Super Reduced String Hacker Rank

 Problem Link:

https://www.hackerrank.com/challenges/reduced-string/problem

Solution:

Stack<Character>stack=new Stack<Character>();

        

        for(int i=0;i<s.length();i++){

            if(stack.isEmpty()){

                stack.push(s.charAt(i));

            }

            else{

                if(stack.peek()==s.charAt(i)){

                    stack.pop();

                }

                else{

                    stack.push(s.charAt(i));

                }

            }

        }

        if(stack.size()!=0){

        StringBuilder sb=new StringBuilder();

        while(!stack.isEmpty()){

            sb.append(stack.pop());

        }

        sb.reverse();

        

        return sb.toString(); 

        }

        else{

            return "Empty String";

        }

        

TIME COMPLEXITY: O(String Length)

SPACE COMPLEXITY: O(1)

AUXILLARY SPACE:O(String Length)

TOTAL TEST CASES: 16

APPROACH:

Using Stacks.

Here we will use the same approach as we do for balanced parenthesis.

First, we will traverse the string and push characters in the stack.

If stacks peek is equal to the incoming character value then we will pop that character

and if not equal then we will push that character into the stack.

Ex :

if the string is aaa then stack ="" (initially)

Traversal start: stack=" a" now string is aaa (first a traversed)

Now stack. peek()=a is equal to incoming character ie a (at index 1) so it will pop

stack="" string traversed=aaa

now last a is left

so stack="a" string travered="aaa" string traversal full. stack result is a return output is a

If stack is empty or stack.size()==0 then we will return output as "Empty string".

Ex : string = aaaa stack=""

string = aaaa stack="a"

string = aaaa stack=""

string = aaaa stack="a"

string = aaaa stack=""

(String traversal is shown in bold).

return "Empty String" as output.

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