Is it a pleasing number?

Question

A pleasing number (according to this definition) is a number that can be passed through the following process (let's use 41802000 as an example):

• Separate the trailing identical numbers (41802000 => [41802, 000])
• Sort the first half of the digits and take the maximum ([41802, 000] => [8, 000])
• Take the length of the final element. Call the resultant elements A and B ([8, 000] => A=8, B=3)
• Is N^B = A for any integer N? (True in this example; 2³ = 8)

If the input results in True from the following process, it is considered pleasing.

Your task is to take an integer x, and output whether x is a pleasing number. You can output any two distinct values for truthy and falsey, but please state what you're using in your answer. x is guaranteed to have at least two distinct digits (e.g. 111 is an invalid input). As with most challenges, you may take x as a string, or a list of digits.

This is a code-golf so shortest code in bytes wins.

Test cases

          Input          Output
       41802000               1
            100               1
      123456788               0
      451111111               0
234543454545444               0
             12               1
       41902000               0
          92599               1

Answer 1

APL (Dyalog), 36 bytes

{(⊢≡⌊)(⌈/⍎¨⍵/⍨~o)*÷≢⍵/⍨o←⌽⌊\1,2=/⌽⍵}

Try it online!

How?

Almost a poem.

⌽⍵ - reverse the input once,

1,2=/ - get the differences list.

⌊\ - keep only the first group of ones,

⌽ - and flip it over to complete.

---

o← - assign to o,

~o - switch ones and zero(s),

⍵/⍨ - filter the input with it,

⍎¨ - turn the result into list of each digit,

• ⌈/ - and get the maximum. (that's A)

---

⍵/⍨o - filter the input with o unalt(ered),

≢ - and take the length, that would be B.

÷ - get one divided by this result,

* - and take A to that power thee.

---

⊢≡⌊ - integer?

Answer 2

05AB1E, 11 bytes

γRćgUZXzm.ï

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Explanation

γRćgUZXzm.ï ~ Full program.

γ           ~ Split into runs of digits.
 R          ~ Reverse.
  ć         ~ Push a[1:], a[0] to the stack.
   g        ~ Length (of a[0]).
    U       ~ Assign this to the integer variable X.
     ZX     ~ Get the maximum, without popping, and push X.
       zm   ~ A1/B.
         .ï ~ Is it an integer?

Emigna saved 1 byte.

Relies on the fact that if A is a positive integer N raised to the power of B, then N=A^1/B, hence it must be an integer.

Answer 3

Haskell, 85 75 72 71 bytes

Edit: -10 bytes by taking a list of digits instead of a string. Thanks to WhatToDo for pointing out that this is allowed. -3 bytes thanks to Ourous' solution in Clean. -1 byte thanks to user28667.

f s|(b,a)<-span(==last s)$reverse s=or[n^length b==maximum a|n<-[1..9]]

Try it online! Takes input as a list of digits. Example usage: f [4,1,8,0,2,0,0,0] yields True.

Explanation:

Given an input s=[4,1,8,0,2,0,0,0], we reverse the list and separate the leading elements with span(==last s): ([0,0,0],[2,0,8,1,4]). The pattern matching on (b,a) yields b=[0,0,0] and a=[2,0,8,1,4].

The list comprehension or[n^length b==maximum a|n<-[1..a]] checks whether any integer n in the range from 1 to 9 satisfies n^length b==maximum a, that is n^3=8.

Answer 4

Haskell, 104 89 bytes

@Laikoni found a shorter solution, but this is the best I could do. Thanks @Laikoni for letting me know that we can also accept lists of digits as input.

import Data.List
(g.length.last<*>maximum.concat.init).group
g b a=any(==a)$(^b)<$>[1..a]

Explanation:

Try it online!

Answer 5

Jelly, 11 bytes

ŒgµṪL9*€fṀL

Takes input as a list of digits.

Try it online!

How it works

ŒgµṪL9*€fṀL  Main link. Argument: D (digit array)

Œg           Group runs of digits, yielding a run array R.
  µ          Begin a new chain with argument D.
   Ṫ         Tail; remove and yield the last element of D.
    L        Take the length. Let's call it b.
     9*€     Compute [1**b, ..., 9**b].
         Ṁ   Take the maximum run in R, yileding [a, ..., a].
        f    Filter, yielding either [a] (if a = n**b for some n) or [].
          L  Take the length. 

Answer 6

R, 80 bytes

function(x)any((0:(m=max((d=rle(rev(utf8ToInt(c(x,''))-48)))$v[-1])))^d$l[1]==m)

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Uses utf8ToInt - 48 to split the number into digits. This throws a warning from the conversion to a string.

Using rle get the count of the trailing digits and the max value of the first digits. Return true if the any of the range 0 to max value to the power of trailing count equals max value.

I think there are further golfing opportunities, but that can wait until tomorrow.

