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 the implementation of this approach :

C++

// 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;
}

Java

// Java to find number of integer 
// less or greater given
// integer queries
import java.util.Arrays;
class GFG
{
// Return the index of integer
// which are not less than x
// (or greater than or equal
// to x)
static 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.
static 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;
}
  
static void query(int arr[], int n, int type, int x)
{
    // Counting number of integer 
    // which are greater than x.
    if (type==1)
        System.out.println(n - upper_bound(arr, 0, n, x));
  
    // Counting number of integer which
    // are not less than x (Or greater
    // than or equal to x)
    else
        System.out.println(n - lower_bound(arr, 0, n, x));
}
  
// Driver code
public static void main (String[] args)
{
    int arr[] = { 1, 2, 3, 4 };
    int n = arr.length;
  
    Arrays.sort(arr);
  
    query(arr, n, 0, 5);
    query(arr, n, 1, 3);
    query(arr, n, 0, 3);
}
}
  
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find number of integer 
# less or greater given integer queries
  
# Return the index of integer  
# which are not less than x
# (or greater than or equal to x)
def lower_bound(arr, start, end, x):
  
    while (start < end):
      
        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.
def upper_bound(arr, start, end, x):
  
    while (start < end):
      
        mid = (start + end) >> 1
        if (arr[mid] <= x):
            start = mid + 1
        else:
            end = mid
      
    return start
  
def query(arr, n, type, x):
  
    # Counting number of integer
    # which are greater than x.
    if (type == 1):
        print(n - upper_bound(arr, 0, n, x))
  
    # Counting number of integer
    # which are not less than x
    # (Or greater tha or equal to x)
    else:
        print(n - lower_bound(arr, 0, n, x))
  
# Driver code
arr = [ 1, 2, 3, 4 ]
n =len(arr)
  
arr.sort()
  
query(arr, n, 0, 5)
query(arr, n, 1, 3)
query(arr, n, 0, 3)
  
# This code is contributed by Anant Agarwal.

C#

// C# to find number of integer less 
// or greater given integer queries
using System;
  
class GFG {
      
// Return the index of integer which are
// not less than x (or greater than or
// equal to x)
static 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.
static 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;
}
  
static void query(int []arr, int n, int type, int x)
{
    // Counting number of integer 
    // which are greater than x.
    if (type==1)
        Console.WriteLine(n - upper_bound(arr, 0, n, x));
  
    // Counting number of integer which
    // are not less than x (Or greater
    // than or equal to x)
    else
        Console.WriteLine(n - lower_bound(arr, 0, n, x));
}
  
// Driver code
public static void Main ()
{
    int []arr = {1, 2, 3, 4};
    int n = arr.Length;
  
    Array.Sort(arr);
  
    query(arr, n, 0, 5);
    query(arr, n, 1, 3);
    query(arr, n, 0, 3);
}
}
  
// This code is contributed by vt_m.

PHP

<?php
// PHP to find number of integer
// less or greater given
// integer queries
  
// Return the index of integer
// which are not less than x
// (or greater than or equal to x)
function lower_bound($arr, $start, $end, $x)
{
    while ($start < $end)
    {
    $mid = ($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.
function upper_bound($arr, $start, $end, $x)
{
    while ($start < $end)
    {
        $mid = ($start + $end) >> 1;
        if ($arr[$mid] <= $x)
            $start = $mid + 1;
        else
            $end = $mid;
    }
  
    return $start;
}
  
function query($arr, $n, $type, $x)
{
      
    // Counting number of integer
    // which are greater than x.
    if ($type)
        echo $n - upper_bound($arr, 0, $n, $x) ,"\n";
  
    // Counting number of integer
    // which are not less than x
    // (Or greater tha or equal to x)
    else
        echo $n - lower_bound($arr, 0, $n, $x) ,"\n";
}
  
    // Driver Code
    $arr = array(1, 2, 3, 4);
    $n = count($arr);
  
    sort($arr);
  
    query($arr, $n, 0, 5);
    query($arr, $n, 1, 3);
    query($arr, $n, 0, 3);
  
// This code is contributed by anuj_67.
?>


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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Improved By : vt_m



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