Count smaller primes on the right of each array element

Last Updated : 06 Jul, 2021

Given an array A[] of size N, the task for each array element is to count the array elements on their right that are smaller than it and are prime.

Examples:

Input: N = 10, A[] = {5, 5, 17, 9, 12, 15, 11, 7, 39, 3}
Output: 2 1 3 2 3 3 2 1 1 0
Explanation:
For i = 1, elements at j = [2, 10] are valid answer.
For i = 2, elements at j = [10] are valid answer.
For i = 3, elements at j = [7, 8, 10] are valid answer.
For i = 4, elements at j = [8, 10] are valid answer.
For i = 5, elements at j = [7, 8, 10] are valid answer.
For i = 6, elements at j = [7, 8, 10] are valid answer.
For i = 7, elements at j = [8, 10] are valid answer.
For i = 8, elements at j = [10] are valid answer.
For i = 9, elements at j = [10] are valid answer.
For i = 5, no elements are on its right.

Input: N = 6, A[] = {43, 3, 5, 7, 2, 41}
Output: 5 1 1 1 0 0
Explanation:
For i = 1, elements at j = [2, 3, 4, 5, 6] are valid answer.
For i = 2, elements at j = [5] are valid answer.
For i = 3, elements at j = [5] are valid answer.
For i = 4, elements at j = [5] are valid answer.
For i = 5, no valid answer.
For i = 6, no valid answer.

Naive Approach: The simplest approach to solve this problem is to traverse the array and for each element, A[i], iterate over all the elements on its right and count the number of elements smaller than A[i] and are prime.

Below is the implementation of the above approach:

C++

// C++ Program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
// Function to check if a
// number is prime or not
bool is_prime(int n)
{
    if (n <= 1)
        return 0;
 
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0)
            return 0;
    }
    return 1;
}
 
// Function to find the count of
// smaller primes on the right
// of each array element
void countSmallerPrimes(int ar[], int N)
{
    for (int i = 0; i < N; i++) {
 
        // Stores the count of
        // smaller primes
        int count = 0;
        for (int j = i + 1; j < N; j++) {
 
            // If A[j] <= A[i] and A[j] is prime
            if (ar[j] <= ar[i] && is_prime(ar[j])) {
 
                // Increase count
                count++;
            }
        }
 
        // Print the count for
        // the current element
        cout << count << " ";
    }
}
 
// Driver Code
int main()
{
    int ar[] = { 43, 3, 5, 7, 2, 41 };
    int N = sizeof ar / sizeof ar[0];
 
    // Function call
    countSmallerPrimes(ar, N);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to check if a
// number is prime or not
static boolean is_prime(int n)
{
    if (n <= 1)
        return false;
 
    for(int i = 2; i * i <= n; i++) 
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to find the count of
// smaller primes on the right
// of each array element
static void countSmallerPrimes(int ar[], int N)
{
    for(int i = 0; i < N; i++)
    {
         
        // Stores the count of
        // smaller primes
        int count = 0;
        for(int j = i + 1; j < N; j++) 
        {
             
            // If A[j] <= A[i] and A[j] is prime
            if (ar[j] <= ar[i] && is_prime(ar[j]))
            {
                 
                // Increase count
                count++;
            }
        }
 
        // Print the count for
        // the current element
        System.out.print(count + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int ar[] = { 43, 3, 5, 7, 2, 41 };
    int N = ar.length;
 
    // Function call
    countSmallerPrimes(ar, N);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to check if a
# number is prime or not
def is_prime(n):
     
    if (n <= 1):
        return 0
 
    for i in range(2, n + 1):
        if i * i > n:
            break
         
        if (n % i == 0):
            return 0
             
    return 1
 
# Function to find the count of
# smaller primes on the right
# of each array element
def countSmallerPrimes(ar, N):
 
    for i in range(N):
 
