Problem 23. Finding Perfect Squares

Created by Cody Team

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Solution Stats

42658 Solutions
11231 Solvers

Last Solution submitted on Aug 03, 2021

Last 200 Solutions

Problem Comments

32 Comments

32 Comments

Show 29 older comments

Alex Feinman on 25 Jan 2012

Another of the classic "check if MATLAB has a function that does this already" problems...

Michael Joslin on 6 Feb 2012

What about a vector like [2 3 1]? The top solution will still return true because 1^2 is still one but it is not the square of one of the OTHER numbers.

Felix on 12 Dec 2012

is the Str2func construct really more effective???
Is it possible that this distorts the metric a bit?

Gregoire on 11 Apr 2014

It is currently impossible to submit a solution for this problem. The web page will never load.

Mattias on 2 Jan 2015

The problem description states that one of the numbers has to be a square of one of the OTHER numbers. Still, the test suite tests if [1] gives true. I think [1] should give false, or that describtion be modified.

Ramoflaple on 4 Jul 2015

Agree with Mattias. the length of input vector should be 2 or larger.

Shamma Sultan on 30 Aug 2016

The difference between test 4 and 5 which returns different answers is the number 36.. but 36 is not a square root of any number of the list in test 4!!

Sagar Rupapara on 4 Nov 2016

great solution

David Verrelli on 19 Mar 2017

Agree with Mattias: the Test Suite is defective, and not consistent with the problem specification.
Having an input of [1] is OK, but result should be false, as there is no OTHER element to compare the 1 to!

ishmam zabir on 26 Apr 2017

can you share the shortest code?

Marco Green on 11 May 2018

easy

Ryan Klots on 4 Dec 2018

Fun question!

asa czylabson on 30 Jan 2019

the word "other" should be ignored in the description...

Harsha Priya Daggubati on 21 Feb 2019

fun

Mark Sanderson on 13 Mar 2019

Good question!

MAHMOUD ABDEL-LATIF on 4 Jun 2019

Cool

Ali on 15 Nov 2019

Clever problem

Jyothsna Chintala on 21 Apr 2020

good

Robison Saalfeld on 7 Jul 2020

The heading reads:
"...a square of one of the OTHER numbers...".

How can the function be expected to return [true] for a = [1]?
It makes no sense.

Lesandu Wenitha Sugathadasa on 10 Aug 2020

Good question

Abhijeet Deshmukh on 11 Aug 2020

This problem is very good

源樹上林 on 13 Aug 2020

いい問題。
sumかnnzか

yu wang on 22 Sep 2020

intersect的返回值为一个向量，即使a和b的交集为空集，返回值仍为向量
x=[20:30]
例如：intersect(x.^2,x)

ans =

空的 1×0 double 行向量
所以isvector(intersect(x.^2,x))是错误的
而isempty是判断数组是不是空的，所以isempty可行

Elizabeth Katulka on 18 Oct 2020

Tough one, but interesting

Leo S on 30 Nov 2020

I solved it with 2 for-loops, 1 function, 1 if-condition.
It works, but there are more efficient ways apparently

Yunpeng Teng on 2 Dec 2020

Hey,the demand of this problem declare that "if one " and "the other",so if the intersect of x and x.^2 only have one value of 0 or 1,it may be return false because their squares are themselves, which is not satisfy the condition implied that c and c^2 is not one.

Nicholas Erimias on 9 Dec 2020

Nice Problem

David Lusk on 20 Dec 2020

The question asks if one number is the square of one of the OTHER numbers. Test 3 should return false.

Arunika Oyshi on 3 Feb 2021

Great problem

Filip Novak on 21 Apr 2021

Nice

Andrea Taveras on 22 Apr 2021

goo problem

Sadullah Avci on 3 Jun 2021

if else

Solution Comments

Solution 5665610

1 Comment

1 Comment

john perkins on 4 May 2021

very challenging problem!!

Solution 5151653

2 Comments

2 Comments

Emoonah McClerklin on 23 Mar 2021

I don't understand why #4 if false. Can someone help me?

goc3 on 24 Mar 2021

Test case #4 is true, not false, because the vector contains a perfect square and its square root.
Your function is not passing because it overwrites the check variable for each vector element, only actually checking the last value to return the true/false variable. It happened to pass the other four cases because their answers (either false or true) happen to correlate with your check for the last (or only) element in the vector.

Solution 5115427

2 Comments

2 Comments

Angelo Ricciardi on 19 Mar 2021

a = [20:30];
assert(isequal(isItSquared(a),false))
Assertion failed.
It should be true because in the array there is 25.
a = [20:30]
a = 20 21 22 23 24 25 26 27 28 29 30

Dyuman Joshi on 19 Mar 2021

The question asks that whether a perfect square and it's root both are present in the array or not. You have to check for both and not one only.

Solution 4378370

Solution 3552480

Solution 3504489

1 Comment

1 Comment

Touhidur Rahman on 1 Nov 2020

Hi,
I verified this code with the given test suite. but the system is not accepting the code as it is.

thanks

Solution 2758927

1 Comment

1 Comment

Shuvo Kumar Paul on 30 Jul 2020

What's wrong with this code? Runs perfectly on my computer.

Solution 2758405

Solution 2585557

Solution 2391632

1 Comment

1 Comment

文博李 on 26 May 2021

I didn't see the leading code,so I put my code here to give some inspiration.
Looking forward to a shorter one
b=~isempty(intersect(a,a.^2));

Solution 2247535

Solution 2026266

1 Comment

1 Comment

Raymond MacNeil on 18 Nov 2019

whoops, don't know how I managed to confuse .^ with .* for a bit there ---I was like, "What's WRONG?"

