Check whether two strings contain same characters in same order
Given two strings s1 and s2, the task is to find whether the two strings contain the same characters that occur in the same order. For example string “Beginner” and string “Geks” contain the same characters in same order.
Examples:
Input: s1 = “Beginner”, s2 = “Geks”
Output: YesInput: s1 = “Arnab”, s2 = “Andrew”
Output: No
Approach: We have two strings now we have to check whether the strings contain the same characters in the same order. So we will replace the contiguous similar element with a single element i.e. if we have “eee”, we will replace it with a single “e”. Now we will check that both the strings are equal or not. If equal then print Yes else No.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> using namespace std; string getString( char x) { // string class has a constructor // that allows us to specify size of // string as first parameter and character // to be filled in given size as second // parameter. string s(1, x); return s; } // Function that returns true if // the given strings contain same // characters in same order bool solve(string s1, string s2) { // Get the first character of both strings string a = getString(s1[0]), b = getString(s2[0]); // Now if there are adjacent similar character // remove that character from s1 for ( int i = 1; i < s1.length(); i++) if (s1[i] != s1[i - 1]) { a += getString(s1[i]); } // Now if there are adjacent similar character // remove that character from s2 for ( int i = 1; i < s2.length(); i++) if (s2[i] != s2[i - 1]) { b += getString(s2[i]); } // If both the strings are equal // then return true if (a == b) return true ; return false ; } // Driver code int main() { string s1 = "Beginner" , s2 = "Geks" ; if (solve(s1, s2)) cout << "Yes" ; else cout << "No" ; return 0; } |
Java
// Java implementation of the approach class temp { static String getString( char x) { // String class has a constructor // that allows us to specify size of // String as first parameter and character // to be filled in given size as second // parameter. String s = String.valueOf(x); return s; } // Function that returns true if // the given Strings contain same // characters in same order static boolean solve(String s1, String s2) { // Get the first character of both Strings String a = getString(s1.charAt( 0 )), b = getString(s2.charAt( 0 )); // Now if there are adjacent similar character // remove that character from s1 for ( int i = 1 ; i < s1.length(); i++) if (s1.charAt(i) != s1.charAt(i - 1 )) { a += getString(s1.charAt(i)); } // Now if there are adjacent similar character // remove that character from s2 for ( int i = 1 ; i < s2.length(); i++) if (s2.charAt(i) != s2.charAt(i - 1 )) { b += getString(s2.charAt(i)); } // If both the Strings are equal // then return true if (a.equals(b)) return true ; return false ; } // Driver code public static void main(String[] args) { String s1 = "Beginner" , s2 = "Geks" ; if (solve(s1, s2)) System.out.print( "Yes" ); else System.out.print( "No" ); } } // This code is contributed by ankush_953 |
Python3
# Python3 implementation of the approach def getString(x): # string class has a constructor # that allows us to specify the size of # string as first parameter and character # to be filled in given size as the second # parameter. return x # Function that returns true if # the given strings contain same # characters in same order def solve(s1, s2): # Get the first character of both strings a = getString(s1[ 0 ]) b = getString(s2[ 0 ]) # Now if there are adjacent similar character # remove that character from s1 for i in range ( 1 , len (s1)): if s1[i] ! = s1[i - 1 ]: a + = getString(s1[i]) # Now if there are adjacent similar character # remove that character from s2 for i in range ( 1 , len (s2)): if s2[i] ! = s2[i - 1 ]: b + = getString(s2[i]) # If both the strings are equal # then return true if a = = b: return True return False # Driver code s1 = "Beginner" s2 = "Geks" if solve(s1, s2): print ( "Yes" ) else : print ( "No" ) # This code is contributed by ankush_953 |
C#
// C# implementation of the approach using System; public class temp { static String getString( char x) { // String class has a constructor // that allows us to specify size of // String as first parameter and character // to be filled in given size as second // parameter. String s = String.Join( "" ,x); return s; } // Function that returns true if // the given Strings contain same // characters in same order static Boolean solve(String s1, String s2) { // Get the first character of both Strings String a = getString(s1[0]), b = getString(s2[0]); // Now if there are adjacent similar character // remove that character from s1 for ( int i = 1; i < s1.Length; i++) if (s1[i] != s1[i - 1]) { a += getString(s1[i]); } // Now if there are adjacent similar character // remove that character from s2 for ( int i = 1; i < s2.Length; i++) if (s2[i] != s2[i - 1]) { b += getString(s2[i]); } // If both the strings are equal // then return true if (a == b) return true ; return false ; } // Driver code public static void Main(String[] args) { String s1 = "Beginner" , s2 = "Geks" ; if (solve(s1, s2)) Console.Write( "Yes" ); else Console.Write( "No" ); } } // This code is contributed by Princi Singh |
