Count numbers in a given range whose count of prime factors is a Prime Number
Last Updated :
26 Jul, 2025
Given a 2D array Q[][] of size N * 2 representing queries of the form {L, R}. For each query, the task is to print the count of numbers in the range [L, R] with a count of prime factors equal to a prime number.
Examples:
Input: Q[][] = {{4, 8}, {30, 32}}
Output: 3 2
Explanation:
Query 1:
Prime factors of 4 = {2, 2} and count of prime factors = 2
Prime factors of 5 = {5} and count of prime factors = 1
Prime factors of 6 = {2, 3} and count of prime factors = 2
Prime factors of 7 = {7} and count of prime factors = 1
Prime factors of 8 = {2, 2, 2} and count of prime factors = 3
Therefore, the total count of numbers in the range [4, 8] having count of prime factors is a prime number is 3.
Query 2:
Prime factors of 30 = {2, 3, 5} and count of prime factors = 3
Prime factors of 31 = {31} and count of prime factors = 1
Prime factors of 32 = {2, 2, 2, 2, 2} and count of prime factors = 5
Therefore, the total count of numbers in the range [4, 8] having count of prime factors is a prime number is 2.
Input: Q[][] = {{7, 12}, {10, 99}}
Output: 4
Naive Approach: The simplest approach to solve this problem is to traverse all the numbers in the range [L, R], and for each number, check if the count of prime factors of the number is a prime number or not. If found to be true, increment the counter by 1. After traversing, print the value of counter for each query.
Time Complexity: O(|Q| * (max(arr[i][1] - arr[i][0] + 1)) * sqrt(max(arr[i][1]))
Auxiliary space: O (1)
Efficient Approach: To optimize the above approach the idea is to precompute the smallest prime factor of each number in the range [Li, Ri] using Sieve of Eratosthenes. Follow the steps below to solve the problem:
- Generate and store the smallest prime factor of each element using Sieve of Eratosthenes.
- Find the count of prime factors for each number in the range [Li, Ri] using the Sieve.
- For each number, check if the total count of prime factors is a prime number or not. If found to be true then increment the counter.
- Create a prefix sum array, say sum[], where sum[i] will store the sum of elements from the range [0, i] whose count of prime factors is a prime number.
- Finally, for each query, print the value sum[arr[i][1]] – sum[arr[i][0] - 1].
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 1001
// Function to find the smallest prime factor
// of all the numbers in range [0, MAX]
vector<int> sieve()
{
// Stores smallest prime factor of all
// the numbers in the range [0, MAX]
vector<int> spf(MAX);
// No smallest prime factor of
// 0 and 1 exists
spf[0] = spf[1] = -1;
// Traverse all the numbers
// in the range [1, MAX]
for (int i = 2; i < MAX; i++) {
// Update spf[i]
spf[i] = i;
}
// Update all the numbers whose
// smallest prime factor is 2
for (int i = 4; i < MAX; i = i + 2) {
spf[i] = 2;
}
// Traverse all the numbers in
// the range [1, sqrt(MAX)]
for (int i = 3; i * i < MAX; i++) {
// Check if i is a prime number
if (spf[i] == i) {
// Update all the numbers whose
// smallest prime factor is i
for (int j = i * i; j < MAX;
j = j + i) {
// Check if j is
// a prime number
if (spf[j] == j) {
spf[j] = i;
}
}
}
}
return spf;
}
// Function to find count of
// prime factor of num
int countFactors(vector<int>& spf, int num)
{
// Stores count of
// prime factor of num
int count = 0;
// Calculate count of
// prime factor
while (num > 1) {
// Update count
count++;
// Update num
num = num / spf[num];
}
return count;
