# Covering a Skyline with brush strokes

Given a non-negative integer skyline height list, answer how many uninterrupted 1-unit-high horizontal brush strokes are needed to cover it.

[1,3,2,1,2,1,5,3,3,4,2], visualised as:

      5
5  4
3    5334
32 2 53342
13212153342


needs nine brush strokes:

      1
2  3
4    5555
66 7 88888
99999999999


### Examples

[1,3,2,1,2,1,5,3,3,4,2]9

[5,8]8

[1,1,1,1]1

[]0

[0,0]0

[2]2

[2,0,2]4

[10,9,8,9]11

• For interested high-rep users: Based on this as per this.
Commented Feb 4, 2019 at 11:22
• So, all brush strokes are horizontal?
– tsh
Commented Feb 4, 2019 at 11:38
Commented Feb 4, 2019 at 11:46
• It wasnt codegolf, but I had this question for an interview code test about a year ago. Commented Feb 6, 2019 at 22:19

# JavaScript (Node.js), 38 bytes

a=>a.map(v=>(n+=v>p&&v-p,p=v),p=n=0)|n


Try it online!

Simply a greedy algorithm which scan from left to right, only draw lines if needed, and draw it as long as possible.

Thanks Arnauld, save 2 3 bytes

• @Arnauld nice catch. totally forgot it.
– tsh
Commented Feb 4, 2019 at 11:56
• How did you realise this?
Commented Feb 4, 2019 at 13:19
• @Adám Nothing magic. First time I read the question I was confused by how to search until i realize all lines are horizontal only. And then this formula just came up to my mind naturally....
– tsh
Commented Feb 4, 2019 at 13:45
• magic seems like a well-fitting word to describe that process.
Commented Feb 4, 2019 at 13:46
• While this is the origin of the now widely used algorithm, it is explained here.
Commented Feb 5, 2019 at 12:14

# 05AB1E,  8 7  5 bytes

Saved 2 bytes thanks to @Adnan

0š¥þO


Try it online!

### How?

This is using the algorithm that was first found by @tsh. If you like this answer, make sure to upvote their answer as well!

Each time a skyscraper is lower than or as high as the previous one, it can be painted 'for free' by simply extending the brushstrokes.

For instance, painting skyscrapers $$\B\$$ and $$\C\$$ in the figure below costs nothing.

On the other hand, we need 2 new brushstrokes to paint skyscraper $$\E\$$, no matter if they're going to be reused after that or not.

For the first skyscraper, we always need as many brushstrokes as there are floors in it.

Turning this into maths:

$$S=h_0+\sum_{i=1}^n \max(h_i-h_{i-1},0)$$

If we prepend $$\0\$$ to the list, this can be simplified to:

$$S=\sum_{i=1}^n \max(h_i-h_{i-1},0)$$

### Commented

0š¥þO     # expects a list of non-negative integers  e.g. [10, 9, 8, 9]
0š        # prepend 0 to the list                    -->  [0, 10, 9, 8, 9]
¥       # compute deltas                           -->  [10, -1, -1, 1]
þ      # keep only values made of decimal digits
# (i.e. without a minus sign)              -->  ["10", "1"]
O     # sum                                      -->  11

• I think 0š¥ʒd}O saves you a byte. Commented Feb 4, 2019 at 12:27
• @Don'tbeax-tripledot I was editing my answer to exactly that when I saw your comment ;) Commented Feb 4, 2019 at 12:29
• Beautiful explanation.
Commented Feb 4, 2019 at 14:32
• Replacing ʒd} with þ should save you two bytes. Commented Feb 4, 2019 at 15:23
• @Adnan Ah, nice. Thanks! Commented Feb 4, 2019 at 15:34

(0%)
p%(h:t)=max(h-p)0+h%t
p%_=0


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An improvement on Lynn's solution that tracks the previous element p instead of looking at the next element. This makes the base case and recursive call shorter in exchange for needing to invoke (0%).

max(h-p)0 could be max h p-p for the same length.

# Python 3, 37 bytes

lambda a:sum(a)-sum(map(min,a[1:],a))


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-5 bytes by switching to Python 3, thanks to Sarien

# Python 2, 4743 42 bytes

lambda a:sum(a)-sum(map(min,a[1:],a[:-1]))


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Alt:

lambda a:sum(a)-sum(map(min,zip(a[1:],a)))


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• In Python 3 you can ditch the [:-1], saving 5 bytes. Commented Feb 4, 2019 at 13:24
• @Sarien Thanks :D, I didn't know map is different in python 2 and 3 Commented Feb 4, 2019 at 13:28

f(a:b:c)=max(a-b)0+f(b:c)
f a=sum a


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Line 2 could be f[a]=a if I didn't also have to handle the case [].

