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kotykmax [81]
2 years ago
12

Why is it now difficult for computer technology to maintain the pace of

Computers and Technology
1 answer:
KatRina [158]2 years ago
6 0

It now difficult for computer technology as manufacturing cannot keep up with overwhelming transistor demand.

<h3>What was Moore's Law ?</h3>

Moore's Law is the one that talks about the fact that the number of transistors on a microchip is one that often goes up about two years always, even if the cost of computers is reduced.

Based on the above, It now difficult for computer technology as manufacturing cannot keep up with overwhelming transistor demand.

Learn more about computer technology  from

brainly.com/question/24912812

#SPJ1

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Programming challenge description: Given a string comprising just of the characters (,),{,},[,] determine if it is well-formed o
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Answer:

Employing Java language,

Explanation:

//for using stack data structure

import java.util.Stack;

//class definition

public class BracketMatch {

   //function to find the string is well formed or not

   public static boolean isWellFormed(String str) {

       //variable to store the character value

       char ch;

       //check if the first character itself is the end of the brackets

       if (str.charAt(0) == ')'||str.charAt(0) =='}'||str.charAt(0) ==']')

           return false;

       //creating a stack of character type

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

       //iterating through the characters in string

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

           //storing the current character from the string in the variable

           ch = str.charAt(i);

           if(ch == '('||ch=='{'||ch=='[')

               //if the character is starting of the brackets push it into stack

               stack.push(ch);

           else if(ch == ')')

               //if it is closing bracket and the stack is empty then the string is not well formatted

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                   return false;

               else if(stack.peek() == '(')

                   stack.pop();

               else

                   return false;

           else if(ch == '}')

               if(stack.empty())

                   return false;

               else if(stack.peek() == '{')

                   stack.pop();

               else

                   return false;

       }

       //returning the stack content, if the string will be well formed the stack would be empty

       //after all the operations

       return stack.empty();

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   //main method

   public static void main(String[] args) {

       //calling function to find if the string is welll formatted or not

       System.out.println(isWellFormed("{}"));

       //calling function to find if the string is welll formatted or not

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}

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Extension: Cash Register Program
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