}
// Function to precalculate the count of
// numbers in the range [0, i] whose count
// of prime factors is a prime number
vector<int> precalculateSum(vector<int>& spf)
{
// Stores the sum of all the numbers
// in the range[0, i] count of
// prime factor is a prime number
vector<int> sum(MAX);
// Update sum[0]
sum[0] = 0;
// Traverse all the numbers in
// the range [1, MAX]
for (int i = 1; i < MAX; i++) {
// Stores count of prime factor of i
int prime_factor
= countFactors(spf, i);
// If count of prime factor is
// a prime number
if (spf[prime_factor] == prime_factor) {
// Update sum[i]
sum[i] = sum[i - 1] + 1;
}
else {
// Update sum[i]
sum[i] = sum[i - 1];
}
}
return sum;
}
// Driver Code
int main()
{
// Stores smallest prime factor of all
// the numbers in the range [0, MAX]
vector<int> spf = sieve();
// Stores the sum of all the numbers
// in the range[0, i] count of
// prime factor is a prime number
vector<int> sum = precalculateSum(spf);
int Q[][2] = { { 4, 8 }, { 30, 32 } };
// int N = sizeof(Q) / sizeof(Q[0]);
for (int i = 0; i < 2; i++) {
cout << (sum[Q[i][1]] - sum[Q[i][0] - 1])
<< " ";
}
return 0;
}
// Java program to implement
// the above approach
import java.util.*;
class GFG{
public static int MAX = 1001;
// Function to find the smallest prime factor
// of all the numbers in range [0, MAX]
public static int[] sieve()
{
// Stores smallest prime factor of all
// the numbers in the range [0, MAX]
int spf[] = new int[MAX];
// No smallest prime factor of
// 0 and 1 exists
spf[0] = spf[1] = -1;
// Traverse all the numbers
// in the range [1, MAX]
for(int i = 2; i < MAX; i++)
{
// Update spf[i]
spf[i] = i;
}
// Update all the numbers whose
// smallest prime factor is 2
for(int i = 4; i < MAX; i = i + 2)
{
spf[i] = 2;
}
// Traverse all the numbers in
// the range [1, sqrt(MAX)]
for(int i = 3; i * i < MAX; i++)
{
// Check if i is a prime number
if (spf[i] == i)
{
// Update all the numbers whose
// smallest prime factor is i
for(int j = i * i; j < MAX; j = j + i)
{
// Check if j is
// a prime number
if (spf[j] == j)
{
spf[j] = i;
}
}
}
}
return spf;
}
// Function to find count of
// prime factor of num
public static int countFactors(int spf[], int num)
{
// Stores count of
// prime factor of num
int count = 0;
// Calculate count of
// prime factor
while (num > 1)
{
// Update count
count++;
// Update num
num = num / spf[num];
}
return count;
}
// Function to precalculate the count of
// numbers in the range [0, i] whose count
// of prime factors is a prime number
public static int[] precalculateSum(int spf[])
{
// Stores the sum of all the numbers
// in the range[0, i] count of
// prime factor is a prime number
int sum[] = new int[MAX];
// Update sum[0]
sum[0] = 0;
// Traverse all the numbers in
// the range [1, MAX]
for(int i = 1; i < MAX; i++)
{
// Stores count of prime factor of i
int prime_factor = countFactors(spf, i);
// If count of prime factor is
// a prime number
if (spf[prime_factor] == prime_factor)
{
// Update sum[i]
sum[i] = sum[i - 1] + 1;
}
else
{
// Update sum[i]
sum[i] = sum[i - 1];
}
}
return sum;
}
// Driver code
public static void main(String[] args)
{
// Stores smallest prime factor of all
// the numbers in the range [0, MAX]
int spf[] = sieve();
// Stores the sum of all the numbers
// in the range[0, i] count of
// prime factor is a prime number
int sum[] = precalculateSum(spf);
int Q[][] = { { 4, 8 }, { 30, 32 } };
// int N = sizeof(Q) / sizeof(Q[0]);
for(int i = 0; i < 2; i++)
{
System.out.print((sum[Q[i][1]] -
sum[Q[i][0] - 1]) + " ");
