Queries for greater than and not less than using C++

In this article, we are given a problem, we are given an array, and there are two types of queries we need to answer.
Type 0 − we have to calculate the number of greater elements than or equal to x(given value).
Type 1 − we have to calculate the number of strictly greater elements than x(given value).
So here is a simple example −
Input : arr[] = { 10, 15, 30 , 40, 45 } and Q = 3
   Query 1: 0 50
   Query 2: 1 40
   Query 3: 0 30
Output :
   0
   1
   3
Explanation:
x = 50, q = 0 : No elements greater than or equal to 50.
x = 40, q = 1 : 45 is greater than 40.
x = 30, q = 0 : three elements 30, 40, 45 are greater than or equal to 30.
We can use two different methods to find the solution. Firstly we will use the brute force solution and then check if it can work for higher constraints or not. If not, then we proceed to optimize our solution.
In this approach, we will traverse through the array for all q queries and find the numbers that satisfy the given conditions.
#include <bits/stdc++.h>
using namespace std;
void query(int *arr, int n, int type, int val) {
   int count = 0; // answer
   if(!type) { // when type 0 query is asked
      for(int i = 0; i < n; i++) {
         if(arr[i] >= val)
            count++;
      }
   } else { // when type 1 query is asked
      for(int i = 0; i < n; i++) {
         if(arr[i] > val)
            count++;
      }
   }
   cout << count << "\n";
}
int main() {
   int ARR[] = { 10, 15, 30, 40, 45 };
   int n = sizeof(ARR)/sizeof(ARR[0]); // size of our array
   query(ARR, n, 0, 50); // query 1
   query(ARR, n, 1, 40); // query 2
   query(ARR, n, 0, 30); // query 3
   return 0;
}
0
1
3
In the above approach, we are simply traversing through the array and calculating the answer for the queries; this approach is working for the given examples, but if we encounter higher constraint, this approach will fail as the overall time complexity of the program is O(N*Q) where N is the size of our array and Q is the number of queries so now we will optimize this approach such that it works for higher constraints as well.
In this approach, we will be using binary search to find the upper bound and lower bound of the given value. We first sort our array using binary search and then apply our lower bound and upper bound functions accordingly.
#include <bits/stdc++.h>

using namespace std;
void lowerbound(int *arr, int n, int val) {
   int l = -1, r = n;
   while(r - l > 1) { // binary searching the answer
      int mid = (l+r)/2;
      if(arr[mid] >= val)
         r = mid;
      else
         l = mid;
   }
   if(r == n) // if r is unmoved then it means there is no element that satisfy the condition
      cout << "0\n";
   else
      cout << n - r << "\n";
}
void upperbound(int *arr, int n, int val) {
   int l = -1, r = n;
   while(r - l > 1) { // binary searching the answer
      int mid = (l+r)/2;
      if(arr[mid] > val)
         r = mid;
      else
         l = mid;
   }
   if(r == n)// if r is unmoved then it means there is no element that satisfy the condition
      cout << "0\n";
   else
      cout << n - r <<"\n";
}
void query(int *arr, int n, int type, int val) {
   if(!type) // if type == 0 we call lower bound function
      lowerbound(arr, n, val);
   else // if type == 1 we call upperbound function
      upperbound(arr, n, val);
}
int main() {
   int arr[] = { 1, 2, 3, 4 };
   int n = sizeof(arr)/sizeof(arr[0]); // size of our array
   sort(arr, arr+n); // sorting the array
   query(arr, n, 0, 5); // query 1
   query(arr, n, 1, 3); // query 2
   query(arr, n, 0, 3); // query 3
   return 0;
}
0
1
2
The above code works on a binary search that decreases our time complexity substantially. Thus our final complexity becomes O(NlogN), where N is the size of our array.
In this approach, we will be using binary search to find the upper bound and lower bound of the given value. Now for binary search, we first sort our array as it works only with sorted arrays. We make our lower bound and upper bound function that helps us find the first number that satisfies the type 0, type 1 conditions, respectively, now as we have sorted the array. We found the first number that helps the condition, so the elements after this element also follow the condition, so we print the difference of the index of this element with N(the size of our array).
In this article, we solve a problem to solve Queries for greater than and not less than using Binary search. We also learned the C++ program for this problem and the complete approach ( Normal and efficient ) by which we solved this problem. We can write the same program in other languages such as C, java, python, and other languages. We hope you find this article helpful.