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Generate elements of the array following given conditions
  • Last Updated : 11 May, 2021

Given an integer N, for every integer i in the range 2 to N, assign a positive integer a_i  such that the following conditions hold : 
 

  • For any pair of indices (i, j), if i and j are coprime then a_i \neq a_j  .
  • The maximum value of all a_i  should be minimized (i.e. max value should be as small as possible).

A pair of integers is called coprime if their gcd(i, j) = 1.
Examples: 
 

Input: N = 4 
Output: 1 2 1 
Input: N = 10 
Output: 1 2 1 3 2 4 1 2 3 
 

 

Approach: We have to keep two things in mind while assigning values at indices from 2 to n. First, for any pair (i, j), if i and j are coprime then we cannot assign the same value to this pair of indices. Second, we have to minimize the maximum value assigned to each index from 2 to n.
This can be achieved if distinct numbers are assigned to each prime because all primes are coprime to each other. Then assign values at all-composite positions the same as any of its prime divisors. This solution works as for any pair (i, j), i is given the same number of a divisor and so is j, so if they are coprime, they cannot be given the same number.
This approach can be implemented by making a small change when building the Sieve of Eratosthenes
When we met a prime for the first time, then assign a unique smallest value to it and all of its multiples. This way, all prime indices should have unique values and all composite indices have valued the same as any of its prime divisors.
Below is the implementation of the above approach: 
 



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to generate the required array
void specialSieve(int n)
{
 
    // Initialize cnt variable for assigning
    // unique value to prime and its multiples
    int cnt = 0;
    int prime[n + 1];
 
    for (int i = 0; i <= n; i++)
        prime[i] = 0;
 
    for (int i = 2; i <= n; i++) {
 
        // When we get a prime for the first time
        // then assign a unique smallest value to
        // it and all of its multiples
        if (!prime[i]) {
            cnt++;
            for (int j = i; j <= n; j += i)
                prime[j] = cnt;
        }
    }
 
    // Print the generated array
    for (int i = 2; i <= n; i++)
        cout << prime[i] << " ";
}
 
// Driver code
int main()
{
    int n = 6;
 
    specialSieve(n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.io.*;
 
class GFG
{
 
// Function to generate the required array
static void specialSieve(int n)
{
 
    // Initialize cnt variable for assigning
    // unique value to prime and its multiples
    int cnt = 0;
    int prime[] = new int[n+1];
 
    for (int i = 0; i <= n; i++)
        prime[i] = 0;
 
    for (int i = 2; i <= n; i++)
    {
 
        // When we get a prime for the first time
        // then assign a unique smallest value to
        // it and all of its multiples
        if (!(prime[i]>0))
        {
            cnt++;
            for (int j = i; j <= n; j += i)
                prime[j] = cnt;
        }
    }
 
    // Print the generated array
    for (int i = 2; i <= n; i++)
        System.out.print(prime[i] + " ");
}
 
// Driver code
public static void main (String[] args)
{
    int n = 6;
 
specialSieve(n);
 
}
}
 
// This code is contrubuted by anuj_67..

Python3




# Python3 implementation of the approach
 
# Function to generate the required array
def specialSieve(n) :
 
    # Initialize cnt variable for assigning
    # unique value to prime and its multiples
    cnt = 0;
    prime = [0]*(n + 1);
 
    for i in range(2, n + 1) :
 
        # When we get a prime for the first time
        # then assign a unique smallest value to
        # it and all of its multiples
        if (not prime[i]) :
            cnt += 1;
            for j in range(i, n + 1, i) :
                prime[j] = cnt;
 
    # Print the generated array
    for i in range(2, n + 1) :
        print(prime[i],end = " ");
 
 
# Driver code
if __name__ == "__main__" :
 
    n = 6;
    specialSieve(n);
     
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to generate the required array
static void specialSieve(int n)
{
 
    // Initialize cnt variable for assigning
    // unique value to prime and its multiples
    int cnt = 0;
    int []prime = new int[n+1];
 
    for (int i = 0; i <= n; i++)
        prime[i] = 0;
 
    for (int i = 2; i <= n; i++)
    {
 
        // When we get a prime for the first time
        // then assign a unique smallest value to
        // it and all of its multiples
        if (!(prime[i] > 0))
        {
            cnt++;
            for (int j = i; j <= n; j += i)
                prime[j] = cnt;
        }
    }
 
    // Print the generated array
    for (int i = 2; i <= n; i++)
        Console.Write(prime[i] + " ");
}
 
// Driver code
public static void Main ()
{
    int n = 6;
 
    specialSieve(n);
 
}
}
 
// This code is contrubuted by anuj_67..

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to generate the required array
function specialSieve(n)
{
 
    // Initialize cnt variable for assigning
    // unique value to prime and its multiples
    let cnt = 0;
    let prime = new Array(n + 1);
 
    for (let i = 0; i <= n; i++)
        prime[i] = 0;
 
    for (let i = 2; i <= n; i++) {
 
        // When we get a prime for the first time
        // then assign a unique smallest value to
        // it and all of its multiples
        if (!prime[i]) {
            cnt++;
            for (let j = i; j <= n; j += i)
                prime[j] = cnt;
        }
    }
 
    // Print the generated array
    for (let i = 2; i <= n; i++)
        document.write(prime[i] + " ");
}
 
// Driver code
    let n = 6;
 
    specialSieve(n);
 
</script>
Output: 
1 2 1 3 2

 

Time Complexity: O(N Log N)
 

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