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# mkozakov/gist:59af0fd5bddbed1a0399Secret Created Oct 29, 2013

 /* package whatever; // don't place package name! */ import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ class Ideone { public static void main (String[] args) throws java.lang.Exception { int[] myIntArray = {2, 5, 1, 2, 3, 4, 7, 7, 6}; System.out.println(calculateVolume(myIntArray)); } public static int calculateVolume(int[] land) { int leftMax = 0; int rightMax = 0; int left = 0; int right = land.length - 1; int volume = 0; while(left < right) { if(land[left] > leftMax) { leftMax = land[left]; } if(land[right] > rightMax) { rightMax = land[right]; } if(leftMax >= rightMax) { volume += rightMax - land[right]; right--; } else { volume += leftMax - land[left]; left++; } } return volume; } }

### igor47 commented Oct 30, 2013

 fun problem! i took a stab at it also: https://gist.github.com/igor47/7228586

### sathya-me commented Oct 30, 2013

 will this work for the following case [2,7,2,7,4,7,1,7,3,7]

### joist commented Oct 30, 2013

 @sathya-me : It does handle it properly. The left and the right pointers rise to 7 and then fill in all the gaps until they meet. I tried my own hand at this problem and came up with a slightly more naive (but clearer) solution : https://gist.github.com/Joist/7229807 edit: I put my own attempt at the single pass method in my code with comments to make it a bit clearer.

### msimpson commented Oct 30, 2013

 I wrote up a solution using JS, here: https://gist.github.com/msimpson/eb0189dd687d696ab70a Mine is based on the point-inside-polygon methodology. It is more complex than what is above, and generates a maximum height at the start. However, it would continue to work with only slight modification on a two, or possibly three, dimensional matrix -- even with gaps inside columns that create enclosed areas.

### dmitry commented Oct 30, 2013

 Looked for the solutions and rewritten them to the ruby: https://gist.github.com/dmitry/287bd6cad9abc70436f4

### shaobohou commented Oct 30, 2013

 Another single pass solution, https://gist.github.com/shaobohou/7231961

### Shruubi commented Oct 30, 2013

 Did up a small Python solution passing both test cases: https://gist.github.com/Shruubi/7232152

### Zibx commented Oct 30, 2013

 JS implementation. 2n complexity https://gist.github.com/Zibx/7235776

### thisvar commented Oct 30, 2013

 Java, n complexity, single straight pass from left to right : https://gist.github.com/anonymous/7236132

### twuttke commented Oct 30, 2013

 Slightly simpler, slightly faster, and handles negative heights: http://jsbin.com/iquQOSo/1/edit ```function maxWater(heights) { var left = 0, right = heights.length - 1, leftMax = heights[left], rightMax = heights[right], volume = 0; while (left < right) { if (leftMax < rightMax) { var height = heights[++left]; leftMax = Math.max(leftMax, height); volume += leftMax - height; } else { var height = heights[--right]; rightMax = Math.max(rightMax, height); volume += rightMax - height; } } return volume; }```

### olajep commented Oct 30, 2013

 Did a quick Haskell solution: https://gist.github.com/olajep/7236551 ``````rainSolver = solve 0 solve acc [] = acc solve acc (x:xs) | xs' == [] = solve acc xs -- no right border -> this x cannot be left border | otherwise = solve (acc+area) xs' where height = minimum [x, maximum xs] (elems,xs') = span (

### Zibx commented Oct 30, 2013

 JS one cycle implementation. https://gist.github.com/Zibx/7235776/726b210cf6b6c646a925c911bb00a5ab68200810

### AndroidHumanoid commented Oct 30, 2013

 public static int mycalculateVolume(int[] land) { int leftMax = 0; int subvolume = 0; int volume = 0; `````` for (int index = 0; index < land.length; index++) { if (land[index] >= leftMax){ leftMax = land[index]; volume = volume + subvolume; subvolume = 0; } else { subvolume = subvolume + leftMax - land[index]; } } return volume; } ``````

### hwmrocker commented Oct 30, 2013

 me too, i used just one iteration https://gist.github.com/hwmrocker/7241002

### pavelfatin commented Oct 30, 2013

 O(n) recursive one-liners (Haskell, Clojure, Scala)

### domluna commented Oct 31, 2013

 Solution in julia, just did a simple linear walkthrough https://gist.github.com/Niessy/7243069

### amuraco commented Oct 31, 2013

 Single Iteration Streaming Java solution. Not that memory friendly though. https://gist.github.com/amuraco/7244151

### jdanielmyers commented Oct 31, 2013

 pretty simple single pass python solution: https://gist.github.com/jdanielmyers/7245316

### leikind commented Oct 31, 2013

 Maybe not the most efficient, but it's only 10 lines of code of functional style Ruby

### abo-abo commented Oct 31, 2013

 My take in Clojure: https://gist.github.com/abo-abo/7247459 . Not the best efficiency, but very straightforward.

### flowolf commented Oct 31, 2013

 here is my Python solution: https://gist.github.com/flowolf/7247585

### KvanTTT commented Oct 31, 2013

 There is my awful but satisfactory solution with all answers generation for predifined count of walls and max wall height (in my case 8 * 8): https://github.com/KvanTTT/Twitter-Puddles-Counter But it's not ideal. Number of answers can be reduced to 4 times due to symmetry along x and y axis. I mean V(x_1, x_2, ..., x_n) = V(x_n, x_n-1, ..., x_1) and V(x_1, x_2, ..., x_n) = n*h - V(h - x_1, h - x_2, ..., h - x_n). Possible there are another dependencies from input heights which are not so clear.

