Find the number of elements greater than k in a sorted array using C++

In this problem, we are given an array arr[] consisting of N sorted integer values and an integer k. Our task is to Find the number of elements greater than k in a sorted array.

Let’s take an example to understand the problem,

Input

arr[] = {1, 2, 5, 7, 8, 9} k = 4

Output

Explanation

Elements greater than k = 4 are
5, 7, 8, 9

Solution Approach

A simple solution to the problem is by using a loop over the array from 0 to N. And then stop at the first element greater than k. Then count the number of values remaining.

Example

Program to illustrate the working of our solution

#include <iostream>
using namespace std;
int findGreaterCount(int arr[], int n, int k){
   for(int i = 0; i < n; i++){
      if(arr[i] > k)
         return (n - i);
   }
   return -1;
}
int main(){
   int arr[] = { 1, 3, 5, 7, 7, 8, 12, 21};
   int n = sizeof(arr) / sizeof(arr[0]);
   int k = 5;
   cout<<"The number of elements greater than k is "<<findGreaterCount(arr, n, k);
   return 0;
}

Output

The number of elements greater than k is 5

The above code works well but the time complexity of the program is of the order O(N).

Another more efficient approach is using binary search to find the elements greater than k. And then return the count of greater elements.

Example

Program to illustrate the working of our solution

#include <iostream>
using namespace std;
int findGreaterCount(int arr[], int n, int k){
   int s = 0;
   int e = n - 1;
   int firstGreterEle = n;
   while (s <= e) {
      int mid = s + (e - s) / 2;
      if (arr[mid] > k) {
         firstGreterEle = mid;
         e = mid - 1;
      }
      else
         s = mid + 1;
   }
   return (n - firstGreterEle);
}
int main(){
   int arr[] = { 1, 3, 5, 7, 7, 8, 12, 21};
   int n = sizeof(arr) / sizeof(arr[0]);
   int k = 5;
   cout<<"The number of elements greater than k is "<<findGreaterCount(arr, n, k);
   return 0;
}

Output

The number of elements greater than k is 5

sudhir sharma

Updated on 14-Feb-2022 08:04:13

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