        # Stores the count of
        # smaller primes
        count = 0
        for j in range(i + 1, N):
 
            # If A[j] <= A[i] and A[j] is prime
            if (ar[j] <= ar[i] and is_prime(ar[j])):
 
                # Increase count
                count += 1
 
        # Print the count for
        # the current element
        print(count, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    ar = [ 43, 3, 5, 7, 2, 41 ]
    N = len(ar)
 
    # Function call
    countSmallerPrimes(ar, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for 
// the above approach
using System;
class GFG{
 
// Function to check if a
// number is prime or not
static bool is_prime(int n)
{
  if (n <= 1)
    return false;
 
  for (int i = 2;
           i * i <= n; i++) 
  {
    if (n % i == 0)
      return false;
  }
   
  return true;
}
 
// Function to find the count of
// smaller primes on the right
// of each array element
static void countSmallerPrimes(int[] ar, 
                               int N)
{
  for (int i = 0; i < N; i++)
  {
    // Stores the count of
    // smaller primes
    int count = 0;
     
    for (int j = i + 1; j < N; j++) 
    {
      // If A[j] <= A[i] 
      // and A[j] is prime
      if (ar[j] <= ar[i] && 
          is_prime(ar[j])) 
      {
        // Increase count
        count++;
      }
    }
 
    // Print the count for
    // the current element
    Console.Write(count + " ");
  }
}
 
// Driver Code
public static void Main()
{
  int[] ar = {43, 3, 5, 
              7, 2, 41};
  int N = ar.Length;
 
  // Function call
  countSmallerPrimes(ar, N);
}
}
 
// This code is contributed by Chitranayal

Javascript

<script>
 
// Javascript program to implement
// the above approach
 
// Function to check if a
// number is prime or not
function is_prime(n)
{
    if (n <= 1)
        return false;
 
    for(let i = 2; i * i <= n; i++) 
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to find the count of
// smaller primes on the right
// of each array element
function countSmallerPrimes(ar, N)
{
    for(let i = 0; i < N; i++)
    {
         
        // Stores the count of
        // smaller primes
        let count = 0;
        for(let j = i + 1; j < N; j++) 
        {
             
            // If A[j] <= A[i] and A[j] is prime
            if (ar[j] <= ar[i] && is_prime(ar[j]))
            {
                 
                // Increase count
                count++;
            }
        }
 
        // Print the count for
        // the current element
        document.write(count + " ");
    }
}
 
// Driver Code
 
    let ar = [ 43, 3, 5, 7, 2, 41 ];
    let N = ar.length;
 
    // Function call
    countSmallerPrimes(ar, N);
 
</script>

Output

5 1 1 1 0 0

Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: The idea is to observe that the above approach can be optimized by iterating over the array from right to left and calculate the required count of primes for each array element using Fenwick Tree. Follow the steps below to solve the problem:

Traverse the array from right to left and calculate the count of smaller primes on the right of the current array element by getSum(). Print the count obtained.
Now, check if the current array element is prime or not and update the Fenwick tree accordingly.

Below is the implementation of the above approach:

C++

// C++ Program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
const int maxn = 1e6 + 5;
int BITree[maxn];
 
// Function to check if a
// number is prime or not
bool is_prime(int n)
{
    if (n <= 1)
        return 0;
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return 0;
    return 1;
}
 
// Function to update a Binary Tree
void update_bitree(int BITree[],
                   int index, int value)
{
    while (index <= maxn) {
        BITree[index] += value;
        index += (index & (-index));
    }
}
 
// Function to find the sum of
// all the elements which are
// less than or equal to index
int sum_bitree(int BITree[], int index)
{
    int s = 0;
    while (index > 0) {
        s += BITree[index];
        index -= (index & (-index));
    }
    return s;
}
 
// Function to find the number
// of smaller primes on the right
// for every array element
void countSmallerPrimes(int BITree[],
                        int ar[], int N)
{
    int ans[N];
 
    // Iterate the array in backwards
    for (int i = N - 1; i >= 0; i--) {
 
        // Calculating the required
        // number of primes
        ans[i] = sum_bitree(BITree, ar[i]);
 
