# Count numbers < = N whose difference with the count of primes upto them is > = K in C++

C++Server Side ProgrammingProgramming

Given two integers N and K, the goal is to find the count of numbers such that they follow below conditions −

• Number<=N

• | Number−count | >=K Where count is the number of prime numbers less than or equal to Number.

For Example

## Input

N = 5, K = 2

## Output

Count of numbers < = N whose difference with the count of primes upto them is > = K are: 2

## Explanation

The numbers that follow the conditions are:
5 ( 5−2>=2 ) and 4 ( 4−2>=2 )

## Input

N = 10, K = 6

## Output

Count of numbers < = N whose difference with the count of primes upto them
is > = K are: 1

## Explanation

The numbers that follow the conditions are:
10 ( 10−4>=6 )

Approach used in the below program is as follows

In this approach we will use binary search to reduce our calculations. If the count of prime numbers upto num is count1 and for number num+1 this count is count2. Then the difference num+1−count2 >= num−count1. So if num is valid, num+1 will also be valid.For the first number found, say ‘num’ using binary search that follows the condition, then ‘num’+1 would also follow the same condition. In this way all the numbers between num to N will be counted.

• Take variables N and K as input.

• Array arr[] is used to store the count of prime numbers upto i will be stored at index i.

• Function set_prime() updates array arr[] for storing the counts of primes.

• Array check[i] stores true if i is prime else stores false.

• Set check[0]=check[1] = false as they are non primes.

• Traverse check from index i=2 to i*i<size(1000001). And if any check[i] is 1, number is prime then set all check[j] with 0 from j=i*2 to j<size.

• Now traverse arr[] using for loop and update it. All counts upto arr[i]=arr[i−1]. If arr[i] itself is prime then that count will increase by 1. Set arr[i]++.

• Function total(int N, int K) takes N and K and returns Count of numbers < = N whose difference with the count of primes upto them is > = K.

• Call set_prime().

• Take temp_1=1 and temp_2=N. Take the initial count as 0.

• Now using binary search, in while loop take set = (temp_1 + temp_2) >> 1 ((first+last) /2 ).

• If set−arr[set] is >=K then condition is met, update count with set and temp_2=set−1.

• Otherwise set temp_1=temp_1+1.

• At the end set count as minimum valid number N−count+1 or 0.

• At the end of all loops return count as result.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
#define size 1000001
int arr[size];
void set_prime(){
bool check[size];
memset(check, 1, sizeof(check));
check[0] = 0;
check[1] = 0;
for (int i = 2; i * i < size; i++){
if(check[i] == 1){
for (int j = i * 2; j < size; j += i){
check[j] = 0;
}
}
}
for (int i = 1; i < size; i++){
arr[i] = arr[i − 1];
if(check[i] == 1){
arr[i]++;
}
}
}
int total(int N, int K){
set_prime();
int temp_1 = 1;
int temp_2 = N;
int count = 0;
while (temp_1 <= temp_2){
int set = (temp_1 + temp_2) >> 1;
if (set − arr[set] >= K){
count = set;
temp_2 = set − 1;
} else {
temp_1 = set + 1;
}
}
count = (count ? N − count + 1 : 0);
return count;
}
int main(){
int N = 12, K = 5;
cout<<"Count of numbers < = N whose difference with the count of primes upto them is > = K are: "<<total(N, K);
return 0;
}

## Output

If we run the above code it will generate the following output −

Count of numbers < = N whose difference with the count of primes upto them is > = K are: 4
Updated on 05-Jan-2021 06:53:39

Advertisements