• Commented Feb 4, 2019 at 19:34

# Jelly, 5 bytes

A port of my 05AB1E answer, which itself is similar to @tsh JS answer.

ŻI0»S


Try it online!

### Commented

ŻI0»S    - main link, expecting a list of non-negative integers  e.g. [10, 9, 8, 9]
Ż        - prepend 0                                             -->  [0, 10, 9, 8, 9]
I       - compute the deltas                                    -->  [10, -1, -1, 1]
0»     - compute max(0, v) for each term v                     -->  [10, 0, 0, 1]
S    - sum                                                   -->  11


# K (oK), 12 7 bytes

-5 bytes thanks to ngn!

A k (oK) port of Arnauld's 05AB1E solution (and tsh's JavaScript solution):

+/0|-':


Try it online!

# J, 15 bytes

A J port of Arnauld's 05AB1E solution (and tsh's JavaScript solution):

1#.0>./2-~/\0,]


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1#.2(1 0-:])\0,@|:@,~1$~"0]  Try it online! • oK: each prior (':) uses an implicit identity element (0 for -) before the list, so 0, is unnecessary. you can omit { x} to make it a composition: +/0|-': – ngn Commented Feb 4, 2019 at 15:11 • @ngn Thanks! Apparently I've forgotten this: Some primitive verbs result in a different special-cased initial value: +, *, - and & are provided with 0, 1, 0 or the first element of the sequence, respectively Commented Feb 4, 2019 at 15:25 # Haskell, 34 32 bytes 2 bytes trimmed by Lynn g x=sum$max 0<$>zipWith(-)x(0:x)  Try it online! So to start we have zipWith(-). This takes two lists and produces a new list of their pairwise differences. We then combine it with x and (0:x). (0:) is a function that adds zero to the front of a list and by combining it with zipWith(-) we get the differences between consecutive elements of that list with a zero at the front. Then we turn all the negative ones to zero with (max 0<$>). This creates a new list where each element is the number of new strokes that has to be started at each tower. To get the total we just sum these with sum.

• g x=sum$max 0<$>zipWith(-)x(0:x) is 32 bytes :)
– lynn
Commented Feb 4, 2019 at 19:39
• As is sum.zipWith((max 0.).(-))<*>(0:)
– lynn
Commented Feb 4, 2019 at 19:41
• @Lynn Your second one is going to need extra parentheses since the . has higher precedence than <*>. Commented Feb 4, 2019 at 19:43

# Dyalog APL, 11 bytes

+/0⌈¯2-/0,⊢


, concatenate, so 0, concatenates 0.

¯2-/
or
¯2 length of window. (minus means the window is rotated)
-/ does a window operation of -(minus) with window size 2

0⌈ the operation is called ceiling, but when it gets an input from both sides it also does "get the larger of the two inputs" so 0⌈1 3 0 ¯2 will give 1 3 0 0.

and finally, a +/ which is a sum.

Try it!

• Here is a link which you could add to your post to help others test your solution. If you click on Postify over there it will also provide you with some formatting for the answer
– ovs
Commented Apr 6, 2022 at 7:41
• I've fixed your formatting and byte count (assuming you use some for of Single Byte Character Set), and included ovs's link.
Commented Apr 6, 2022 at 7:44
• Thanks for the help! Commented Apr 6, 2022 at 8:35
• Somehow this works for negative towers as well. Very neat. Commented Apr 6, 2022 at 9:34
• Welcome to Code Golf, by the way! Commented Apr 6, 2022 at 13:26

# Japt, 8 bytes

-2 bytes from @Shaggy

mîT Õ¸¸è


Explanation

mîT Õ¸¸è      Full program. Implicit input U
e.g. U = [2,0,2]
mîT             Map each item X and repeat T(0) X times
U = ["00","","00"]
Õ           Transpose rows with columns
U = ["0 0","0 0"]
U = ["0","0","0","0"]
è       Return the count of the truthy values


Try it online!

• 8 bytes: mîT Õ¸¸è Commented Feb 4, 2019 at 12:27
• Nice use of A.y()'s padding, by the way. Commented Feb 4, 2019 at 20:21

# MATL, 8 bytes

0whd3Y%s


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Pretty much @Arnauld's algorithm. Saved one byte (thanks @LuisMendo) by casting to uint64 rather than selecting ) positive entries.

