Given N and Standard Deviation, find N elements

Given N and Standard deviation find the N elements.

Mean is average of element.
Mean of arr[0..n-1] = Σ(arr[i]) / n
where 0 <= i < n

Variance is sum of squared differences from the mean divided by number of elements.

Variance = Σ(arr[i] – mean)² / n

Standard Deviation is square root of variance
Standard Deviation = Σ(variance)

Please refer Mean, Variance and Standard Deviation for details.

Examples:

Input: 6 0 
Output: 0 0 0 0 0 0  
Explanation: 
The standard deviation of 0, 0, 0, 0, 0, 0 is 0. 
Also the standard deviation of 4, 4, 4, 4, 4, 4 
is 0, we print any of the possible N elements.

Input: 3 3
Output: 0 -3.67423 3.67423 
Explanation: 
On calculating SD of these N elements,
we get standard deviation to be 3.

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

Approach:
If we look at the formula, we have two unknown terms one is xi and the other is mean. The main motive is to make the mean 0 so that we can get the formula for X elements. There will be two cases, one for even and one for odd.

When N is even:
To make mean of N elements 0, best way is to express N elements as -X +X -X +X …. Formula will be sqrt(summation of (x^2)/n), x2+x^2+x^2+………N terms, so formula turns out to be sqrt (N*(x^2)/N). N cancel out each other, so sqrt (x^2) turns out to be SD. So, we get the N elements as -SD +SD -SD +SD…… to get the mean 0. We need to print -SD +SD -SD +SD……

When N is odd:
The mean of N elements will be 0. So, one element will be 0 and other N-1 elements will be -X +X -X …. Formula will be sqrt(summation of (x^2)/n), x2+x^2+x^2+………N-1 terms, so formula turns out to be sqrt((N-1)*(x^2)/N), so
X= SD * sqrt(n/(n-1)). The n elements are 0 -X +X -X +X …

When SD is 0 then all elements will be same, so we can print 0 for it.

Below is the implementation of the above approach:

C++

// CPP program to find n elements 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to print series of n elements 
void series(int n, int d) 
{ 
  
    // if S.D. is 0 then print all 
    // elements as 0. 
    if (d == 0) { 
  
        // print n 0's 
        for (int i = 0; i < n; i++) 
            cout << "0 "; 
  
        cout << endl; 
        return; 
    } 
  
    // if S.D. is even 
    if (n % 2 == 0) { 
  
        // print -SD, +SD, -SD, +SD 
        for (int i = 1; i <= n; i++) { 
            cout << pow(-1, i) * d << " "; 
        } 
        cout << endl; 
    } 
    else // if odd 
    { 
        // convert n to a float integer 
        float m = n; 
        float r = (m / (m - 1)); 
        float g = (float)(d * (float)sqrtf(r)); 
  
        // print one element to be 0 
        cout << "0 "; 
  
        // print (n-1) elements as xi derived 
        // from the formula 
        for (int i = 1; i < n; i++) { 
            cout << pow(-1, i) * g << " "; 
        } 
        cout << endl; 
    } 
} 
  
// driver program to test the above function 
int main() 
{ 
    int n = 3, d = 3; 
    series(n, d); 
  
    return 0; 
} 

Java

// Java program to find n elements 
import java.util.*; 
import java.lang.*; 
  
public class GfG { 
  
    // function to print series of n elements 
    public static void series(int n, int d) 
    { 
  
        // if S.D. is 0 then print all 
        // elements as 0. 
        if (d == 0) { 
  
            // print n 0's 
            for (int i = 0; i < n; i++) 
                System.out.print("0 "); 
            System.out.println(); 
            return; 
        } 
  
        // if S.D. is even 
        if (n % 2 == 0) { 
  
            // print -SD, +SD, -SD, +SD 
            for (int i = 1; i <= n; i++) { 
                System.out.print(Math.pow(-1, i) * d + " "); 
            } 
            System.out.println(); 
        } 
        else // if odd 
        { 
            // convert n to a float integer 
            float m = n; 
            float r = (m / (m - 1)); 
            float g = (float)(d * (float)(Math.sqrt(r))); 
  
            // print one element to be 0 
            System.out.print("0 "); 
  
            // print (n-1) elements as xi 
            // derived from the formula 
            for (int i = 1; i < n; i++) { 
                System.out.print(Math.pow(-1, i) * g + " "); 
            } 
            System.out.println(); 
        } 
    } 
  
