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 “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 

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 = "Geeks", 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 = "Geeks", 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 = "Geeks"
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 = "Geeks", 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 = "Geeks", s2 = "Geks";
if (solve(s1, s2))
    document.write( "Yes");
else
    document.write( "No");


// This code is contributed by rutvik_56.
</script>

Output:

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 = "Geeks", 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 = "Geeks", 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 = "Geeks"
    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 = "Geeks", 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 = "Geeks", s2 = "Geks";

if (checkSequence(s1, s2))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by mukesh07

</script>

Output

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.