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krek1111 [17]
3 years ago
8

Which of the following is not an algorithm?

Computers and Technology
1 answer:
Kipish [7]3 years ago
7 0

Answer:

The correct answer is: "D. shapoo instructions (lather, rinse, repeat)".

Explanation:

Among the options given, letter D is the only one which does not apply and  may not function as an algorithm. An algorithm is a set of rules and/or instructions which aims at solving a problem and/or task, therefore, "shapoo instructions" do not classify as an algorithm because they are only written informations to explain how to use a product (lather, rinse, repeat), rather than setting a procedure for solving a problem in terms of technology tools. All the other options are examples of algorithms because they already exist as such and also function as technology tools.

(ps: mark as brainliest, please?!)

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comparing the cost of care provided to medicare beneficiaries assigned to primary care nurse practitioners and physicians
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Go to explaination for the program code

Explanation:

import java.util.Stack;

public class Lab3 {

public static void main(String[] args) {

String s1="DataStructuresIssss###Fun";

String s2="DataStructuresIszwp###Fun";

boolean ans=backspaceCompare(s1,s2);

System.out.println(ans);

/*String s1="abc##";

String s2="wc#d#";

boolean ans=backspaceCompare(s1,s2);

System.out.println(ans);*/

}

public static boolean backspaceCompare(String s1, String s2) {

Stack<Character> s1_stack=new Stack<Character>();

Stack<Character> s2_stack=new Stack<Character>();

//backspaceCount is a variable to count back space

int backspaceCount=0;

//logic is that if '#' encountered we are putting pop else push

for(int i=0;i<s1.length();i++){

if(s1.charAt(i)=='#'){

backspaceCount++;

s1_stack.pop();

}

else

{

s1_stack.push(s1.charAt(i));

}

}

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for(int i=0;i<s2.length();i++){

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}

//here is the main logic first we are adding based upon # means we pop up the string while adding the string if any # character found

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for(int i=0;i<s1.length()-2*backspaceCount;i++){

if(s1_stack.pop()!=s2_stack.pop()) return false;

}

return true;

}

}

6 0
3 years ago
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