Given a positive integer **N**, the task is to count the number of integers from the range **[1, N]** having exactly 5 divisors.

**Examples:**

Input:N = 18Output:1Explanation:

From all the integers over the range [1, 18], 16 is the only integer that has exactly 5 divisors, i.e. 1, 2, 8, 4 and 16.

Therefore, the count of such integers is 1.

Input:N = 100Output:2

**Naive Approach:** The simplest approach to solve the given problem is to iterate over the range **[1, N]** and count those integers in this range having the count of divisors as **5**. **Time Complexity:** O(N^{4/3})**Auxiliary Space:** O(1)

**Efficient Approach:** The above approach can also be optimized by observing a fact that the numbers that have exactly **5 divisors** can be expressed in the form of **p ^{4}**, where

**p**is a prime number as the count of divisors is exactly

**5**. Follow the below steps to solve the problem:

- Generate all primes such that their fourth power is less than
**10**by using Sieve of Eratosthenes and store it in vector, say^{18}**A[]**. - Initialize two variables, say
**low**as**0**and**high**as**A.size() – 1**. - For performing the Binary Search iterate until
**low**is less than**high**and perform the following steps:- Find the value of
**mid**as the**(low + high)/2**. - Find the value of fourth power of element at indices
**mid****(mid – 1)**and store it in a variable, say**current**and**previous**respectively. - If the value of
**current**is**N**, then print the value of**A[mid]**as the result. - If the value of
**current**is greater than**N**and**previous**is**at most N**, then print the value of**A[mid]**as the result. - If the value of
**current**is greater than**N**then update the value of**high**as**(mid – 1)**. Otherwise, update the value of**low**as**(mid + 1)**.