}
}
}
// This code is contributed by divyeshrabadiya07
# Python3 program to implement
# the above approach
MAX = 1001
# Function to find the smallest
# prime factor of all the numbers
# in range [0, MAX]
def sieve():
# Stores smallest prime factor of all
# the numbers in the range [0, MAX]
global MAX
spf = [0] * MAX
# No smallest prime factor of
# 0 and 1 exists
spf[0] = spf[1] = -1
# Traverse all the numbers
# in the range [1, MAX]
for i in range(2, MAX):
# Update spf[i]
spf[i] = i
# Update all the numbers whose
# smallest prime factor is 2
for i in range(4, MAX, 2):
spf[i] = 2
# Traverse all the numbers in
# the range [1, sqrt(MAX)]
for i in range(3, MAX):
# Check if i is a prime number
if (spf[i] == i):
# Update all the numbers whose
# smallest prime factor is i
for j in range(i * i, MAX):
# Check if j is
# a prime number
if (spf[j] == j):
spf[j] = i
return spf
# Function to find count of
# prime factor of num
def countFactors(spf, num):
# Stores count of
# prime factor of num
count = 0
# Calculate count of
# prime factor
while (num > 1):
# Update count
count += 1
# Update num
num = num // spf[num]
return count
# Function to precalculate the count of
# numbers in the range [0, i] whose count
# of prime factors is a prime number
def precalculateSum(spf):
# Stores the sum of all the numbers
# in the range[0, i] count of
# prime factor is a prime number
sum = [0] * MAX
# Traverse all the numbers in
# the range [1, MAX]
for i in range(1, MAX):
# Stores count of prime factor of i
prime_factor = countFactors(spf, i)
# If count of prime factor is
# a prime number
if (spf[prime_factor] == prime_factor):
# Update sum[i]
sum[i] = sum[i - 1] + 1
else:
# Update sum[i]
sum[i] = sum[i - 1]
return sum
# Driver code
if __name__ == '__main__':
# Stores smallest prime factor of all
# the numbers in the range [0, MAX]
spf = sieve()
# Stores the sum of all the numbers
# in the range[0, i] count of
# prime factor is a prime number
sum = precalculateSum(spf)
Q = [ [ 4, 8 ], [ 30, 32 ] ]
sum[Q[0][1]] += 1
# N = sizeof(Q) / sizeof(Q[0]);
for i in range(0, 2):
print((sum[Q[i][1]] -
sum[Q[i][0]]), end = " ")
# This code is contributed by Princi Singh
// C# program to implement
// the above approach
using System;
class GFG{
public static int MAX = 1001;
// Function to find the smallest
// prime factor of all the numbers
// in range [0, MAX]
public static int[] sieve()
{
// Stores smallest prime factor
// of all the numbers in the
// range [0, MAX]
int []spf = new int[MAX];
// No smallest prime factor
// of 0 and 1 exists
spf[0] = spf[1] = -1;
// Traverse all the numbers
// in the range [1, MAX]
for(int i = 2; i < MAX; i++)
{
// Update spf[i]
spf[i] = i;
}
// Update all the numbers whose
// smallest prime factor is 2
for(int i = 4; i < MAX; i = i + 2)
{
spf[i] = 2;
}
// Traverse all the numbers in
// the range [1, sqrt(MAX)]
for(int i = 3; i * i < MAX; i++)
{
// Check if i is a prime number
if (spf[i] == i)
{
// Update all the numbers
// whose smallest prime
// factor is i
for(int j = i * i;
j < MAX; j = j + i)
{
// Check if j is
// a prime number
if (spf[j] == j)
{
spf[j] = i;
}
}
}
}
return spf;
}
// Function to find count of
// prime factor of num
public static int countFactors(int []spf,
int num)
{
// Stores count of
// prime factor of num
int count = 0;
// Calculate count of
// prime factor
while (num > 1)
{
// Update count
count++;
// Update num
num = num / spf[num];
}
return count;
}
// Function to precalculate the count of
// numbers in the range [0, i] whose count