Answer 7

R, 66 bytes

This answer is more or less a medley of MickyT and NofP's answers, and on their request, here it is:

function(x,R=rle(rev(utf8ToInt(x)-48)))!max(R$v[-1])^(1/R$l[1])%%1

It takes x as a string.

> f=function(x,R=rle(rev(utf8ToInt(x)-48)))!max(R$v[-1])^(1/R$l[1])%%1
> f("41802000")
[1] TRUE
> f("100")
[1] TRUE
> f("123456788")
[1] FALSE
> f("451111111")
[1] FALSE
> f("234543454545444")
[1] FALSE
> f("12")
[1] TRUE
> f("41902000")
[1] FALSE

Answer 8

JavaScript (ES7), 66 bytes

Takes input as either a string or an array of characters. Returns a boolean.

f=([c,...s],m)=>s.some(n=>n-c)?f(s,c<m?m:c):!(m**(1/-~s.length)%1)

Test cases

f=([c,...s],m)=>s.some(n=>n-c)?f(s,c<m?m:c):!(m**(1/-~s.length)%1)

console.log(f(       '41802000')) // 1
console.log(f(            '100')) // 1
console.log(f(      '123456788')) // 0
console.log(f(      '451111111')) // 0
console.log(f('234543454545444')) // 0
console.log(f(             '12')) // 1

Answer 9

Clean, 130 128 118 93 bytes

import StdEnv
@l#(b,a)=span((==)(last l))(reverse l)
=or[n^length b==last(sort a)\\n<-[0..9]]

Defines the function @, taking a list of integer digits.

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Answer 10

Python 2, 95 78 bytes

• Saved seventeen bytes thanks to Rod.

def f(s,i=~0):
	while s[i]==s[~0]:i-=1
	return int(max(s[:-~i]))**(1./~i)%1==0

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Answer 11

C (gcc), 144 126 117 bytes

• Saved eighteen bytes thanks to Dennis.
• Saved nine bytes thanks to ceilingcat.

j,k,m;f(char*N){N+=j=strlen(N);k=~j;for(j=-1;m=N[j--]==N[j];);for(;k++<j;)m=fmax(N[k],m);j=!fmod(pow(m-48,1./~j),1);}

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Answer 12

R, 93 bytes

function(x){n=nchar(x)
d=x%/%10^(n:1-1)%%10
i=max(which(d!=d[n]))
max(d[1:i])^(1/(n-i))%%1>0}

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The code takes an integer as input and returns FALSE if the number is pleasing, and TRUE otherwise.

Answer 13

Python 3, 88 85 bytes

def f(n):p=n.rstrip(n[-1]);a=int(max(p));b=len(n)-len(p);return round(a**(1/b))**b==a

Ungolfed:

def is_pleasing_number( n ):
    prefix = n.rstrip(n[-1])
    a = int(max(prefix))
    b = len(n) - len(prefix)
    return round(a ** (1 / b)) ** b == a

• The input argument is expected to be a digit string
• The output is either True or False.
• Similar to yet developed independently of Halvard’s answer but uses floating point arithmetic in a way that doesn't suffer from rounding errors until a ** (1 / b) is off by at least 0.5 from ^b√a which requires a value above 2⁵³ (or whatever floating point radix and mantissa length Python happens to use, see sys.float_info).
• Can be modified trivially to still work with arbitrary number bases between 2 and 36.

Answer 14

Ruby, 64 bytes

->a{!([a[/(\d)\1*$/].size,$`.chars.max]*?x!~/x1$|^2x[49]|^3x8/)}

Input as a string, returns true if:

• B==1 (no need to check A)
• A==4 and B==2
• A==9 and B==2
• A==8 and B==3

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Answer 15

Perl 6, 55 bytes

{m/(\d+?)((\d)$0*)$/;so any(^10)**$1.comb==$0.comb.max}

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After the evaluation of the initial regex--which can only succeed if the input is a positive integer--$0 contains the initial part of the number, and $1 contains the trailing repeated digits.

The comb method without arguments, applied to a string, returns a list of the characters, which in numeric context evaluates to the length of the list. So $0.comb.max is the largest of the digits in the prefix, and $1.comb is the length of the suffix.

We then check whether any(^10) (ie, the or-junction of the numbers from 0-9), when raised to the power of the length of the suffix, is equal to the largest digit in the prefix. The so forces boolean evaluation of the resulting junction, which would otherwise be just fine on its own as a truthy value, but the challenge calls for just two distinct values to be returned.

Answer 16

Kotlin, 106 bytes

fun String.p()=split(Regex("(?=(.)\\1*$)")).let{Math.pow((it[0].max()?:'0')-'0'+.0,1.0/(it.size-1))}%1==.0

Output: true/false

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Answer 17

C# (.NET Core), 132 bytes

n=>{int A=0,B=1,s=1,i=n.Length-1;for(;i-->0;)if(n[i]==n[i+1]&s>0)B++;else{A=n[i]>A?n[i]:A;s=0;}return System.Math.Pow(A,1d/B)%1==0;}

Try it online!