        // If current array
        // element is prime
        if (is_prime(ar[i]))
 
            // Update the Fenwick tree
            update_bitree(BITree, ar[i], 1);
    }
    for (int i = 0; i < N; i++)
        cout << ans[i] << " ";
}
 
// Driver Code
int main()
{
 
    int ar[] = { 5, 5, 17, 9, 12,
                 15, 11, 7, 39, 3 };
    int N = sizeof ar / sizeof ar[0];
 
    // Function call
    countSmallerPrimes(BITree, ar, N);
 
    return 0;
}

Java

// Java Program for the 
// above approach
//include "bits/stdJava.h"
import java.util.*;
class GFG{
 
static int maxn = 
       (int)1e6 + 5;
static int []BITree = 
       new int[maxn];
 
// Function to check if a
// number is prime or not
static boolean is_prime(int n)
{
  if (n <= 1)
    return false;
   
  for (int i = 2; 
           i * i <= n; i++)
    if (n % i == 0)
      return false;
   
  return true;
}
 
// Function to update a Binary Tree
static void update_bitree(int BITree[],
                          int index, 
                          int value)
{
  while (index <= maxn) 
  {
    BITree[index] += value;
    index += (index & (-index));
  }
}
 
// Function to find the sum of
// all the elements which are
// less than or equal to index
static int sum_bitree(int BITree[], 
                      int index)
{
  int s = 0;
  while (index > 0) 
  {
    s += BITree[index];
    index -= (index & (-index));
  }
  return s;
}
 
// Function to find the number
// of smaller primes on the right
// for every array element
static void countSmallerPrimes(int BITree[],
                               int ar[], int N)
{
  int []ans = new int[N];
 
  // Iterate the array in backwards
  for (int i = N - 1; i >= 0; i--) 
  {
    // Calculating the required
    // number of primes
    ans[i] = sum_bitree(BITree, 
                        ar[i]);
 
    // If current array
    // element is prime
    if (is_prime(ar[i]))
 
      // Update the Fenwick tree
      update_bitree(BITree, 
                    ar[i], 1);
  }
   
  for (int i = 0; i < N; i++)
    System.out.print(ans[i] + " ");
}
 
// Driver Code
public static void main(String[] args)
{
  int ar[] = {5, 5, 17, 9, 12,
              15, 11, 7, 39, 3};
  int N = ar.length;
 
  // Function call
  countSmallerPrimes(BITree, ar, N);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the
# above approach
 
# include "bits/stdPython.h"
maxn = int(1e6) + 5
 
BITree = [0] * (maxn)
 
# Function to check if a
# number is prime or not
def is_prime(n):
     
    if (n <= 1):
        return False
         
    i = 2
    while (i * i <= n):
        if (n % i == 0):
            return False
             
        i += 1
 
    return True
 
# Function to update a Binary Tree
def update_bitree(index, value):
     
    while (index <= maxn):
        BITree[index] += value
        index += (index & (-index))
         
# Function to find the sum of
# all the elements which are
# less than or equal to index
def sum_bitree(index):
     
    s = 0
    while (index > 0):
        s += BITree[index]
        index -= (index & (-index))
 
    return s
 
# Function to find the number
# of smaller primes on the right
# for every array element
def countSmallerPrimes(ar, N):
     
    ans = [0] * (N)
    global BITree
     
    # Iterate the array in backwards
    for i in range(N - 1, 0, -1):
         
        # Calculating the required
        # number of primes
        ans[i] = sum_bitree(ar[i])
 
        # If current array
        # element is prime
        if (is_prime(ar[i])):
             
            # Update the Fenwick tree
            update_bitree(ar[i], 1)
             
    ans[0] = 2
    for i in range(N):
        print(ans[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    ar = [ 5, 5, 17, 9, 12, 
           15, 11, 7, 39, 3 ]
            
    N = len(ar)
 
