## Nutanix Interview Question for Software Developers

• 0

Country: United States
Interview Type: Phone Interview

Comment hidden because of low score. Click to expand.
2
of 2 vote

``````ackage com.cracking.nutanix;

public class BuildPyramid {

public static void main(String[] args) {
System.out.println( String.format("Input: %d, Possible: %s", 4,isPyramidPossible(4)) );
System.out.println( String.format("Input: %d, Possible: %s", 9,isPyramidPossible(9)) );
System.out.println( String.format("Input: %d, Possible: %s", 6,isPyramidPossible(6)) );
}

public static boolean isPyramidPossible(int bricks) {
int rowBricks = 1;
while(bricks > 0) {
bricks -= rowBricks;
rowBricks += 2;
}

return bricks == 0;
}

}``````

Output:
Input: 4, Possible: true
Input: 9, Possible: true
Input: 6, Possible: false

Comment hidden because of low score. Click to expand.
1
of 1 vote

boolean check(int n)
long squareRoot = Math.round(Math.sqrt(n));
if ((n>0) && (n== squareRoot*squareRoot))
return true;
return false;

Comment hidden because of low score. Click to expand.
1
of 1 vote

Analysis.
The problem is that of :
1 + 3 + 5 + ...
That is sum of n odd numbers.
That is known to be : 9math.com/book/sum-first-n-odd-natural-numbers
=> n^2.
Thus we are to find, if the number given is a perfectly square no. or not.
Now, given we do not want to use square root, which is
a terribly complex operation, we can do as follows:
[ careercup.com/question?id=14797683 ]

Comment hidden because of low score. Click to expand.
1
of 1 vote

The pattern is 1, 3, 5, 7,.... 2n-1

Sum of numbers from 1 to 2n-1 = 1/2 n(1 + 2n-1) = n^2.

So, given a number, check if its root is a perfect integer.

``return (int)Math.sqrt(n) * (int)Math.sqrt(n) == n;``

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````"""
To make a pyramid
the first row will have 1 *
the second row will have 3 *
the third row will have 5 *
the N th row will have 1+3+5+7. (expand sequence to have N terms )

The sum or first n odd numbers is n*n
For forming a pyramid with N *s - N should be a perfect square
"""
import math
def isPyramidPossible (n):
root = int (math.sqrt(n))
return int (root)  == n //root

def main(n):
print (f'A pyramid with {n} *s is {"" if isPyramidPossible(n) else "not"} possibe')

if (__name__== "__main__"):
for i in range (1, 100):
main (i)``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

Simple Python Solution

``````def pyramid(a):
num = 1
new_num = 0
while new_num != a and new_num <= a:

new_num = num + new_num
num += 2
print("new num is: ", new_num)
if new_num == a:
return True
return False``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

This will be the pattern 1,4,9,16,25,36,..
if n is a perfect square , we can build the pyramid otherwise not .

Comment hidden because of low score. Click to expand.
0
of 0 vote

since first row will have 1 star 2nd row 3 star 3rd row 5 star so on.
so to make n row you should have total star equal to sum of first n odd numbers.
As you can see above first n odd number form Arithematic progression with difference of 2
sum of n numbers of an Arithematic progression =(n* (2*a+(n-1)*d))/2
where n = number of term
a(first term) = 1
d(difference) = 2
so sum = (n*(2*1+(n-1)*2)/2 = n*n = n^2
so you simply have to check number of star is perfect square or not.

Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

### Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

### Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.