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Check if two strings after processing backspace character are equal or not
  • Difficulty Level : Easy
  • Last Updated : 17 Nov, 2020

Given two strings s1 and s2, let us assume that while typing the strings there were some backspaces encountered which are represented by #. The task is to determine whether the resultant strings after processing the backspace character would be equal or not.

Examples:  

Input: s1= geee#e#ks, s2 = gee##eeks 
Output: True 
Explanation: Both the strings after processing the backspace character
becomes "geeeeks". Hence, true.

Input: s1 = equ#ual, s2 = ee#quaal#
Output:  False
Explanation: String 1 after processing the backspace character
becomes "equal" whereas string 2 is "eequaal". Hence, false.


Approach:
To solve the problem mentioned above we have to observe that if the first character is ‘#’, that is there is certainly no character typed initially and hence we perform no operation. When we encounter any character other than ‘#’, then we add the character just after the current index. When we encounter a ‘#’, we move one index back, so instead of deleting the character, we just ignore it. Then finally compare the two strings by comparing each character from start to end.

Below is the implementation of the above approach: 

C++




/* C++ implementation to Check if
two strings after processing
backspace character are equal or not*/
 
#include <bits/stdc++.h>
using namespace std;
 
// function to compare the two strings
bool compare(string s, string t)
{
    int ps, pt, i;
 
    /* the index of character in string which
       would be removed when
       backspace is encountered*/
    ps = -1;
 
    for (i = 0; i < s.size(); i++) {
 
        /* checks if a backspace is encountered or not.
            In case the first character is #,
            no change in string takes place*/
        if (s[i] == '#' && ps != -1)
 
            ps -= 1;
 
        /* the character after the # is added
         after the character at position rem_ind1 */
        else if (s[i] != '#') {
 
            s = s[i];
            ps += 1;
        }
    }
 
    pt = -1;
 
    for (i = 0; i < t.size(); i++) {
        /* checks if a backspace is encountered or not */
        if (t[i] == '#' && pt != -1)
            pt -= 1;
 
        else if (t[i] != '#') {
 
            t[pt + 1] = t[i];
 
            pt += 1;
        }
    }
 
    /* check if the value of
        rem_ind1 and rem_ind2
        is same, if not then
        it means they have different
        length */
    if (pt != ps)
 
        return false;
 
    /* check if resultant strings are empty */
    else if (ps == -1 && pt == -1)
 
        return true;
 
    /* check if each character in the resultant
       string is same */
    else {
 
        for (i = 0; i <= pt; i++) {
 
            if (s[i] != t[i])
 
                return false;
        }
        return true;
    }
}
 
// Driver code
int main()
{
    // initialise two strings
    string s = "geee#e#ks";
    string t = "gee##eeks";
 
    if (compare(s, t))
 
        cout << "True";
    else
 
        cout << "False";
}

Java




// Java implementation to Check if 
// two strings after processing 
// backspace character are equal or not
import java.util.*;
import java.lang.*;
 
class GFG{
  
// Function to compare the two strings
static boolean compare(StringBuilder s,
                       StringBuilder t)
{
    int ps, pt, i;
   
    // The index of character in string
    // which would be removed when
    // backspace is encountered
    ps = -1;
   
    for(i = 0; i < s.length(); i++)
    {
         
        // Checks if a backspace is encountered
        // or not. In case the first character
        // is #, no change in string takes place
        if (s.charAt(i) == '#' && ps != -1)
            ps -= 1;
   
        // The character after the # is added
        // after the character at position rem_ind1
        else if (s.charAt(i) != '#')
        {
            s.setCharAt(ps + 1, s.charAt(i));
            ps += 1;
        }
    }
   
    pt = -1;
   
    for(i = 0; i < t.length(); i++)
    {
         
        // Checks if a backspace is
        // encountered or not
        if (t.charAt(i) == '#' && pt != -1)
            pt -= 1;
   
        else if (t.charAt(i) != '#')
        {
            t.setCharAt(pt + 1, t.charAt(i));
             
            pt += 1;
        }
    }
   
    // Check if the value of rem_ind1 and
    // rem_ind2 is same, if not then it
    // means they have different length
    if (pt != ps)
        return false;
   
    // Check if resultant strings are empty
    else if (ps == -1 && pt == -1)
        return true;
   
    // Check if each character in the
    // resultant string is same
    else
    {
        for(i = 0; i <= pt; i++)
        {
            if (s.charAt(i) != t.charAt(i))
                return false;
        }
        return true;
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // Initialise two strings
    StringBuilder s = new StringBuilder("geee#e#ks");
    StringBuilder t = new StringBuilder("gee##eeks");
     
    if (compare(s, t))
        System.out.print("True");
    else
        System.out.print("False");
}
}
 
// This code is contributed by offbeat
Output: 



True




 

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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