### Daily Challenge #12 - Next Larger Number

#### dev.to staff on July 10, 2019

Apologies on missing yesterday's challenge. We have a fun one to hop in on today. If you haven't been following the series, feel free to give thi... [Read Full] Read this, sat down, just typed the following into `irb`, seems to work. Not recommended, but I do like Ruby's standard library:

``````a = n.to_s.chars.permutation(n.digits.size).sort.map { |n| n.join.to_i }
(i = a.find_index(n)) && a[i + 1]

``````

## Perl 5

``````#!/usr/bin/env perl

use strict;
use warnings;
use utf8;
use feature qw{ postderef say signatures state switch };
no warnings
qw{ experimental::postderef experimental::smartmatch experimental::signatures };

use Algorithm::Permute;
use JSON;

my \$json = JSON->new->pretty->canonical;

my \$base = 2019;
my @base = split m{},\$base;
my \$iter = Algorithm::Permute->new(\@base);
my @list;
while ( my @num = \$iter->next ) {
push @list, join '', @num;
}
for my \$n ( sort @list ) {
next if \$n <= \$base;
say \$n;
exit;
}
``````

Here's mine.

First, a function for permuting the digits (using recursion):

``````const permute = arr => {
if (arr.length < 2) {
return (typeof arr === "string" ? [arr] : arr);
}

if (typeof arr === "string") {
return permute(arr.split("")).map(arr => arr.join(""));
}

const result = [];
for (let i = 0; i < arr.length; i++) {
const unused = arr.map(i => i);
const candidate = unused.splice(i, 1);
const permuteUnused = permute(unused);
for (let r of permuteUnused) {
result.push([...candidate, ...Array.isArray(r) ? r : [r]]);
};
}
return result;
}
``````

Next, a sorting function:

``````const numericalOrder = (a, b) => {
const nA = Number(a);
const nB = Number(b);
return nA < nB ? -1
: nA > nB ? 1
: 0;
}
``````

And finally, the function to calculate the next largest number itself, relying on the idea that if we sort in numerical ascending order and filter out all the items less than the requested number, then the first one remaining must be our next largest number.

``````const nextLargerNumber = n => {
const digits = n.toString();
const nextValidNumbers = permute(digits)
.sort(numericalOrder)
.filter(a => Number(a) > n);
if (nextValidNumbers.length < 1) {
return null;
}
return Number(nextValidNumbers);
}
``````

Full code and tests in gist: gist.github.com/kerrishotts/a0a96d...

Ruby solution

Trying out `Object#then`, which was introduced recently in Ruby 2.6, just for the fun of it.

``````require "minitest/autorun"

class NextBiggerNumber
def self.from number
number.to_s.split('')
.then { |result| result.permutation }
.then { |result| result.map(&:join) }
.then { |result| result.map(&:to_i) }
.then { |result| result.sort }
.then { |result| result[result.index(number) + 1] }
end
end

class NextBiggerNumberTest < MiniTest::Test
def test_next_bigger_number_with_two_digits
assert_equal 21, NextBiggerNumber.from(12)
end

def test_next_bigger_number_with_three_digits
assert_equal 531, NextBiggerNumber.from(513)
end

def test_next_bigger_number_with_four_digits
assert_equal 2091, NextBiggerNumber.from(2019)
end
end
``````

Took a long drive up for vacation yesterday so I wasn't able to get this types out! I think I know what I'm gonna implement so just gotta see if I can't type it out today! Can't get oo far behind on these challenges! I owe two now !

Got it done! Was able to write out what I was thinking last night without too much hassle, pretty happy with how it came out.

Could have tried not converting to chars in the middle, but it worked out pretty nicely still I think!

``````pub fn next_largest(n: u32) -> Option<u32> {
let s = n.to_string();

let mut chars: Vec<_> = s.chars().collect();

for i in (0..chars.len() - 1).rev() {
let first = chars[i].to_digit(10).unwrap();
let second = chars[i + 1].to_digit(10).unwrap();

if second > first {
chars.swap(i, i + 1);
let s: String = chars.iter().collect();

return Some(s.parse().unwrap());
}
}

None
}

#[cfg(test)]
mod tests {
use crate::*;

#[test]
fn it_works_for_examples_that_have_no_largest() {
assert_eq!(next_largest(4), None);
assert_eq!(next_largest(100), None);
assert_eq!(next_largest(9876), None);
}

#[test]
fn it_works_for_the_examples() {
assert_eq!(next_largest(12), Some(21));
assert_eq!(next_largest(2019), Some(2091));
assert_eq!(next_largest(513), Some(531));
}

#[test]
fn it_works_for_large_numebrs() {
assert_eq!(next_largest(36852367), Some(36852376));
assert_eq!(next_largest(123456789), Some(123456798));
assert_eq!(next_largest(5010), Some(5100));
}
}
``````

I notice a lot of people did theirs differently so I thought I'd explain what I did!

One of the things I noticed that led to my solution, was the fast that the next largest number, was 1 'sort' away from the number we had. What I mean is if we imagine our number as an array of its digits, the number we wanted was 1 swap away AND would make our 'array' more sorted than it was before!

This made me realize that a modified bubble sort was exactly what I was looking for! So below I conconted something loosely based on a bubble sort. It starts at the end of the number and moves backward seeing if it can make a swap. If it does, it returns the swapped value. If we make it to the beggining of the list we know there wasn't a larger number and simply return `None`!

I wanted to do it like this (in Erlang), but didn’t figure out how to. Nice.