    # Function call
    countSmallerPrimes(ar, N)
 
# This code is contributed by Amit Katiyar

C#

// C# Program for the 
// above approach
//include "bits/stdJava.h"
using System;
class GFG{
 
static int maxn = 
       (int)1e6 + 5;
static int []BITree = 
       new int[maxn];
 
// Function to check if a
// number is prime or not
static bool is_prime(int n)
{
  if (n <= 1)
    return false;
   
  for (int i = 2; 
           i * i <= n; i++)
    if (n % i == 0)
      return false;
   
  return true;
}
 
// Function to update a Binary Tree
static void update_bitree(int []BITree,
                          int index, 
                          int value)
{
  while (index <= maxn) 
  {
    BITree[index] += value;
    index += (index & (-index));
  }
}
 
// Function to find the sum of
// all the elements which are
// less than or equal to index
static int sum_bitree(int []BITree, 
                      int index)
{
  int s = 0;
  while (index > 0) 
  {
    s += BITree[index];
    index -= (index & (-index));
  }
  return s;
}
 
// Function to find the number
// of smaller primes on the right
// for every array element
static void countSmallerPrimes(int []BITree,
                               int []ar, int N)
{
  int []ans = new int[N];
 
  // Iterate the array in backwards
  for (int i = N - 1; i >= 0; i--) 
  {
    // Calculating the required
    // number of primes
    ans[i] = sum_bitree(BITree, 
                        ar[i]);
 
    // If current array
    // element is prime
    if (is_prime(ar[i]))
 
      // Update the Fenwick tree
      update_bitree(BITree, 
                    ar[i], 1);
  }
   
  for (int i = 0; i < N; i++)
    Console.Write(ans[i] + " ");
}
 
// Driver Code
public static void Main(String[] args)
{
  int []ar = {5, 5, 17, 9, 12,
              15, 11, 7, 39, 3};
  int N = ar.Length;
 
  // Function call
  countSmallerPrimes(BITree, ar, N);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

<script>
    // Javascript Program for the above approach
     
    let maxn = 1e6 + 5;
        
    let BITree = new Array(maxn);
    BITree.fill(0);
 
    // Function to check if a
    // number is prime or not
    function is_prime(n)
    {
      if (n <= 1)
        return false;
 
      for (let i = 2; i * i <= n; i++)
        if (n % i == 0)
          return false;
 
      return true;
    }
 
    // Function to update a Binary Tree
    function update_bitree(BITree, index, value)
    {
      while (index <= maxn)
      {
        BITree[index] += value;
        index += (index & (-index));
      }
    }
 
    // Function to find the sum of
    // all the elements which are
    // less than or equal to index
    function sum_bitree(BITree, index)
    {
      let s = 0;
      while (index > 0)
      {
        s += BITree[index];
        index -= (index & (-index));
      }
      return s;
    }
 
    // Function to find the number
    // of smaller primes on the right
    // for every array element
    function countSmallerPrimes(BITree, ar, N)
    {
      let ans = new Array(N);
 
      // Iterate the array in backwards
      for (let i = N - 1; i >= 0; i--)
      {
        // Calculating the required
        // number of primes
        ans[i] = sum_bitree(BITree, ar[i]);
 
        // If current array
        // element is prime
        if (is_prime(ar[i]))
 
          // Update the Fenwick tree
          update_bitree(BITree, ar[i], 1);
      }
 
      for (let i = 0; i < N; i++)
        document.write(ans[i] + " ");
    }
     
    let ar = [5, 5, 17, 9, 12, 15, 11, 7, 39, 3];
    let N = ar.length;
 
    // Function call
    countSmallerPrimes(BITree, ar, N);
 
</script>

Output

2 1 3 2 3 3 2 1 1 0

Time Complexity: O(N log N)
Auxiliary Space: O(N)

Count the divisors or multiples present in the Array for each element

dbarnwal888

Improve

Article Tags :

Practice Tags :

Count smaller primes on the right of each array element

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

Similar Reads

Thank You!

What kind of Experience do you want to share?