# Generate elements of the array following given conditions

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

• For any pair of indices (i, j), if i and j are coprime then .
• The maximum value of all 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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 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 values same as any of its prime divisors.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `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.. `

Output:

```1 2 1 3 2
```

Time Complexity: O(N Log N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : vt_m, AnkitRai01