- Find the value of

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `#define ll long long int` `const` `int` `MAX = 1e5;` `using` `namespace` `std;` `// Function to calculate the value of` `// (x^y) using binary exponentiation` `ll power(ll x, unsigned ll y)` `{` ` ` `// Stores the value of x^y` ` ` `ll res = 1;` ` ` `// Base Case` ` ` `if` `(x == 0)` ` ` `return` `0;` ` ` `while` `(y > 0) {` ` ` `// If y is odd multiply` ` ` `// x with result` ` ` `if` `(y & 1)` ` ` `res = (res * x);` ` ` `// Otherwise, divide y by 2` ` ` `y = y >> 1;` ` ` `x = (x * x);` ` ` `}` ` ` `return` `res;` `}` `// Function to perform the Sieve Of` `// Eratosthenes to find the prime` `// number over the range [1, 10^5]` `void` `SieveOfEratosthenes(` ` ` `vector<pair<ll, ll> >& v)` `{` ` ` `bool` `prime[MAX + 1];` ` ` `memset` `(prime, ` `true` `, ` `sizeof` `(prime));` ` ` `prime[1] = ` `false` `;` ` ` `for` `(` `int` `p = 2; p * p <= MAX; p++) {` ` ` `// If prime[p] is not changed` ` ` `// then it is a prime` ` ` `if` `(prime[p] == ` `true` `) {` ` ` `// Set all the multiples of` ` ` `// p to non-prime` ` ` `for` `(` `int` `i = p * 2;` ` ` `i <= MAX; i += p)` ` ` `prime[i] = ` `false` `;` ` ` `}` ` ` `}` ` ` `int` `num = 1;` ` ` `// Iterate over the range [1, MAX]` ` ` `for` `(` `int` `i = 1; i <= MAX; i++) {` ` ` `// Store all the prime number` ` ` `if` `(prime[i]) {` ` ` `v.push_back({ i, num });` ` ` `num++;` ` ` `}` ` ` `}` `}` `// Function to find the primes having` `// only 5 divisors` `int` `countIntegers(ll n)` `{` ` ` `// Base Case` ` ` `if` `(n < 16) {` ` ` `return` `0;` ` ` `}` ` ` `// First value of the pair has the` ` ` `// prime number and the second value` ` ` `// has the count of primes till that` ` ` `// prime numbers` ` ` `vector<pair<ll, ll> > v;` ` ` `// Precomputing all the primes` ` ` `SieveOfEratosthenes(v);` ` ` `int` `low = 0;` ` ` `int` `high = v.size() - 1;` ` ` `// Perform the Binary search` ` ` `while` `(low <= high) {` ` ` `int` `mid = (low + high) / 2;` ` ` `// Calculate the fourth power of` ` ` `// the curr and prev` ` ` `ll curr = power(v[mid].first, 4);` ` ` `ll prev = power(v[mid - 1].first, 4);` ` ` `if` `(curr == n) {` ` ` `// Return value of mid` ` ` `return` `v[mid].second;` ` ` `}` ` ` `else` `if` `(curr > n and prev <= n) {` ` ` `// Return value of mid-1` ` ` `return` `v[mid - 1].second;` ` ` `}` ` ` `else` `if` `(curr > n) {` ` ` `// Update the value of high` ` ` `high = mid - 1;` ` ` `}` ` ` `else` `{` ` ` `// Update the value of low` ` ` `low = mid + 1;` ` ` `}` ` ` `}` ` ` `return` `0;` `}` `// Driver Code` `int` `main()` `{` ` ` `ll N = 100;` ` ` `cout << countIntegers(N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.Vector;` `public` `class` `GFG {` ` ` `static` `int` `MAX = (` `int` `)1e5;` ` ` `public` `static` `class` `pair {` ` ` `long` `first;` ` ` `long` `second;` ` ` `pair(` `long` `first, ` `long` `second)` ` ` `{` ` ` `this` `.first = first;` ` ` `this` `.second = second;` ` ` `}` ` ` `}` ` ` `// Function to calculate the value of` ` ` `// (x^y) using binary exponentiation` ` ` `static` `long` `power(` `long` `x, ` `long` `y)` ` ` `{` ` ` ` ` `// Stores the value of x^y` ` ` `long` `res = ` `1` `;` ` ` `// Base Case` ` ` `if` `(x == ` `0` `)` ` ` `return` `0` `;` ` ` `while` `(y > ` `0` `) ` ` ` `{` ` ` `// If y is odd multiply` ` ` `// x with result` ` ` `if` `((y & ` `1` `) == ` `1` `)` ` ` `res = (res * x);` ` ` `// Otherwise, divide y by 2` ` ` `y = y >> ` `1` `;` ` ` `x = (x * x);` ` ` `}` ` ` `return` `res;` ` ` `}` ` ` `// Function to perform the Sieve Of` ` ` `// Eratosthenes to find the prime` ` ` `// number over the range [1, 10^5]` ` ` `static` `void` `SieveOfEratosthenes(Vector<pair> v)` ` ` `{` ` ` `boolean` `prime[] = ` `new` `boolean` `[MAX + ` `1` `];` ` ` `for` `(` `int` `i = ` `0` `; i < prime.length; i++) {` ` ` `prime[i] = ` `true` `;` ` ` `}` ` ` `prime[` `1` `] = ` `false` `;` ` ` `for` `(` `int` `p = ` `2` `; p * p <= MAX; p++) {` ` ` `// If prime[p] is not changed` ` ` `// then it is a prime` ` ` `if` `(prime[p] == ` `true` `) {` ` ` `// Set all the multiples of` ` ` `// p to non-prime` ` ` `for` `(` `int` `i = p * ` `2` `; i <= MAX; i += p)` ` ` `prime[i] = ` `false` `;` ` ` `}` ` ` `}` ` ` `int` `num = ` `1` `;` ` ` `// Iterate over the range [1, MAX]` ` ` `for` `(` `int` `i = ` `1` `; i <= MAX; i++) {` ` ` `// Store all the prime number` ` ` `if` `(prime[i]) {` ` ` `v.add(` `new` `pair(i, num));` ` ` `num++;` ` ` `}` ` ` `}` ` ` `}` ` ` `// Function to find the primes having` ` ` `// only 5 divisors` ` ` `static` `long` `countIntegers(` `long` `n)` ` ` `{` ` ` `// Base Case` ` ` `if` `(n < ` `16` `) {` ` ` `return` `0` `;` ` ` `}` ` ` `// First value of the pair has the` ` ` `// prime number and the second value` ` ` `// has the count of primes till that` ` ` `// prime numbers` ` ` `Vector<pair> v = ` `new` `Vector<>();` ` ` `// Precomputing all the primes` ` ` `SieveOfEratosthenes(v);` ` ` `int` `low = ` `0` `;` ` ` `int` `high = v.size() - ` `1` `;` ` ` `// Perform the Binary search` ` ` `while` `(low <= high) {` ` ` `int` `mid = (low + high) / ` `2` `;` ` ` `// Calculate the fourth power of` ` ` `// the curr and prev` ` ` `long` `curr = power(v.get(mid).first, ` `4` `);` ` ` `long` `prev = power(v.get(mid - ` `1` `).first, ` `4` `);` ` ` `if` `(curr == n) {` ` ` `// Return value of mid` ` ` `return` `v.get(mid).second;` ` ` `}` ` ` `else` `if` `(curr > n && prev <= n) {` ` ` `// Return value of mid-1` ` ` `return` `v.get(mid - ` `1` `).second;` ` ` `}` ` ` `else` `if` `(curr > n) {` ` ` `// Update the value of high` ` ` `high = mid - ` `1` `;` ` ` `}` ` ` `else` `{` ` ` `// Update the value of low` ` ` `low = mid + ` `1` `;` ` ` `}` ` ` `}` ` ` `return` `0` `;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `long` `N = ` `100` `;` ` ` `System.out.println(countIntegers(N));` ` ` `}` `}` `// This code is contributed by abhinavjain194` |

**Output:**

2

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

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