# Maximize the size of array by deleting exactly k sub-arrays to make array prime in C++

C++Server Side ProgrammingProgramming

Given the task is to delete exactly K sub-arrays from a given array Arr[] with N positive elements such that all the remaining elements in the array are prime and the size of the remaining array is maximum.

Input

Arr[]={4, 3, 3, 4, 3, 4, 3} , K=2

Output

3

Explanation − K=2, this means only 2 sub-arrays must be deleted.

The sub-arrays deleted are Arr[0] and Arr[3…5] which leaves the array Arr[]={3,3,3} with all the prime elements and the maximum size possible.

Input

Arr[]={7, 6, 2, 11, 8, 3, 12}, K=2

Output

3

Explanation − The sub-arrays deleted are Arr[1] and Arr[4…6] and the remaining prime array is Arr[]={7,2,11}.

## Approach used in the below program as follows

• First store all the primes into another array prime[] using the sieve of Eratosthenes by calling the sieve() function

• In function MaxSize()run a loop from i=0 till i<N and store the index numbers of all the composite numbers into a vector vect of type int.

• Then run another loop from i=1 till i< vect.size to compute the number of primes that lie between two consecutive composites and store it into another vector diff of type int.

• Then sort the vector diff using the sort() function.

• Now run another loop from i=1 to i<diff.size() and compute the prefix sum of this vector which will help us to know how many primes need to be deleted.

• Using an if statement check for impossible case, that is when K=0 and there are no composites.

• If K is greater than or equal to the number of composites, then delete all the composites including extra primes and the size of these sub-arrays to be deleted should be 1, in order to get optimal answer

• If K is less than the number of composites, then we will have to delete those sub-arrays that contain composites and prime sub-arrays should not fall in this category.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
const int Num = 1e5;
bool prime[Num];
//Sieve of Eratosthenes
void sieve(){
for (int i = 2; i < Num; i++) {
if (!prime[i]){
for (int j = i + i; j < Num; j += i){
prime[j] = 1;
}
}
}
prime[1] = 1;
}
int MaxSize(int* arr, int N, int K){
vector<int> vect, diff;
//Inserting the indices of composite numbers
for (int i = 0; i < N; i++){
if (prime[arr[i]])
vect.push_back(i);
}
/*Computing the number of prime numbers between
two consecutive composite numbers*/
for (int i = 1; i < vect.size(); i++){
diff.push_back(vect[i] - vect[i - 1] - 1);
}
//Sorting the diff vector
sort(diff.begin(), diff.end());
//Computing the prefix sum of diff vector
for (int i = 1; i < diff.size(); i++){
diff[i] += diff[i - 1];
}
//Impossible case
if (K > N || (K == 0 && vect.size())){
return -1;
}
//Deleting sub-arrays of length 1
else if (vect.size() <= K){
return (N - K);
}
/*Finding the number of primes to be deleted
when deleting the sub-arrays*/
else if (vect.size() > K){
int tt = vect.size() - K;
int sum = 0;
sum += diff[tt - 1];
int res = N - (vect.size() + sum);
return res;
}
}
//Main function
int main(){
sieve();
int arr[] = { 7, 2, 3, 4, 3, 6, 3, 3 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 2;
cout<< MaxSize(arr, N, K);
return 0;
}

## Output

If we run the above code we will get the following output −

6
Published on 17-Aug-2020 09:01:06