    // driver function 
    public static void main(String args[]) 
    { 
        int n = 3, d = 3; 
        series(n, d); 
    } 
} 
  
/* This code is contributed by Sagar Shukla */

Python3

# Python program to find n elements 
import math 
  
# function to print series of n elements 
def series( n, d): 
  
    # if S.D. is 0 then print all 
    # elements as 0. 
    if d == 0: 
      
        # print n 0's 
        for i in range(n): 
            print("0", end = ' ') 
        return 1
          
    # if S.D. is even 
    if n % 2 == 0: 
      
        # print -SD, +SD, -SD, +SD 
        i = 1
        while i <= n: 
            print("%.5f"%((math.pow(-1, i) * d)), 
                  end =' ') 
            i += 1
    else: 
        # if odd 
        # convert n to a float integer 
        m = n 
        r = (m / (m - 1)) 
        g = (float)(d * float(math.sqrt(r))) 
          
        # print one element to be 0 
        print("0 ", end = ' ') 
          
        # print (n-1) elements as xi derived 
        # from the formula 
        i = 1
        while i < n: 
            print("%.5f"%(math.pow(-1, i) * g), 
                  end = ' ') 
            i = i + 1
    print("\n") 
  
# driver code to test the above function 
n = 3
d = 3
series(n, d) 
  
# This code is contributed by "Sharad_Bhardwaj". 

C#

// C# program to find n elements 
using System; 
  
public class GfG { 
  
    // function to print series of n 
    // elements 
    public static void series(int n, int d) 
    { 
  
        // if S.D. is 0 then print all 
        // elements as 0. 
        if (d == 0) { 
  
            // print n 0's 
            for (int i = 0; i < n; i++) 
                Console.Write("0"); 
                  
            Console.WriteLine(); 
              
            return; 
        } 
  
        // if S.D. is even 
        if (n % 2 == 0) { 
  
            // print -SD, +SD, -SD, +SD 
            for (int i = 1; i <= n; i++) { 
                Console.Write(Math.Pow(-1, i) 
                                   * d + " "); 
            } 
              
            Console.WriteLine(); 
        } 
        else // if odd 
        { 
              
            // convert n to a float integer 
            float m = n; 
            float r = (m / (m - 1)); 
            float g = (float)(d *  
                       (float)(Math.Sqrt(r))); 
  
            // print one element to be 0 
            Console.Write("0 "); 
  
            // print (n-1) elements as xi 
            // derived from the formula 
            for (int i = 1; i < n; i++) { 
                Console.Write(Math.Pow(-1, i) 
                                   * g + " "); 
            } 
              
            Console.WriteLine(); 
        } 
    } 
  
    // driver function 
    public static void Main() 
    { 
        int n = 3, d = 3; 
          
        series(n, d); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find n elements 
  
// function to print  
// series of n elements 
function series( $n, $d) 
{ 
  
    // if S.D. is 0 then print all 
    // elements as 0. 
    if ($d == 0) 
    { 
  
        // print n 0's 
        for ($i = 0; $i < $n;$i++) 
            echo "0 "; 
  
        echo "\n"; 
        return; 
    } 
  
    // if S.D. is even 
    if ($n % 2 == 0)  
    { 
  
        // print -SD, +SD, -SD, +SD 
        for ( $i = 1; $i <= $n; $i++)  
        { 
            echo pow(-1, $i) * $d , " "; 
        } 
        echo"\n"; 
    } 
      
    // if odd 
    else 
    { 
          
        // convert n to a  
        // float integer 
        $m = $n; 
        $r = ($m / ($m - 1)); 
        $g = ($d * sqrt($r)); 
  
        // print one element 
        // to be 0 
        echo "0 "; 
  
        // print (n-1) elements  
        // as xi derived 
        // from the formula 
        for ($i = 1; $i < $n; $i++)  
        { 
            echo pow(-1, $i) * $g , " "; 
        } 
        echo"\n"; 
    } 
} 
  
    // Driver Code 
    $n = 3; $d = 3; 
    series($n, $d); 
  
// This code is contributed by anuj_67. 
?> 

Output:

0 -3.67423 3.67423

My Personal Notes

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

C++

Java

Python3

C#

PHP

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