Queries for greater than and not less than

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 3
Output :0
        1
        2
Explanation:
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).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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