I think you have a mistake: `next_largest(351)` should return `513`, and you return `531`.

That’s where I gave up with the swapping approach.

This was my solution on CodeWars (JS):

``````//helper functions
const nearestUp = (input, lookup) => lookup.sort((a, b) => a - b).find(n => n > input);
const isArrayDecreasing = (array) => !(array.reduce((n, item) => n !== false && item >= n && item))

//main function
const nextBigger = n =>  {
const numberSplitted = String(n).split('');
for (let i = numberSplitted.length-1; i > 0; i--) {

if (isArrayDecreasing([numberSplitted[i], numberSplitted[i-1]])){
let nextHigherNumber = numberSplitted.splice(0, i-1);
const firstToReplace = nearestUp(numberSplitted, numberSplitted);

numberSplitted.sort((a , b) => a - b);
nextHigherNumber.push(firstToReplace);
numberSplitted.splice(numberSplitted.indexOf(firstToReplace), 1);

nextHigherNumber = nextHigherNumber.concat(...numberSplitted) === '0' ? -1 : Number(nextHigherNumber.concat(...numberSplitted).join(''));
return nextHigherNumber;
}
}
return -1;
}
``````

I’m learning Erlang.

At first, I wanted to use digit swapping, like @coreyja , but it was harder than I initially thought (e.g. `351` needs two swaps, because the next larger number is `513`, not `531`.)

To validate the swapping method, I wrote a blunt version that converts a number into an array of digits, lists all the permutations, sorts them and finds the next occurence. For that I searched for a permutation function. It uses generators and list comprehensions, and I haven’t taken the time to understand it yet, but I’ll definitely try that in future challenges.

So my current solution is the blunt one (and I removed the unit tests for the intermediate functions, that are useful for development but not interesting for the solution).

``````-module( next ).

-include_lib("eunit/include/eunit.hrl").

% convert an integer to an array of digits
digits( N ) when N >= 0 ->
digits( N, [] ).
digits( N, Digits ) when N < 10 ->
[ N | Digits ];
digits( N, Digits ) ->
digits( N div 10, [ N rem 10 | Digits ] ).

% convert an array of digits to an integer
number( Digits ) ->
number( Digits, 0 ).
number( [ D | Rest ], N ) ->
number( Rest, N * 10 + D );
number( [], N ) ->
N.

% list all permutations of an array, taken from
% http://erlang.org/doc/programming_examples/list_comprehensions.html#permutations
perms( [] ) -> [ [] ];
perms( L )  -> [ [ H | T ] || H <- L, T <- perms( L -- [ H ] ) ].

% return the first element occuring after N in a list
find_next( N, [ N, M | _ ] ) ->
M;
find_next( N, [ _ | Rest ] ) ->
find_next( N, Rest );
find_next( _, [] ) ->
none.

next( N ) ->
Digits = digits( N ),
Next = find_next( Digits, lists:sort( perms( digits( N ) ) ) ),
case Next of
none -> none;
_ -> number( Next )
end.

next_test_() -> [
?_assertEqual( none, next( 5 ) ),
?_assertEqual( 51, next( 15 ) ),
?_assertEqual( 536, next( 365 ) ),
?_assertEqual( 21, next( 12 ) ),
?_assertEqual( 531, next( 513 ) ),
?_assertEqual( 513, next( 351 ) ),
?_assertEqual( none, next( 531 ) ),
?_assertEqual( 2091, next( 2019 ) )
].
``````

I know I am late.
This what I did in C#

``````static void Main()
{
Console.WriteLine("Enter any integer: ");
var referrer = new string(read.ToCharArray().OrderBy(i => i).ToArray());
var input = read.ToCharArray().Select(x => int.Parse(x.ToString())).ToList();
var last = int.Parse(string.Join("", input.OrderByDescending(i => i).ToArray()));
var permutations = Enumerable.Range(first, last).ToList();
var container = new List<int>();
foreach (var item in permutations)
{
var reference = new string(item.ToString().ToCharArray().OrderBy(i => i).ToArray());

int c = string.Compare(referrer, reference);
if (c != 0) continue;
}

if (inputNumberIndex >= 0)
{
if (container.Count == 0) Console.WriteLine(container.FirstOrDefault());

var findIndex = inputNumberIndex + 1;
Console.WriteLine(container[findIndex]);
}
else
{
Console.WriteLine("-1");
}

}
``````

## Perl 6

``````#!/usr/bin/env perl6

my \$base = 2019;
my @numbers = split('',\$base);

my @perm;
for @numbers.permutations -> @n { @perm.push( @n.join('') )}

for @perm.unique.sort -> \$p {
next unless \$p > \$base;
say \$p ;
exit;
}
``````

Ruby

Put this together on my phone sorry for formatting.

This one doesn’t generate all the permutations of all the digits.

``````[9999, 12, 513, 2019].each do |initial|
next_larger = initial.to_s.reverse
next_larger.length.times do |i|
next_larger.length.times do |j|
# luckily for us, ascii numbers are stored
# in the same order as actual numbers
if next_larger[j] > next_larger[i]
tmp = next_larger[j]
next_larger[j] = next_larger[i]
next_larger[i] = tmp
end
end
end
next_larger.reverse!

puts "#{initial}\t#{next_larger}" if initial != next_larger.to_i
end
``````

repl.it/@daxyq/DailyChallenge12

Ooops. Today I caught the challenge a bit late... I sketched something, hopefully I'll finish it by tomorrow before the next challenge :-/

I didn't get to this one till today too, and I still own one from a few days ago! So don't feel too bad lol

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