1); Arrays.sort(t2); copy_S1 = new String(t1); copy_S2 = new String(t2); if (!copy_S1.equals(copy_S2)) { return false ; } // make key of type string for search in map String key = (S1 + " " + S2); // checking if both string are before calculated or not // if calculated means find in map then return it's // value if (mp.containsKey(key)) { return mp.get(key); } // declaring flag variable to store result boolean flag = false ; for ( int i = 1 ; i < n; i++) { // Check if S2[0...i] is a scrambled // string of S1[0...i] and if S2[i+1...n] // is a scrambled string of S1[i+1...n] if (isScramble(S1.substring( 0 , i), S2.substring( 0 , i)) && isScramble(S1.substring(i, n), S2.substring(i, n))) { flag = true ; mp.put(key, flag); //store for future use return true ; } // Check if S2[0...i] is a scrambled // string of S1[n-i...n] and S2[i+1...n] // is a scramble string of S1[0...n-i-1] if (isScramble(S1.substring( 0 , i), S2.substring(n - i, n)) && isScramble(S1.substring(i, n), S2.substring( 0 , n - i))) { flag = true ; mp.put(key, flag); //store for future use return true ; } } // add key & flag value to map (store for future use) // so next time no required to calculate it again mp.put(key, flag); // If none of the above conditions are satisfied return false ; } public static void main(String[] args) { String S1 = "coder" ; String S2 = "ocred" ; if (isScramble(S1, S2)) { System.out.print( "Yes" ); } else { System.out.print( "No" ); } } } // This code is contributed by divyesh072019. |
Python3
# Declaring unordered map globally map = {} # Python3 program to check if a # given string is a scrambled # form of another string def isScramble(S1: str , S2: str ): # Strings of non-equal length # cant' be scramble strings if len (S1) ! = len (S2): return False n = len (S1) # Empty strings are scramble strings if not n: return True # Equal strings are scramble strings if S1 = = S2: return True # Check for the condition of anagram if sorted (S1) ! = sorted (S2): return False # Checking if both Substrings are in # map or are already calculated or not if (S1 + ' ' + S2 in map ): return map [S1 + ' ' + S2] # Declaring a flag variable flag = False for i in range ( 1 , n): # Check if S2[0...i] is a scrambled # string of S1[0...i] and if S2[i+1...n] # is a scrambled string of S1[i+1...n] if (isScramble(S1[:i], S2[:i]) and isScramble(S1[i:], S2[i:])): flag = True return True # Check if S2[0...i] is a scrambled # string of S1[n-i...n] and S2[i+1...n] # is a scramble string of S1[0...n-i-1] if (isScramble(S1[ - i:], S2[:i]) and isScramble(S1[: - i], S2[i:])): flag = True return True # Storing calculated value to map map [S1 + " " + S2] = flag # If none of the above # conditions are satisfied return False # Driver Code if __name__ = = "__main__" : S1 = "great" S2 = "rgate" if (isScramble(S1, S2)): print ( "Yes" ) else : print ( "No" ) |
C#
// C# Program to check if a // given string is a scrambled // form of another string using System; using System.Collections.Generic; class GFG { // map declaration for storing key value pair // means for storing subproblem result static Dictionary< string , bool > mp = new Dictionary< string , bool >(); static bool isScramble( string S1, string S2) { // Strings of non-equal length // cant' be scramble strings if (S1.Length != S2.Length) { return false ; } int n = S1.Length; // Empty strings are scramble strings if (n == 0) { return true ; } // Equal strings are scramble strings if (S1 == S2) { return true ; } // Check for the condition of anagram string copy_S1 = S1, copy_S2 = S2; char [] t1 = copy_S1.ToCharArray(); char [] t2 = copy_S2.ToCharArray(); Array.Sort(t1); Array.Sort(t2); copy_S1 = new string (t1); copy_S2 = new string (t2); if (copy_S1 != copy_S2) { return false ; } // make key of type string for search in map string key = (S1 + " " + S2); // checking if both string are before calculated or not // if calculated means find in map then return it's // value if (mp.ContainsKey(key)) { return mp[key]; } // declaring flag variable to store result bool flag = false ; for ( int i = 1; i < n; i++) { // Check if S2[0...i] is a scrambled // string of S1[0...i] and if S2[i+1...n] // is a scrambled string of S1[i+1...n] if (isScramble(S1.Substring(0, i), S2.Substring(0, i)) && isScramble(S1.Substring(i, n - i), S2.Substring(i, n - i))) { flag = true ; return true ; } // Check if S2[0...i] is a scrambled // string of S1[n-i...n] and S2[i+1...n] // is a scramble string of S1[0...n-i-1] if (isScramble(S1.Substring(0, i), S2.Substring(n - i, i)) && isScramble(S1.Substring(i, n - i), S2.Substring(0, n - i))) { flag = true ; return true ; } } // add key & flag value to map (store for future use) // so next time no required to calculate it again mp[key] = flag; // If none of the above conditions are satisfied return false ; } static void Main() { string S1 = "coder" ; string S2 = "ocred" ; if (isScramble(S1, S2)) { Console.Write( "Yes" ); } else { Console.Write( "No" ); } } } // This code is contributed by rameshtravel07. |
Javascript
<script> // Javascript Program to check if a // given string is a scrambled // form of another string // map declaration for storing key value pair // means for storing subproblem result let mp = new Map(); function isScramble(S1, S2) { // Strings of non-equal length // cant' be scramble strings if (S1.length != S2.length) { return false ; } let n = S1.length; // Empty strings are scramble strings if (n == 0) { return true ; } // Equal strings are scramble strings if (S1 == S2) { return true ; } // Check for the condition of anagram let copy_S1 = S1, copy_S2 = S2; let t1 = copy_S1.split(' ') let t2 = copy_S2.split(' ') t1.sort(); t2.sort(); copy_S1 = t1.join(""); copy_S2 = t2.join(""); if (copy_S1 != copy_S2) { return false; } // make key of type string for search in map let key = (S1 + " " + S2); // checking if both string are before calculated or not // if calculated means find in map then return it' s // value if (mp.has(key)) { return mp[key]; } // declaring flag variable to store result let flag = false ; for (let i = 1; i < n; i++) { // Check if S2[0...i] is a scrambled // string of S1[0...i] and if S2[i+1...n] // is a scrambled string of S1[i+1...n] if (isScramble(S1.substring(0, i), S2.substring(0, i)) && isScramble(S1.substring(i, n), S2.substring(i, n))) { flag = true ; return true ; } // Check if S2[0...i] is a scrambled // string of S1[n-i...n] and S2[i+1...n] // is a scramble string of S1[0...n-i-1] if (isScramble(S1.substring(0, i), S2.substring(n - i, n)) && isScramble(S1.substring(i, n), S2.substring(0, n - i))) { flag = true ; return true ; } } // add key & flag value to map (store for future use) // so next time no required to calculate it again mp[key] = flag; // If none of the above conditions are satisfied return false ; } let S1 = "coder" ; let S2 = "ocred" ; if (isScramble(S1, S2)) { document.write( "Yes" ); } else { document.write( "No" ); } // This code is contributed by suresh07. </script> |
Yes
Time Complexity: O(N^2), where N is the length of the given strings.
Auxiliary Space: O(N^2), As we need to store O(N^2) string in our mp map.
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