Given a string Str, the task is to remove first adjacent pairs of similar characters until we can.
Note: Remove adjacent characters to get a new string and then again remove adjacent duplicates from the new string and keep repeating this process until all similar adjacent character pairs are removed.
Examples:
Input: str = “keexxllx”
Output: kx
Step 0: Remove ee to get “kxxllx”
Step 1: Remove xx to get “kllx”
Step 2: Remove ll to get “kx”
Input: str = “abbaca”
Output: ca
Approach:
Use string’s back() and pop_back() method STL in C++ to solve the above problem. Iterate for every character in the string, and if the adjacent characters are same, then remove the adjacent characters using pop_back() function. At the end, return the final string.
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to remove adjacent duplicates string removeDuplicates(string S) { string ans = "" ;
// Iterate for every character
// in the string
for ( auto it : S) {
// If ans string is empty or its last
// character does not match with the
// current character then append this
// character to the string
if (ans.empty() or ans.back() != it)
ans.push_back(it);
// Matches with the previous one
else if (ans.back() == it)
ans.pop_back();
}
// Return the answer
return ans;
} // Driver Code int main()
{ string str = "keexxllx" ;
cout << removeDuplicates(str);
} |
// Java implementation of the above approach class GFG
{ // Function to remove adjacent duplicates
static String removeDuplicates(String S)
{
String ans = "" ;
// Iterate for every character
// in the string
for ( int i = 0 ; i < S.length(); i++)
{
// If ans string is empty or its last
// character does not match with the
// current character then append this
// character to the string
if (ans.isEmpty() ||
ans.charAt(ans.length() - 1 ) != S.charAt(i))
ans += S.charAt(i);
// Matches with the previous one
else if (ans.charAt(ans.length() - 1 ) == S.charAt(i))
ans = ans.substring( 0 , ans.length() - 1 );
}
// Return the answer
return ans;
}
// Driver Code
public static void main(String[] args)
{
String str = "keexxllx" ;
System.out.println(removeDuplicates(str));
}
} // This code is contributed by // sanjeev2552 |
# Python3 implementation of the above approach # Function to remove adjacent duplicates def removeDuplicates(S) :
ans = "";
# Iterate for every character
# in the string
for it in S :
# If ans string is empty or its last
# character does not match with the
# current character then append this
# character to the string
if (ans = = "" or ans[ - 1 ] ! = it) :
ans + = it ;
# Matches with the previous one
elif (ans[ - 1 ] = = it) :
ans = ans[: - 1 ];
# Return the answer
return ans;
# Driver Code if __name__ = = "__main__" :
string = "keexxllx" ;
print (removeDuplicates(string));
# This code is contributed by AnkitRai01 |
// C# implementation of the above approach using System;
class GFG
{ // Function to remove adjacent duplicates
static String removeDuplicates(String S)
{
String ans = "" ;
// Iterate for every character
// in the string
for ( int i = 0; i < S.Length; i++)
{
// If ans string is empty or its last
// character does not match with the
// current character then append this
// character to the string
if (ans == "" ||
ans[ans.Length - 1] != S[i])
ans += S[i];
// Matches with the previous one
else if (ans[ans.Length - 1] == S[i])
ans = ans.Substring(0, ans.Length - 1);
}
// Return the answer
return ans;
}
// Driver Code
public static void Main(String[] args)
{
String str = "keexxllx" ;
Console.WriteLine(removeDuplicates(str));
}
} // This code is contributed by Rajput-Ji |
<script> // JavaScript implementation of the above approach // Function to remove adjacent duplicates function removeDuplicates( S) {
var ans = "" ;
// Iterate for every character
// in the string
for (i = 0; i < S.length; i++) {
// If ans string is empty or its last
// character does not match with the
// current character then append this
// character to the string
if (ans.length==0 ||
ans.charAt(ans.length - 1) != S.charAt(i))
ans += S.charAt(i);
// Matches with the previous one
else if (ans.charAt(ans.length - 1) == S.charAt(i))
ans = ans.substring(0, ans.length - 1);
}
// Return the answer
return ans;
}
// Driver Code
var str = "keexxllx" ;
document.write(removeDuplicates(str));
// This code contributed by Rajput-Ji </script> |
kx
Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the length of the string.
Auxiliary Space: O(N), as we are using extra space for the ans string. Where N is the length of the string.