// of prime factors is a prime number
public static int[] precalculateSum(int []spf)
{
// Stores the sum of all the numbers
// in the range[0, i] count of
// prime factor is a prime number
int []sum = new int[MAX];
// Update sum[0]
sum[0] = 0;
// Traverse all the numbers in
// the range [1, MAX]
for(int i = 1; i < MAX; i++)
{
// Stores count of prime factor of i
int prime_factor = countFactors(spf, i);
// If count of prime factor is
// a prime number
if (spf[prime_factor] == prime_factor)
{
// Update sum[i]
sum[i] = sum[i - 1] + 1;
}
else
{
// Update sum[i]
sum[i] = sum[i - 1];
}
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
// Stores smallest prime factor
// of all the numbers in the
// range [0, MAX]
int []spf = sieve();
// Stores the sum of all the
// numbers in the range[0, i]
// count of prime factor is a
// prime number
int []sum = precalculateSum(spf);
int [,]Q = {{4, 8}, {30, 32}};
// int N = sizeof(Q) / sizeof(Q[0]);
for(int i = 0; i < 2; i++)
{
Console.Write((sum[Q[i, 1]] -
sum[Q[i, 0] - 1]) +
" ");
}
}
}
// This code is contributed by shikhasingrajput
<script>
// Javascript program to implement
// the above approach
let MAX = 1001
// Function to find the smallest prime factor
// of all the numbers in range [0, MAX]
function sieve()
{
// Stores smallest prime factor of all
// the numbers in the range [0, MAX]
let spf = new Array(MAX);
// No smallest prime factor of
// 0 and 1 exists
spf[0] = spf[1] = -1;
// Traverse all the numbers
// in the range [1, MAX]
for (let i = 2; i < MAX; i++) {
// Update spf[i]
spf[i] = i;
}
// Update all the numbers whose
// smallest prime factor is 2
for (let i = 4; i < MAX; i = i + 2) {
spf[i] = 2;
}
// Traverse all the numbers in
// the range [1, sqrt(MAX)]
for (let i = 3; i * i < MAX; i++) {
// Check if i is a prime number
if (spf[i] == i) {
// Update all the numbers whose
// smallest prime factor is i
for (let j = i * i; j < MAX;
j = j + i) {
// Check if j is
// a prime number
if (spf[j] == j) {
spf[j] = i;
}
}
}
}
return spf;
}
// Function to find count of
// prime factor of num
function countFactors(spf, num)
{
// Stores count of
// prime factor of num
let count = 0;
// Calculate count of
// prime factor
while (num > 1) {
// Update count
count++;
// Update num
num = num / spf[num];
}
return count;
}
// Function to precalculate the count of
// numbers in the range [0, i] whose count
// of prime factors is a prime number
function precalculateSum(spf)
{
// Stores the sum of all the numbers
// in the range[0, i] count of
// prime factor is a prime number
let sum = new Array(MAX);
// Update sum[0]
sum[0] = 0;
// Traverse all the numbers in
// the range [1, MAX]
for (let i = 1; i < MAX; i++) {
// Stores count of prime factor of i
let prime_factor
= countFactors(spf, i);
// If count of prime factor is
// a prime number
if (spf[prime_factor] == prime_factor) {
// Update sum[i]
sum[i] = sum[i - 1] + 1;
}
else {
// Update sum[i]
sum[i] = sum[i - 1];
}
}
return sum;
}
// Driver Code
// Stores smallest prime factor of all
// the numbers in the range [0, MAX]
let spf = sieve();
// Stores the sum of all the numbers
// in the range[0, i] count of
// prime factor is a prime number
let sum = precalculateSum(spf);
let Q = [ [ 4, 8 ], [ 30, 32 ] ];
// let N = sizeof(Q) / sizeof(Q[0]);
for (let i = 0; i < 2; i++) {
document.write((sum[Q[i][1]] - sum[Q[i][0] - 1]) +
" ");
}
// This code is contributed by gfgking
</script>
Time Complexity: O(|Q| + (MAX *log(log(MAX))))
Auxiliary Space: O(MAX)
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