Given an array of **N** integers. There will be **Q** queries, each include two integer of form **q** and **x**, 0 <= q <= 1. Queries are of two types:

- In first query (q = 0), the task is to find count of integers which are not less than x (OR greater than or equal to x).
- In second query (q = 1), the task is to find count of integers greater than x.

Examples:

Input :arr[] = { 1, 2, 3, 4 } and Q = 3 Query 1: 0 5 Query 2: 1 3 Query 3: 0 3Output :0 1 2Explanation:x = 5, q = 0 : There are no elements greater than or equal to it. x = 3, q = 1 : There is one element greater than 3 which is 4. x = 3, q = 0 : There are two elements greater than or equal to 3.

**Method 1: **A **Naive approach** can be for each query, traverse the whole array and count integers less or greater than x, depending on q. Time Complexity for this approach will be **O(Q*N)**.

**Method 2:** An **efficient** approach can be sort the array and use binary search for each query. This will take **O(NlogN + QlogN)**.

Below is C++ implementation of this approach:

// C++ to find number of integer less or greater given // integer queries #include<bits/stdc++.h> using namespace std; // Return the index of integer which are not less than x // (or greater than or equal to x) int lower_bound(int arr[], int start, int end, int x) { while (start < end) { int mid = (start + end)>>1; if (arr[mid] >= x) end = mid; else start = mid + 1; } return start; } // Return the index of integer which are greater than x. int upper_bound(int arr[], int start, int end, int x) { while (start < end) { int mid = (start + end)>>1; if (arr[mid] <= x) start = mid + 1; else end = mid; } return start; } void query(int arr[], int n, int type, int x) { // Counting number of integer which are greater than x. if (type) cout << n - upper_bound(arr, 0, n, x) << endl; // Counting number of integer which are not less than x // (Or greater tha or equal to x) else cout << n - lower_bound(arr, 0, n, x) << endl; } // Driven Program int main() { int arr[] = { 1, 2, 3, 4 }; int n = sizeof(arr)/sizeof(arr[0]); sort(arr, arr + n); query(arr, n, 0, 5); query(arr, n, 1, 3); query(arr, n, 0, 3); return 0; }

Output:

0 1 2

**Time Complexity : **O( (N + Q) * logN).

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