Question

C++, and comments and explanation, please.

Input: the program should read the followings: First line, an integer . Second line, an integer n

C++, and comments and explanation, please.

Answer 1

For this program, we will calculate the lower bound of the given number 'x' (explained later in the code). Lower bound is calculated by expanding and modifying the code for binary search. Unlike binary search where we search for a number (inturn return an index specifying that number) here we will search for the first number (inturn the index) where the number is greater than or equal to the given number. This is done in log(n) time just like binary search. For this program we will use in-built C++ function lower bound function(explained later in the code)

NOTE: Incase you dont like the in-built implementation of the program using vectors(similar to array), just tell me I can also implement lower_bound function using own written code which will work on arrays as well.

C++ CODE (With STL lower_bound) :

#include<bits/stdc++.h>
using namespace std;

int main() {
   int x, n;
   cin >> x >> n;
   vector<int> a(n);
  
   /*
   vectors are just like arrays with added advantages. There dynamically
   allocated like lists(whose size can change) and also have indexes like arrays.
   */
   for (int i = 0; i < n; ++i) {
       cin >> a[i];
   }
   vector<int> :: iterator it; // unlike array, iterators can also be used to traverse an vector.

   /*
   lower_bound() returns an iterator to the elements in the given range which does no compare less(i.e which means is greater than or equal to :))
   than the given value. The range given should be already sorted for lower_bound() to work. In plain words it
   returns an iterator to the lower bound of the given element in a sorted range.
   */
  
   it = lower_bound(a.begin(), a.end(), x);
   /*
   lower bound has returned the position of the element which is greater than or equal to the x.
   Therefore, the numbers less than x will be till that position excluding that number.
   */
   for (int i = 0; i < (it - a.begin()); ++i) {
   /*
   it - a.begin because 'it' is a pointer and its value corresponds to an address.
   Therefore subtracting the starting address of the vector will give us the distance of that pointer from the starting address,
   which would be needed to display the numbers smaller than the given number ('x')
   */
       cout << a[i] << " ";
   }  
}

C++ Code (Without STL lower_bound):

#include<bits/stdc++.h>
using namespace std;
int Lower_bound(int a[], int n, int x) {
    int left = 0;
    int right = n;
    while (left < right) {
        int middle = (left + right) / 2;
        if (x <= a[middle]) {
            right = middle;
        } else {
            left = middle + 1;
        }
    }
    return left;
}

int main() {
   int x, n;
   cin >> x >> n;
   int a[n];
   for (int i = 0; i < n; ++i) {
       cin >> a[i];
   }
   int limit = Lower_bound(a, n, x);
   for (int i = 0; i < limit; ++i) {
       cout << a[i] << " ";
   }
  
}

OUTPUT:

1)

25 1 4 5 30 35 48 68 98 1 4 5 Process exited after 16.87 seconds with return value e Press any key to continue..

2)

31 1 4 5 18 16 48 68 98 1 4 5 10 16 Process exited after 12.61 seconds with return value Press any key to continue .. .