Count elements such that there are exactly X elements with values greater than or equal to X in C++
We are given with an array of integers. The goal is to find the count of elements in the array that
satisfy the following condition −
For each element the count of numbers greater than or equal to it present in the array should be
exactly equal to it. Excluding the element itself. If element is X then array has exactly X numbers
which are greater or equal to X. (Excluding the element).
Input
Arr[]= { 0,1,2,3,4,9,8 }Output
Elements exactly greater than equal to itself : 1
Explanation − Elements and numbers >= to it −
Arr[0]: 6 elements are >= 0 , 6!=0 count=0
Arr[1]: 5 elements are >= 1 , 5!=1 count=0
Arr[2]: 4 elements are >= 2 , 4!=2 count=0
Arr[3]: 3 elements are >= 3 , 3==3 count=1
Arr[4]: 2 elements are >= 4 , 2!=4 count=1
Arr[4]: 0 elements are >= 9 , 0!=9 count=1
Arr[6]: 1 element is >= 8 , 1!=8 count=1
3 is the only element such that exactly 3 elements are >= to it (4,8,9)
Input
Arr[]= { 1,1,1,1,1 }Output
Elements exactly greater than equal to itself : 0
Explanation − All elements are equal and count !=1
Approach used in the below program is as follows
The integer array Arr[] is used to store the integers.
Integer ‘n’ stores the length of the array.
Function findcount(int arr[],int n) takes an array and its size as input and returns the count
of numbers like X explained earlier.
Variable count is used to store the count of numbers like X.
Initialize ans=0, which will count such numbers.
Traverse the array starting from the first element( index=0 ) using for loop.
Inside for loop traverse from the starting element again, if any arr[j]>=arr[j] such that
i!=j, increment the count.
After the end of j loop, compare count with arr[i]. If count==arr[i] (exactly arr[i]
elements are >=arr[i], increment the answer ‘ans’
After the end of both for loops return the result present in ‘ans’.
Example
Live Demo
#include <iostream>
#include <algorithm>
using namespace std;
int findcount(int arr[],int n){
sort(arr,arr+n);
int count=0;
int ans=0;
for(int i=0;i<n;i++){
count=0;
for(int j=0;j<n;j++){
if(arr[j]>=arr[i] && i!=j)
count++;
}
if(count==arr[i])
ans++;
}
return ans;
}
int main(){
int Arr[]= { 0,1,2,3,4,5,6 };
int k=7;
int n=sizeof(Arr)/sizeof(Arr[0]);
std::cout<<"Elements exactly greater than equal to itself : "<<findcount(Arr,n);
return 0;
}Output
Elements exactly greater than equal to itself : 1
Published on 28-Jul-2020 14:37:05