Javascript
<script> // Javascript implementation of the approach function getString(x) { // string class has a constructor // that allows us to specify size of // string as first parameter and character // to be filled in given size as second // parameter. return x } // Function that returns true if // the given strings contain same // characters in same order function solve(s1, s2) { // Get the first character of both strings var a = getString(s1[0]), b = getString(s2[0]); // Now if there are adjacent similar character // remove that character from s1 for ( var i = 1; i < s1.length; i++) if (s1[i] != s1[i - 1]) { a += getString(s1[i]); } // Now if there are adjacent similar character // remove that character from s2 for ( var i = 1; i < s2.length; i++) if (s2[i] != s2[i - 1]) { b += getString(s2[i]); } // If both the strings are equal // then return true if (a == b) return true ; return false ; } // Driver code var s1 = "Beginner" , s2 = "Geks" ; if (solve(s1, s2)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by rutvik_56. </script> |
Yes
Time Complexity: O(m + n)
Auxiliary Space: O(m + n), where m and n are the length of the given strings s1 and s2 respectively.
Using Recursion
C++
#include <iostream> using namespace std; bool checkSequence(string a, string b) { // if length of the b = 0 // then we return true if (b.size() == 0) return true ; // if length of a = 0 // that means b is not present in // a so we return false if (a.size() == 0) return false ; if (a[0] == b[0]) return checkSequence(a.substr(1), b.substr(1)); else return checkSequence(a.substr(1), b); } int main() { string s1 = "Beginner" , s2 = "Geks" ; if (checkSequence(s1, s2)) cout << "Yes" ; else cout << "No" ; } // This code is contributed by SoumikMondal |
Java
/*package whatever //do not write package name here */ import java.io.*; class GFG { public static boolean checkSequence(String a, String b) { //if length of the b = 0 //then we return true if (b.length()== 0 ) return true ; //if length of a = 0 //that means b is not present in //a so we return false if (a.length() == 0 ) return false ; if (a.charAt( 0 ) == b.charAt( 0 )) return checkSequence(a.substring( 1 ), b.substring( 1 )); else return checkSequence(a.substring( 1 ), b); } public static void main(String[] args) { String s1 = "Beginner" , s2 = "Geks" ; if (checkSequence(s1, s2)) System.out.print( "Yes" ); else System.out.print( "No" ); } } |
Python
# Python3 implementation of approach def checkSequence(a, b): # if length of the b = 0 # then we return true if len (b) = = 0 : return True # if length of a = 0 # that means b is not present in # a so we return false if len (a) = = 0 : return False if (a[ 0 ] = = b[ 0 ]): return checkSequence(a[ 1 :], b[ 1 :]) else : return checkSequence(a[ 1 :], b) if __name__ = = '__main__' : s1 = "Beginner" s2 = "Geks" if (checkSequence(s1, s2)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by nirajgusain5 |
C#
// C# implementation of the approach using System; public class temp { public static bool checkSequence(String a, String b) { // if length of the b = 0 // then we return true if (b.Length == 0) return true ; // if length of a = 0 // that means b is not present in // a so we return false if (a.Length == 0) return false ; if (a[0] == b[0]) return checkSequence(a.Substring(1), b.Substring(1)); else return checkSequence(a.Substring(1), b); } // Driver code public static void Main(String[] args) { String s1 = "Beginner" , s2 = "Geks" ; if (checkSequence(s1, s2)) Console.Write( "Yes" ); else Console.Write( "No" ); } } // This code is contributed by Dharanendra L V. |
Javascript
<script> function checkSequence(a, b) { // If length of the b = 0 // then we return true if (b.length == 0) return true ; // If length of a = 0 // that means b is not present in // a so we return false if (a.length == 0) return false ; if (a[0] == b[0]) return checkSequence(a.substring(1), b.substring(1)); else return checkSequence(a.substring(1), b); } // Driver code let s1 = "Beginner" , s2 = "Geks" ; if (checkSequence(s1, s2)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by mukesh07 </script> |
Yes
Time complexity: O(max(M, N)) for strings of length M and N respectively.
Auxiliary Space: O(max(M, N)), due to recursive call stacks.