- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# 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.

## Approach to find The Solution

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.

## Brute Force Approach

In this approach, we will traverse through the array for all q queries and find the numbers that satisfy the given conditions.

## Example

#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; }

## Output

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.

## Efficient Approach

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.

## Example

#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; }

## Output

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.

## Explanation of the above code

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).

## Conclusion

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.

- Related Questions & Answers
- Count of alphabets having ASCII value less than and greater than k in C++
- Count the number of words having sum of ASCII values less than and greater than k in C++
- Largest number less than N with digit sum greater than the digit sum of N in C++
- Check if any value in an R vector is greater than or less than a certain value.
- How to create a subset of matrix in R using greater than or less than a certain value of a column?
- How to select values less than or greater than a specific percentile from an R data frame column?
- How to check whether a column value is less than or greater than a certain value in R?
- How to change the color of bars in histogram for values that are greater than 0 or less than 0 in R?
- Query MongoDB for a datetime value less than NOW?
- Largest even digit number not greater than N in C++
- Count values greater and less than a specific number and display count in separate MySQL columns?
- How to find numbers in an array that are greater than, less than, or equal to a value in java?
- How do I write not greater than in a MySQL query?
- Two Sum Less Than K in Python
- Subarray Product Less Than K in C++