# R, 30 bytes

sum(pmax(0,diff(c(0,scan()))))


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Porting of @Arnauld solution which in turn derives from @tsh great solution

# Common Lisp, 88 87 bytes

(lambda(s)(let((o 0))(dolist(c s)(incf o(max 0 c))(mapl(lambda(y)(decf(car y)c))s))o))


## non-minified

(lambda (skyline)
(let ((output 0))
(dolist (current-skyscraper-height skyline)
(incf output (max 0 current-skyscraper-height))
(mapl (lambda (skyscraper)
(decf (car skyscraper) current-skyscraper-height))
skyline))
output)))


Test it

When one tower is painted, it takes a number of brushstrokes equal to it's height. These brushstrokes translate to all the following ones, which is indicated here by subtracting the current tower's height from all other towers (and itself, but that doesn't matter). If a following tower is shorter, then it will be pushed to a negative number, and this negative number will subsequently subtracted from the towers that follow (indicating brushstrokes that could not be translated from a previous tower to the next ones). It actually just subtracts the number from all tower heights, including previous ones, but this doesn't matter because we don't look at the previous ones again.

• Welcome to PPCG. Could you possibly provide a link to an online testing environment for ease of verification? Commented Feb 7, 2019 at 0:52
• Yes, absolutely. rextester.com/TKBU14782 The answer will be updated shortly Commented Feb 7, 2019 at 0:55
• Well done. +1 for a nice, working first post. Have fun golfing. Commented Feb 7, 2019 at 0:58

# Husk, 4 bytes

ΣẊ>Θ


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Uses the same algorithm as everybody else (see here for an explanation), it just does so with fewer bytes than anybody else :D

### Explanation

ΣẊ>Θ
Θ    Prepend a 0 to the values
Ẋ      For each pair of adjacent values (a,b)
>     Return b-a if b>a else 0
Σ       Sum all of these values


> in Husk does a great job for this task: it does return a truthy value iff b>a, but this truthy value also happens to be the difference b-a, accomplishing two tasks in one go!

# Japt-x, 7 6 bytes

änT f¬


Try it

1 byte saved thanks to Oliver.

änT f¬     :Implicit input of integer array
ä          :Consecutive pairs
n         :Reduced by inverse subtraction
T        :After first prepending 0
f      :Filter elements by
¬     :Square root (The square root of a negative being NaN, which is falsey)
:Implicitly reduce by addition and output

• 6 bytes Commented Feb 4, 2019 at 18:46
• Nice one, @Oliver; wouldn't have thought of that. Commented Feb 4, 2019 at 20:20

# 05AB1E, 13 10 bytes

Z>Lε@γPO}O


Explanation:

Z            # Get the maximum of the (implicit) input-list
>           # Increase it by 1 (if the list only contains 0s)
L          # Create a list in the range [1, max]
ε         # Map each value to:
@        #  Check if this value is >= for each value in the (implicit) input
γ       #  Split into chunks of adjacent equal digits
P      #  Take the product of each inner list
O     #  Take the sum
}O   # And after the map: take the sum (which is output implicitly)


# Retina 0.8.2, 21 bytes

\d+
$* (1+)(?=,\1) 1  Try it online! Link includes test cases. Explanation: \d+$*


Convert to unary.

(1+)(?=,\1)


Delete all overlaps with the next tower, which don't need a new stroke.

1


Count the remaining strokes.

# Nekomata, 5 bytes

ç-P‼∑


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A port of @Arnauld's 05AB1E answer.

ç-P‼∑
ç           Prepend zero
-          Minus to get the delta
P‼        Filter by positive
∑       Sum


# Thunno 2S, 6 5 bytes

0ƤṢ0§


Attempt This Online!

-1 thanks to @alephalpha

#### Explanation

0ƤṢ0§   # Implicit input
0Ƥ      # Prepend a 0
Ṣ     # Compute deltas
0§   # Maximum with 0
# S flag sums the list
# Implicit output

• -1 byte: 0ƤṢ0§. Commented Jun 14, 2023 at 5:41
• @alephalpha thanks, updated Commented Jun 14, 2023 at 8:18

# Perl 5, 21 bytes

$\+=$_>$'&&$_-$';//}{  TIO How • -p + }{ + $\ trick
• // matches empty string so that for next line postmatch $' will contain previous line • $\+=$_>$'&&$_-$' to accumulate difference between current line and previous if current is greater than previous, (could also be written $\+=$_-$' if$_>$', but perl doesn't parse $\+=$_-$'if$_>$' the same)

# C# (Visual C# Interactive Compiler) with flag /u:System.Math, 47 bytes

n=>n.Select((a,i)=>i<1?a:Max(a-n[i-1],0)).Sum()


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## Old version, with flag /u:System.Math, 63 bytes

n=>n.Aggregate((0,0),(a,b)=>(a.Item1+Max(0,b-a.Item2),b)).Item1


I feel like this solution is more elegant than the first one. It goes through the array with a two-value tuple as a starting value, picking up values, and stores the value before it in the second part of the tuple.

