# Check whether two strings contain same characters in same order

Given two strings s1 and s2, the task is to find whether the two string contain the same characters that occur in the same order. For example: string “Geeks” and string “Geks” contain the same characters in same order.

Examples:

Input: s1 = “Geeks”, s2 = “Geks”
Output: Yes

Input: s1 = “Arnab”, s2 = “Andrew”
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of the approach #include 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 = "Geeks", s2 = "Geks";        if (solve(s1, s2))         cout << "Yes";     else         cout << "No";        return 0; }

 // 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 = "Geeks", s2 = "Geks";        if (solve(s1, s2))         System.out.print("Yes");     else         System.out.print("No"); } }    // This code is contributed by ankush_953

 # 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 = "Geeks" s2 = "Geks" if solve(s1, s2):     print("Yes") else:     print("No")    # This code is contributed by ankush_953

 // 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 = "Geeks", s2 = "Geks";        if (solve(s1, s2))         Console.Write("Yes");     else         Console.Write("No"); } }    // This code is contributed by Princi Singh

Output:
Yes

Third year Department of Information Technology Jadavpur University

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