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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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 than 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);        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 than 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

 > 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 than 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, ManasChhabra2