Solution 1971445

1 Comment

1 Comment

Keerty dev Golhani on 11 Oct 2019

nice

Solution 1883399

2 Comments

2 Comments

Hyeokjin Jho on 2 Mar 2021

You cheater!

Dyuman Joshi on 2 Mar 2021

This type of cheat/hack answers are no longer allowed.

Solution 1879664

2 Comments

2 Comments

Michael Price on 20 Jul 2019

Good question

Gustavo Morales on 18 Nov 2019

0 and 1 elements in a will show the bug in your code

Solution 1877827

2 Comments

2 Comments

Rudolf Csikja on 18 Jul 2019

nice one-liner

Gustavo Morales on 18 Nov 2019

a vector 'a' with elements 0 or 1 will show the bug in your code

Solution 1790644

1 Comment

1 Comment

Yuliano Moscoso Yarin on 16 Apr 2019

cool!

Solution 1719302

2 Comments

2 Comments

Natasa on 1 Feb 2019

Wow! This took me a couple of days to solve, but I did it and I’m very proud! :D

Gustavo Morales on 18 Nov 2019

Hi. A vector with ones or zeroes as v = [3, 1] will show the bug in your code.

Solution 1639989

1 Comment

1 Comment

Steven Belcher on 29 Jun 2019

Lol

Solution 1569750

3 Comments

3 Comments

Myeongseon Choi on 27 Jun 2018

Test case : a = [20:30] contains a square. but its expected error is false. It should be fixed

Jann B on 22 Jul 2019

hello,
where do you find a perfect square in [20:30]?

Matthias Kindler on 4 Feb 2020

Hi Jann B,
25 is a perfect square, no?

Solution 1416225

1 Comment

1 Comment

Jafar Mortadha on 14 Jan 2018

How is the output true if the vector contains only one number? People posting such meaningless solutions: Please have some common sense!

Solution 1335186

1 Comment

1 Comment

Oliviero Allan Ceresa Genet on 11 Nov 2017

easy easy

Solution 1230037

2 Comments

2 Comments

Dustin Jones on 12 Jul 2017

That's better...

Gustavo Morales on 18 Nov 2019

a vector as v = [3,2,7,0,5] will show you your code's bug

Solution 1230036

2 Comments

2 Comments

Dustin Jones on 12 Jul 2017

a little smaller now

DilRavsh on 20 Aug 2017

Nice ☺

Solution 1230035

1 Comment

1 Comment

Dustin Jones on 12 Jul 2017

Works

Solution 1229999

1 Comment

1 Comment

Dustin Jones on 12 Jul 2017

just kidding...

Solution 1101504

1 Comment

1 Comment

SIVAPRAKASH DANABALASINGAM on 10 Jan 2017

very easy

Solution 1090789

1 Comment

1 Comment

Noriko Hounoki on 27 Dec 2016

I want to make this code shorter!!

Solution 1088763

1 Comment

1 Comment

Noriko Hounoki on 24 Dec 2016

Case2,a = [20:30],fails.
Others pass.
Why???

Solution 1088350

1 Comment

1 Comment

Habeeb Ali on 23 Dec 2016

in 2 test suite in a=[20:30] , contains 5^2 =25 so it has to be true not false

Solution 927012

1 Comment

1 Comment

Zhang Liang on 25 Jul 2016

But there seems to be a better choice. Why forgetting ismember?

Solution 906195

2 Comments

2 Comments

Gary Ong on 10 Jun 2016

Test 2 in the test suite is wrong!
25 is clearly a perfect square.

Matthew Stuckey on 15 Jun 2016

the problem is to see if a number is the square of another number in the vector. Although 25 is a perfect square, 5 is not included in the vector

Solution 881091

Solution 858098

2 Comments

2 Comments

Michael on 28 Mar 2016

Though it passes the test suite, I feel like I could make this a better way but I am drawing a blank. Any feedback would be appreciated.

Seung Yub Lee on 3 Dec 2016

good

Solution 855996

3 Comments

3 Comments

prakirt on 24 Mar 2016

if a = [20:30] the output should be true as 25 is a perfect square, while in the test case output is given to be false.

Michael on 29 Mar 2016

25 is not a square of any other members of the group. If the group included 5 it would be true.

Dan Po on 24 Oct 2016

hur dur durrr durr. sqrt(25)==5, not included in givens

Solution 731498

1 Comment

1 Comment

Gouda on 8 Sep 2015

Yeah, I did it

Solution 708141

2 Comments

2 Comments

Alexander Wickstrom on 29 Jul 2015

My solution returns false for [20:30]; why not for [6 10 12 14 101]?

Sebastian on 8 Sep 2015

Instead of a(1):a(end) you should just write a.

Solution 607934

2 Comments

2 Comments

Zhao Dennis on 29 Mar 2015

Where did I do wrong？

Tim on 31 Mar 2015

The desired results are not the character strings 'true' and 'false' but the logical values true and false.

Solution 490598

1 Comment

1 Comment

kunal on 26 Aug 2014

hint ; any and ismember.

Solution 480125

1 Comment

1 Comment

Chandan singh on 29 Jul 2014

submitted tested solution but still showing error......
L 17 (C 1-3): Parse error at END: usage might be invalid MATLAB syntax.

Solution 326746

2 Comments

2 Comments

Beshir Aman on 25 Aug 2014

interesting solution... I like it :)

Sebastian on 8 Sep 2015

I disagree. Almost every problem has a solution wrapped in this function. It's cheating as it hides the complexity of the code from the scoring algorithm. That's why "eval" is not allowed ...

Solution 295014