Smallest integer having at least K prime divisors with difference between each factor at least D

Given two integers D and K. The task is to find the smallest number N which has at least K prime divisors and the difference between each pair of divisors is at least D

Examples

Input: D = 3, K = 2
Output: 55
Explanation: It is smallest number which has 4 divisors 1 and 2 prime divisors  5, 11 and their difference between any of the pair is D.

Input: D = 1, K = 4
Output: 210
Explanation: It is the smallest number which has 5 divisors 1 and 4 prime divisors 2, 3, 5, 7, and their difference between any of the pair is D.

Approach: This problem can be solved by using the Sieve of Eratosthenes Follow the steps below to solve the given problem.

• Make a sieve of Eratosthenes.
• Initialize a variable firstDivisor and store D + 1.
• Iterate firstDivisor by 1 until it becomes prime.
• Initialize SmallestNumber = FirstDivisor + D.
• Now iterate the loop and increment SmallestNumber until we get K -1 primes.
• And, the product with all divisors and return the product.

Below is the implementation of the above approach.

C++

 `// C++ program for above approach` `#include ` `using` `namespace` `std;` `const` `int` `N = 1000000;`   `// Function of Sieve of Eratosthenes` `void` `SieveOfEratosthenes(vector<``bool``>& prime)` `{` `    ``for` `(``int` `p = 2; p * p <= N; p++) {` `        ``if` `(prime[p] == ``true``) {` `            ``for` `(``int` `i = p * p; i <= N; i += p)` `                ``prime[i] = ``false``;` `        ``}` `    ``}` `}`   `// Function to find smallest` `// number with given conditions` `int` `SmallestNumber(vector<``bool``> prime,` `                   ``int` `D, ``int` `K)` `{`   `    ``// Initialize first with D + 1` `    ``// because 1 is also a divisor` `    ``int` `FirstDivisor = D + 1;`   `    ``while` `(FirstDivisor < N` `           ``and !prime[FirstDivisor]) {` `        ``++FirstDivisor;` `    ``}`   `    ``// Now value of K is decrement by 1` `    ``K--;`   `    ``// Initialize Divisor with First + D` `    ``// to maintain a difference D` `    ``// We get Remaining divisor` `    ``int` `SmallestNumber = FirstDivisor;` `    ``int` `Divisor = FirstDivisor + D;`   `    ``// Maintain previous divisor` `    ``// to maintain difference` `    ``int` `prevDivisor = FirstDivisor;` `    ``while` `(K > 0 and SmallestNumber < N) {` `        ``if` `(prime[Divisor]` `            ``and Divisor - D >= prevDivisor) {` `            ``SmallestNumber *= Divisor;` `            ``prevDivisor = Divisor;` `            ``K--;` `        ``}` `        ``Divisor++;` `    ``}`   `    ``// Return the final answer` `    ``return` `SmallestNumber;` `}`   `// Driver Code` `int` `main()` `{` `    ``vector<``bool``> prime(N, ``true``);`   `    ``SieveOfEratosthenes(prime);`   `    ``int` `D = 1;` `    ``int` `K = 4;`   `    ``// Function Call` `    ``cout << SmallestNumber(prime, D, K);`   `    ``return` `0;` `}`

Java

 `// Java program to implement` `// the above approach` `import` `java.util.*;`   `class` `GFG` `{`   `static` `int` `N = ``1000000``;`   `// Function of Sieve of Eratosthenes` `static` `void` `SieveOfEratosthenes(``boolean``[] prime)` `{` `    ``for` `(``int` `p = ``2``; p * p < N; p++) {` `        ``if` `(prime[p] == ``true``) {` `            ``for` `(``int` `i = p * p; i < N; i += p)` `                ``prime[i] = ``false``;` `        ``}` `    ``}` `}`   `// Function to find smallest` `// number with given conditions` `static` `int` `SmallestNumber(``boolean``[] prime,` `                   ``int` `D, ``int` `K)` `{`   `    ``// Initialize first with D + 1` `    ``// because 1 is also a divisor` `    ``int` `FirstDivisor = D + ``1``;`   `    ``while` `(FirstDivisor < N` `           ``&& !prime[FirstDivisor]) {` `        ``++FirstDivisor;` `    ``}`   `    ``// Now value of K is decrement by 1` `    ``K--;`   `    ``// Initialize Divisor with First + D` `    ``// to maintain a difference D` `    ``// We get Remaining divisor` `    ``int` `SmallestNumber = FirstDivisor;` `    ``int` `Divisor = FirstDivisor + D;`   `    ``// Maintain previous divisor` `    ``// to maintain difference` `    ``int` `prevDivisor = FirstDivisor;` `    ``while` `(K > ``0` `&& SmallestNumber < N) {` `        ``if` `(prime[Divisor]` `            ``&& Divisor - D >= prevDivisor) {` `            ``SmallestNumber *= Divisor;` `            ``prevDivisor = Divisor;` `            ``K--;` `        ``}` `        ``Divisor++;` `    ``}`   `    ``// Return the final answer` `    ``return` `SmallestNumber;` `}`   `// Driver Code` `public` `static` `void` `main(String args[])` `{` `    ``boolean``[] prime = ``new` `boolean``[N];` `    ``Arrays.fill(prime, ``true``);`   `    ``SieveOfEratosthenes(prime);`   `    ``int` `D = ``1``;` `    ``int` `K = ``4``;`   `    ``// Function Call` `    ``System.out.println(SmallestNumber(prime, D, K));` `}` `}`   `// This code is contributed by sanjoy_62.`

