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There is a structure called employee that holds information like employee code,
name, date of joining. Write a program to create an array of the structure and enter
some data into it. Then ask the user to enter current date. Display the names of those
employees whose tenure is 3 or more than 3 years according to the given current
date.
The placement season has begun in a college. There are N number of students standing outside an interview room in a line. It is given that a person who goes in first has higher chances of getting selected.
Input1- an integer N, which denotes the number of students present
Input2- An array of size N, denoting the problem solving capacity of the students.

Each student has a number associated with them known as the problem-solving capability (PSC). The higher the capability, the higher the chances of selection. Now, each student wants to know the number of students ahead of him/her who have more problem-solving capability than him/her.

Find this number for each student.

Prepare a c++ program of the sample run below:

SAMPLE RUN:

                    Name         Q1      Q2     Q3      Q4     Q5   

           

                  Von                     20

                  


input 1) an integer N,which denotes the number of students present
Input 2) an array of size N denoting the problem solving capacity of Students
Explain in your own words the Analysis of Linked List operations (add,remove,find,back).

Explain in your own words the Analysis of each Array list operation (add, remove and find)?


Consider a Dynamic Array declared inside the code using int* z = new int[10]; . How we will free the memory after we are done with the z array?

Consider a Dynamic Array declared inside the code using int* z = new int[10]; . How we will free the memory after we are done with the z array?

Explain in your own words the Analysis of each Array list operation (add, remove and find)?


Q: Explain the given code in your own words.

 find(X): traverse the array until X is located.
 int find(int X)
 {
    int j;
    for(j=1; j < size+1; j++ )
    if( A[j] == X ) break;
    if( j < size+1 ) {  // found X
    current = j;       // current points to where X found
    return 1;  // 1 for true
  }
    return 0; // 0 (false) indicates not found






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