Try it online!

# Pyth, 8 bytes

s>#0.++0


Yet another port of @tsh's marvellous answer. Takes the sum (s) of the positive values (>#0) of the deltas (.+) of the input with 0 prepended (+0Q, trailing Q inferred).

Try it online here, or verify all the test cases at once here.

## String joining method, 10 bytes

This was the solution I wrote before browsing the other answers.

lcj.t+d*LN


Test suite.

lcj.t+d*LNQ   Implicit: Q=eval(input()), b=<newline>, N=<quote mark>
Trailing Q inferred
L Q   Map each element of Q...
* N    ... to N repeated that many times
+b       Prepend a newline
j           Join on newlines
c            Split on whitespace
l             Take the length, implicit print


# Wolfram Language (Mathematica), 34 bytes

A port of @Arnauld's solution.

Tr@Select[{##,0}-{0,##},#>0&]&@@#&


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# Clojure, 50 bytes

#((reduce(fn[[s n]i][(+(max(- i n)0)s)i])[0 0]%)0)


Try it online! (Why doesn't this print anything?)

#( ; begin anonymous function
(reduce
(fn [[s n] i] ; internal anonymous reducing function, destructures accumulator argument into a sum and the previous item
[(+ (max (- i n) 0) s ; the sum part of the accumulator becomes the previous sum plus the larger of zero and the difference between the current number and the last one, which is how many new strokes need to be started at this point
i]) ; ...and the previous item part becomes the current item
[0 0] ; the initial value of the accumulator gives no strokes yet, and nothing for them to cover yet
%) ; reduce over the argument to the function
0) ; and get the sum element of the last value of the accumulator.

• Welcome to PPCG! I don't know anything about Clojure, but a quick search shows that you'll need to evaluate the lazy for loop. Try it Online! (Tip: you can use the link button to auto-format your answer). Hope you stick around and have some fun!
– Jo King
Commented Feb 6, 2019 at 8:47

# Java (JDK), 57 bytes

a->{int s=0,p=0;for(int x:a)s-=(x>p?p:x)-(p=x);return s;}


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Another port of tsh's Javascript answer. So make sure you've upvoted them.

Note that I used subtraction instead of addition because it allowed me to write (p=x) as right operand in the same statement.

# Factor + math.unicode, 24 bytes

[ dup 0 prefix [v-] Σ ]


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## Explanation

         ! { 10 9 8 9 }
dup      ! { 10 9 8 9 } { 10 9 8 9 }
0        ! { 10 9 8 9 } { 10 9 8 9 } 0
prefix   ! { 10 9 8 9 } { 0 10 9 8 9 }
[v-]     ! { 10 0 0 1 }  (subtract two vectors, clamping negative numbers to 0)
Σ        ! 11


# Kotlin

fun f(test:IntArray):Int{var s=0;var p=0;test.forEach{val f=(if(it>p)p else it);p=it;s-=f-p};return s}


Try it yourself

• Since this is a code-golf question, you are required to golf your answer i.e. Make it as short as possible , removing whitespace and inlining variable values. You can check the tips for Kotlin here to improve your answer further. Commented May 3, 2022 at 12:32

# Arturo, 35 33 bytes

\$=>[l:0sum map&'x[max@[0x-l]l:x]]


Try it

# Uiua, 11 bytes

/+↥0-↘¯1⊂0.


Try it!

/+↥0-↘¯1⊂0.
.  # duplicate
⊂0   # prepend a zero
↘¯1     # drop last element
-        # subtract the two arrays
↥0         # change negative numbers to zero
/+           # sum


# MATL, 1514 13 bytes

ts:<~"@Y'x]vs


Input is a column vector, using ; as separator.

### Explanation

t       % Implicit input: column vector. Duplicate
s       % Sum
:       % Range from 1 to that. Gives a row vector
<~      % Greater or equal? Element-wise with broadcast
"       % For each column
@     %   Push current columnn
Y'    %   Run-length encoding. Gives vector of values (0, 1) and vector of lengths
x     %   Delete vector of lengths
]       % End
v       % Vertically concatenate. May give an empty array
s       % Sum. Implicit display