Given N and Standard Deviation, find N elements

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Given N and Standard deviation find the N elements.

Mean is average of element.
Mean of arr[0..n-1] = ?(arr[i]) / n
where 0 <= i < n
Variance is sum of squared differences from the mean divided by number of elements.
Variance = ?(arr[i] – mean)² / n
Standard Deviation is square root of variance
Standard Deviation = ?(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.

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.
?>

Javascript

<script>
 
// JavaScript 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(let i = 0; i < n; i++)
            document.write("0 ");
             
        document.write();
        return;
    }
 
    // If S.D. is even
    if (n % 2 == 0) 
    {
         
        // Print -SD, +SD, -SD, +SD
        for(let i = 1; i <= n; i++) 
        {
            document.write(Math.pow(-1, i) * d + "  ");
        }
        document.write();
    }
     
    // If odd
    else
    {
         
        // Convert n to a float integer
        let m = n;
        let r = (m / (m - 1));
        let g = (d * (Math.sqrt(r)));
 
        // Print one element to be 0
        document.write("0 ");
 
        // Print (n-1) elements as xi
        // derived from the formula
        for(let i = 1; i < n; i++)
        {
            document.write(Math.pow(-1, i) * g + "  ");
        }
        document.write();
    }
}
 
// Driver Code
let n = 3, d = 3;
 
series(n, d);
 
// This code is contributed by susmitakundugoaldanga
 
</script>

Output:

0 -3.67423 3.67423

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(1), as we are not using any extra space.

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