Acknowledgements

-12 bytes thanks to @KevinCruijssen

DeGolfed

n=>{
    int A=0, // maximum digit
        B=1, // count of trailing identical numbers
        s=1, // 1 if dealing with trailing numbers, 0 otherwise
        i=n.Length-1;

    for(; i-- > 0;)
        if(n[i] == n[i+1] & s > 0)
            B++;
        else
        {
            A = n[i] > A? n[i] : A;
            s = 0;
        }

    return Math.Pow(A, 1d/B) % 1 == 0;
}

Answer 18

Add++, 21 bytes

L,BGubLV@¦+bMG1$/^1%!

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It's been 3 and a half months, I hope I'm not ninja'ing anyone.

Answer 19

Japt, 26 18 bytes

ó¶
o l
ñ o n qV v1

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---

Takes input as a string, returns 1 for pleasing numbers, 0 otherwise.

Short explanation:

ó¶

Take the first input and split it by values where (x,y) => x===y is true. For example '41802000' to ['4','1','8','0','2','000'].

o l

Take the array from the first step, remove the last element and get its length, yielding B.

ñ o n qV v1

Find the largest element in the remaining array, yielding A, take it to the power 1/B and then return if the result is divisible by one.

---

First time working with Japt, very open to any recommendations.
Shaved off 8 bytes thanks to ETHproductions.

Answer 20

Clojure, 168 bytes

(fn[x](let[y(reverse(vec x))z(count(take-while #(= %(first y))y))a(apply max(map #(-(int %)48)(drop z y)))b(Math/pow a(/ 1 z))](<(Math/abs(- b(Math/round b)))0.00001)))

Try it online!

Answer 21

Charcoal, 33 bytes

≔ＥＳιθ⊞υ⊟θＷ⁼§υ⁰§θ±¹⊞υ⊟θ¬﹪ＸＩ⌈θ∕¹Ｌυ¹

Try it online! Link is to verbose version of code. Outputs a - for pleasing numbers. Explanation:

≔ＥＳιθ

Split the input q into characters.

⊞υ⊟θ

Remove the last character from q and push it to u (predefined to an empty list).

Ｗ⁼§υ⁰§θ±¹⊞υ⊟θ

Repeatly pop and push while the last character of q is the first character of u.

¬﹪ＸＩ⌈θ∕¹Ｌυ¹

Take the maximum digit of q and raise it to the power of the reciprocal of the length of u, then check whether the result is an integer.

Answer 22

Python 2, 91 85 bytes

s=`input()`;d=0
p=s.rstrip(s[-1])
exec"`d**(len(s)-len(p))`==max(p)>exit(1);d+=1;"*10

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Answer 23

Perl 5, 73 + 1 (-p) = 74 bytes

s/(.)\1*$//;$/=length$&;$_=(sort/./g)[-1];$q=$"++**$/while$q<$a;$_=$q==$a

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Answer 24

Java 8, 125 bytes

a->{int A=0,B=1,f=1,i=a.length-1;for(;i-->0;)if(a[i]==a[i+1]&f>0)B++;else{A=a[i]>A?a[i]:A;f=0;}return Math.pow(A,1d/B)%1==0;}

Port of @Ayb4btu's C# .NET answer.

Try it online.

Explanation:

a->{                       // Method with digit-array parameter and boolean return-type
  int A=0,                 //  Maximum digit `A` as specified in the challenge description
      B=1,                 //  `B` as specified in the challenge description
      f=1,                 //  Flag-integer, starting at 1
      i=a.length-1;        //  Index integer `i`
  for(;i-->0;)             //  Loop `i` backwards over the digits (excluding the last)
    if(a[i]==a[i+1]        //   If the current and next digits are the same,
       &f>0)               //   and the flag is still 1
      B++;                 //    Increase `B` by 1
    else{                  //   Else:
      A=a[i]>A?            //    If the current digit is larger than `A` 
         a[i]              //     Replace `A` with the current digit
        :                  //    Else:
         A;                //     Leave `A` the same
      f=0;}                //    Set the flag-integer to 0
  return Math.pow(A,1d/B)  //  Return if `A` ^ (1/`B`)
    %1==0;}                //  is an exact integer

Answer 25

Pip, 32 22 21 bytes

'.N(#a-#Ya<|a@v)RTMXy

Uses 1 for falsey and 0 for truthy, saving a byte. Try it online!

Answer 26

Pyth, 29 bytes

JezWqezJ=Pz=hZ)[e-mail]

Test suite

Python 3 translation:

Z=0
z=input()
J=z[-1]
while z[-1]==J:
    z=z[:-1]
    Z+=1
K=max(map(int,z))**(1/Z)
print(K==int(K))