Remove all duplicate adjacent characters from a string using Stack
Last Updated :
01 Sep, 2021
Given a string, str, the task is to remove all the duplicate adjacent characters from the given string.
Examples:
Input: str= “azxxzy”
Output: ay
Removal of "xx" modifies the string to “azzy”.
Now, the removal of "zz" modifies the string to “ay”.
Since the string “ay” doesn't contain duplicates, the output is ay.
Input: "aaccdd"
Output: Empty String
Recursive Approach: Refer to the article Recursively remove all adjacent duplicates to solve this problem recursively.
Time Complexity: O(N)
Auxiliary Space: O(N)
String Functions-based Approach: Refer to this article Remove first adjacent pairs of similar characters until possible to solve this problem using inbuilt functions pop_back() and back() methods of string.
Time Complexity: O(N)
Auxiliary Space: O(N)
Stack-based Approach: The problem can be solved using Stack to use the property of LIFO. The idea is to traverse the string from left to right and check if the stack is empty or the top element of the stack is not equal to the current character of str, then push the current character into the stack. Otherwise, pop the element from the top of the stack. Follow the steps below to solve the problem:
- Create a stack, st to remove the adjacent duplicate characters in str.
- Traverse the string str and check if the stack is empty or the top element of the stack not equal to the current character. If found to be true, push the current character into st.
- Otherwise, pop the element from the top of the stack.
- Finally, print all the remaining elements of the stack.
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to remove adjacent
// duplicate elements
string ShortenString(string str1)
{
// Store the string without
// duplicate elements
stack<char> st;
// Store the index of str
int i = 0;
// Traverse the string str
while (i < str1.length())
{
// Checks if stack is empty or top of the
// stack is not equal to current character
if (st.empty() || str1[i] != st.top())
{
st.push(str1[i]);
i++;
}
// If top element of the stack is
// equal to the current character
else
{
st.pop();
i++;
}
}
// If stack is empty
if (st.empty())
{
return ("Empty String");
}
// If stack is not Empty
else
{
string short_string = "";
while (!st.empty())
{
short_string = st.top() +
short_string;
st.pop();
}
return (short_string);
}
}
// Driver Code
int main()
{
string str1 ="azzxzy";
cout << ShortenString(str1);
return 0;
}
// This code is contributed by divyeshrabadiya07
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to remove adjacent
// duplicate elements
static String ShortenString(String str1)
{
// Store the String without
// duplicate elements
Stack<Character> st =
new Stack<Character>();
// Store the index of str
int i = 0;
// Traverse the String str
while (i < str1.length())
{
// Checks if stack is empty
// or top of the stack is not
// equal to current character
if (st.isEmpty() ||
str1.charAt(i) != st.peek())
{
st.add(str1.charAt(i));
i++;
}
// If top element of the stack is
// equal to the current character
else
{
st.pop();
i++;
}
}
// If stack is empty
if (st.isEmpty())
{
return ("Empty String");
}
// If stack is not Empty
else
{
String short_String = "";
while (!st.isEmpty())
{
short_String = st.peek() +
short_String;
st.pop();
}
return (short_String);
}
}
// Driver Code
public static void main(String[] args)
{
String str1 ="azzxzy";
System.out.print(ShortenString(str1));
}
}
// This code is contributed by Rajput-Ji
# Python3 program to implement
# the above approach
# Function to remove adjacent
# duplicate elements
def ShortenString(str1):
# Store the string without
# duplicate elements
st = []
# Store the index of str
i = 0
# Traverse the string str
while i < len(str1):
# Checks if stack is empty or top of the
# stack is not equal to current character
if len(st)== 0 or str1[i] != st[-1]:
st.append(str1[i])
i += 1
# If top element of the stack is
# equal to the current character
else:
st.pop()
i += 1
# If stack is empty
if len(st)== 0:
return("Empty String")
# If stack is not Empty
else:
short_string = ""
for i in st:
short_string += str(i)
return(short_string)
# Driver Code
if __name__ == "__main__":
str1 ="azzxzy"
print(ShortenString(str1))
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to remove adjacent
// duplicate elements
static String ShortenString(String str1)
{
// Store the String without
// duplicate elements
Stack<char> st = new Stack<char>();
// Store the index of str
int i = 0;
// Traverse the String str
while (i < str1.Length)
{
// Checks if stack is empty
// or top of the stack is not
// equal to current character
if (st.Count == 0 || (st.Count != 0 &&
str1[i] != st.Peek()))
{
st.Push(str1[i]);
i++;
}
// If top element of the stack is
// equal to the current character
else
{
if (st.Count != 0)
st.Pop();
i++;
}
}
// If stack is empty
if (st.Count == 0)
{
return ("Empty String");
}
// If stack is not Empty
else
{
String short_String = "";
while (st.Count != 0)
{
short_String = st.Peek() +
short_String;
st.Pop();
}
return (short_String);
}
}
// Driver Code
public static void Main(String[] args)
{
String str1 ="azzxzy";
Console.Write(ShortenString(str1));
}
}
// This code is contributed by Amit Katiyar
<script>
// JavaScript program to implement
// the above approach
// Function to remove adjacent
// duplicate elements
function ShortenString(str1)
{
// Store the string without
// duplicate elements
var st = [];
// Store the index of str
var i = 0;
// Traverse the string str
while (i < str1.length)
{
// Checks if stack is empty or top of the
// stack is not equal to current character
if (st.length==0 || str1[i] != st[st.length-1])
{
st.push(str1[i]);
i++;
}
// If top element of the stack is
// equal to the current character
else
{
st.pop();
i++;
}
}
// If stack is empty
if (st.length==0)
{
return ("Empty String");
}
// If stack is not Empty
else
{
var short_string = "";
while(st.length!=0)
{
short_string = st[st.length-1] +
short_string;
st.pop();
}
return (short_string);
}
}
// Driver Code
var str1 ="azzxzy";
document.write( ShortenString(str1));
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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