# Generate Array such that max is minimized and arr[i] != arr[j] when j is a multiple of i

• Last Updated : 01 Feb, 2023

Given an integer N, the task is to generate an array arr[] having N positive integers such that arr[i] â‰  arr[j] if j is divisible by i (1-based indexing is considered) such that the maximum value of the sequence is minimum among all possible sequences.

Examples:

Input: N = 3
Output: 1 2 2
Explanation: In this sequence 2 is divisible by 1. arr[2] = 2, arr[1] = 1 and arr[2] â‰  arr[1].
In this sequence 3 is divisible by 1. arr[3] = 2, arr[1] = 1 and arr[3] â‰  arr[1].
The maximum value is 2 which is the minimum possible among all sequences.

Input: N = 1
Output: 1

Approach: This problem can be solved based on the following idea:

For each index i, the value of arr[i] must be at least same as the number of divisors it has (including the number itself).
Instead of directly finding divisors for each number, one can also find multiples (say j) of each index i. and increment count of divisors of j.

Follow the illustration given below for a better understanding.

Say N = 3.
Create arr[] = {0, 0, 0} to store number of divisors of each value.

For i = 1:
=> Multiples of 1 are 1, 2 and 3.
=> Increment values of arr[1], arr[2] and arr[3].
=> So arr[] = {1, 1, 1}

For i = 1:
=> Multiples of 2 is 2.
=> Increment value of arr[2].
=> So arr[] = {1, 2, 1}

For i = 3:
=> Multiples of 3 is 3.
=> Increment value of arr[3].
=> So arr[] = {1, 2, 2}

So the final array is arr = {1, 2, 2} where the maximum value (i.e. 2) is minimum among all possible sequences.

Follow the steps mentioned below to implement the above observation:

• Create an N sized array (say arr[]) with all elements initially filled with 0.
• Use Sieve of Eratosthenes to find all multiples of all numbers from i = 1 to N.
• For this traverse from i = 1 to N:
• Run a loop from j = i to N and increment j by i in each step:
• Increment the value of arr[j] by 1.
• Return the final array.

Below is the implementation of the above approach.

## C++

 // C++ code to implement the approach #include using namespace std; // Function to generate the arrayvector goodSequence(int N){    vector res(N);    res[0] = 1;     // Loop to implement    // sieve of Eratosthenes    for (int i = 1; i <= N; i++) {        for (int j = 2 * i; j <= N; j += i)            res[j - 1] = res[i - 1] + 1;    }    return res;} // Driver codeint main(){    int N = 3;     // Function call    vector ans = goodSequence(N);    for (int x : ans)        cout << x << " ";    return 0;}

## Java

 // Java code to implement the approachimport java.io.*; class GFG {   // Function to generate the array  static int[] goodSequence(int N)  {    int[] res = new int[N];    res[0] = 1;     // Loop to implement    // sieve of Eratosthenes    for (int i = 1; i <= N; i++) {      for (int j = 2 * i; j <= N; j += i)        res[j - 1] = res[i - 1] + 1;    }    return res;  }   // Driver code  public static void main (String[] args) {    int N = 3;     // Function call    int ans[] = goodSequence(N);    for (int x = 0; x < ans.length; x++)      System.out.print(ans[x] + " ");  } } // This code is contributed by hrithikgarg03188.

## Python3

 # python3 code to implement the approach import math # Function to generate the array  def goodSequence(N):     res = [0 for _ in range(N)]    res[0] = 1     # Loop to implement    # sieve of Eratosthenes    for i in range(1, N + 1):        for j in range(2*i, N+1, i):            res[j - 1] = res[i - 1] + 1     return res  # Driver codeif __name__ == "__main__":     N = 3     # Function call    ans = goodSequence(N)    for x in ans:        print(x, end=" ")     # This code is contributed by rakeshsahni

## C#

 // C# code to implement the approachusing System;class GFG {     // Function to generate the array    static int[] goodSequence(int N)    {        int[] res = new int[N];        res[0] = 1;         // Loop to implement        // sieve of Eratosthenes        for (int i = 1; i <= N; i++) {            for (int j = 2 * i; j <= N; j += i)                res[j - 1] = res[i - 1] + 1;        }        return res;    }     // Driver code    public static void Main()    {        int N = 3;         // Function call        int[] ans = goodSequence(N);        for (int x = 0; x < ans.Length; x++)            Console.Write(ans[x] + " ");    }} // This code is contributed by Samim Hossain Mondal.

## Javascript



Output

1 2 2

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

My Personal Notes arrow_drop_up