Python3

 `# Python program to implement` `# the above approach` `N ``=` `1000000``;`   `# Function of Sieve of Eratosthenes` `def` `SieveOfEratosthenes(prime):` `    ``for` `p ``in` `range``(``2``, N``/``/``2``):` `        ``if` `(prime[p] ``=``=` `True``):` `            ``for` `i ``in` `range``(p``*``p,N,p):` `                ``prime[i] ``=` `False``;` `        `  `# Function to find smallest` `# number with given conditions` `def` `SmallestNumber(prime, D, K):`   `    ``# Initialize first with D + 1` `    ``# because 1 is also a divisor` `    ``FirstDivisor ``=` `D ``+` `1``;`   `    ``while` `(FirstDivisor < N ``and` `prime[FirstDivisor]!``=``True``):` `        ``FirstDivisor ``+``=` `1``;` `    `  `    ``# Now value of K is decrement by 1` `    ``K ``-``=` `1``;`   `    ``# Initialize Divisor with First + D` `    ``# to maintain a difference D` `    ``# We get Remaining divisor` `    ``SmallestNumber ``=` `FirstDivisor;` `    ``Divisor ``=` `FirstDivisor ``+` `D;`   `    ``# Maintain previous divisor` `    ``# to maintain difference` `    ``prevDivisor ``=` `FirstDivisor;` `    ``while` `(K > ``0` `and` `SmallestNumber < N):` `        ``if` `(prime[Divisor] ``and` `Divisor ``-` `D >``=` `prevDivisor):` `            ``SmallestNumber ``*``=` `Divisor;` `            ``prevDivisor ``=` `Divisor;` `            ``K ``-``=` `1``;` `        `  `        ``Divisor ``+``=` `1``;`   `    ``# Return the final answer` `    ``return` `SmallestNumber;`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``prime ``=` `[``True` `for` `i ``in` `range``(N)];` `    `  `    ``SieveOfEratosthenes(prime);`   `    ``D ``=` `1``;` `    ``K ``=` `4``;`   `    ``# Function Call` `    ``print``(SmallestNumber(prime, D, K));`   `# This code is contributed by Rajput-Ji `

C#

 `// C# program to implement` `// the above approach` `using` `System;`   `public` `class` `GFG` `{`   `  ``static` `int` `N = 1000000;`   `  ``// Function of Sieve of Eratosthenes` `  ``static` `void` `SieveOfEratosthenes(``bool``[] prime)` `  ``{` `    ``for` `(``int` `p = 2; p * p < N; p++) {` `      ``if` `(prime[p] == ``true``) {` `        ``for` `(``int` `i = p * p; i < N; i += p)` `          ``prime[i] = ``false``;` `      ``}` `    ``}` `  ``}`   `  ``// Function to find smallest` `  ``// number with given conditions` `  ``static` `int` `SmallestNumber(``bool``[] prime,` `                            ``int` `D, ``int` `K)` `  ``{`   `    ``// Initialize first with D + 1` `    ``// because 1 is also a divisor` `    ``int` `FirstDivisor = D + 1;`   `    ``while` `(FirstDivisor < N` `           ``&& !prime[FirstDivisor]) {` `      ``++FirstDivisor;` `    ``}`   `    ``// Now value of K is decrement by 1` `    ``K--;`   `    ``// Initialize Divisor with First + D` `    ``// to maintain a difference D` `    ``// We get Remaining divisor` `    ``int` `SmallestNumber = FirstDivisor;` `    ``int` `Divisor = FirstDivisor + D;`   `    ``// Maintain previous divisor` `    ``// to maintain difference` `    ``int` `prevDivisor = FirstDivisor;` `    ``while` `(K > 0 && SmallestNumber < N) {` `      ``if` `(prime[Divisor]` `          ``&& Divisor - D >= prevDivisor) {` `        ``SmallestNumber *= Divisor;` `        ``prevDivisor = Divisor;` `        ``K--;` `      ``}` `      ``Divisor++;` `    ``}`   `    ``// Return the readonly answer` `    ``return` `SmallestNumber;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main(String []args)` `  ``{` `    ``bool``[] prime = ``new` `bool``[N];` `    ``for``(``int` `i = 0;i

Javascript

 ``

Output

`210`

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

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