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 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[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 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
<?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 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. ?> |
Javascript
<script> // JavaScript 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) function lower_bound(arr, start, end, x) { while (start < end) { let 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) { let 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==1) document.write(n - upper_bound(arr, 0, n, x) + "<br/>"); // Counting number of integer which // are not less than x (Or greater // than or equal to x) else document.write(n - lower_bound(arr, 0, n, x) + "<br/>"); } // Driver code let arr = [ 1, 2, 3, 4 ]; let n = arr.length; arr.sort(); query(arr, n, 0, 5); query(arr, n, 1, 3); query(arr, n, 0, 3); </script> |
Output:
0 1 2
Time Complexity : O( (N + Q) * logN).
Auxiliary Space : O(1)
This article is contributed by Anuj Chauhan. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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