### Areks commented Oct 31, 2013

 Hello guys, let me show my decision )) function calc(input){ if (input.length < 3) return 0; var v = 0, sort = input.slice(0).sort(function(x,y){return y-x}), a = input.indexOf(sort), b = input.indexOf(sort, (sort==sort) ? a+1 : 0), last, first = ( a > b ) ? (last=a,b) : (last=b,a); for (var j = first+1; j < last; j++ ) v+=sort-input[j]; return v + calc(input.filter(function(z,y){return (y < first-1) || (y >= last) })) }

### ghost commented Oct 31, 2013

 PHP solution: public function findVolume(\$arr) { \$v = 0; \$l = 0; \$r = count(\$arr) - 1; ``````while (\$l < count(\$arr) && \$arr[\$l + 1] >= \$arr[\$l]) { \$l++; } while (\$r > 0 && \$arr[\$r - 1] >= \$arr[\$r]) { \$r--; } \$pointer = \$l; while (++\$pointer < \$r) { \$v += \$arr[\$l] - \$arr[\$pointer]; } \$return \$v; `````` }

### fergara commented Oct 31, 2013

 The Groovy version: https://gist.github.com/fergara/7250590#file-calculaterainvolume-groovy

### ghost commented Oct 31, 2013

 @AndroidHumanoid I like your solution, but unfortunately it doesn't work for [2, 5, 1, 2, 3, 4, 7, 5, 6] which should give 11.

### oralokan commented Oct 31, 2013

 I did a python implementation based on a simple observation. For a square to be filled with water, it has to fulfill these two conditions: It has to be a free-space square (as opposed to a solid block) to begin with. There has to be a solid block at the same height on either side of it (not necessarily adjacently) https://gist.github.com/oralokan/107958531b3e33062bc5

### ubermajestix commented Oct 31, 2013

 Here's my solution in ruby that prints the board: https://gist.github.com/ubermajestix/7253875

### DrewFitz commented Oct 31, 2013

 Here's mine which solves it in one pass. https://gist.github.com/DrewFitz/7254644 It moves through the array keeping a list of unresolved "left walls" and resolves the guaranteed water blocks when hitting a right wall. Each time it resolves, it only adds the new blocks gained by "raising" the puddle's water level.

### Ruszrok commented Oct 31, 2013

 I have solution - https://gist.github.com/Ruszrok/3adc23baa83b97de7145 It works in many cases. Can break it anyone?

### HollyGurza commented Nov 1, 2013

 my Python solution - https://gist.github.com/HollyGurza/7261085 please test this

### benjaminliu commented Nov 1, 2013

 Here is a test case, please try it 6, 1, 4, 6, 7, 5, 1, 6, 4 and here is my solution, double loop, much slower https://gist.github.com/benjaminliu/7261468

### jianchenglee commented Nov 1, 2013

 my lua solution: https://gist.github.com/jianchenglee/7262518

### bxt commented Nov 1, 2013

 I have created a python solution that makes random landscapes like this one: `````` # ### #~~~~~~~~######~~~~~~~# #~~~~~~~~~~~###~~~~~~#######~~~~~##~#~~~~~~~##~~~~~~~~~~~~~~~~~~# #~~~~~~~~~~####~~~~##########~~~~#####~~##~~##~~~~~~~~~~#~~~~~~## #~###~~~~~~~#######~~~##########~~~###############~~~~~~~~~#~~~~#####~#~## #~###~~~############~#############~###############~~~~~~~~####~~########### ######~#############~##############################~#~#~~~################# ######~#################################################~################### ########################################################~###################~~## ``````

### AndroidHumanoid commented Nov 1, 2013

 Decker108 "I like your solution, but unfortunately it doesn't work for [2, 5, 1, 2, 3, 4, 7, 5, 6] which should give 11." Yeees, what about that (it seems work, but how): ``````public static int mycalculateVolume(int[] land) { int leftMax = 0; int rightMax = 0; int volumeLeft = 0; int volumeRight = 0; int volume = 0; for (int index = 0; index < land.length; index++) { if (land[index] >= leftMax){ leftMax = land[index]; volume = volume + volumeLeft; volumeLeft = 0; } else { volumeLeft = volumeLeft + leftMax - land[index]; } if (land[land.length - 1 - index] >= rightMax){ rightMax = land[land.length - 1 - index]; volume = volume + volumeRight; volumeRight = 0; } else { volumeRight = volumeRight + rightMax - land[land.length - 1 - index]; } } return volume; } ``````

### BorzdeG commented Nov 2, 2013

 my version of the solution with 22 test models: https://github.com/BorzdeG/info.alenkov.demo.puddle_rain

### philipnilsson commented Nov 5, 2013

 Probably the most elegant solution to this problem is using scanl/r and zipWith. Corresponds very closely to the mathematical breakdown of this problem. Here's a solution in Haskell. ```water hs = sum \$ zipWith (flip (-)) hs \$ zipWith min (scanl1 max hs) (scanr1 max hs)```

### skariel commented Nov 6, 2013

 Mot solutions are actually wrong...

### yannick1974 commented Nov 6, 2013

 Hi all, Another solution in Python using numpy (and matplotlib): https://gist.github.com/yannick1974/7334581 Yannick

### celwell commented Nov 7, 2013

 So does this work for: 4,0,0,4,0,4,0,0,4,2 ?

### ghost commented Nov 8, 2013

 Yet another solution: https://gist.github.com/syfou/7364514 It uses recursion: I tried to go for a readable and a painless, quickly implemented solution rather than something more clever.

### est commented Nov 11, 2013

 my dumb but working version ``````def get_volumn(seq): total = 0 for level in xrange(max(seq)-1, 0, -1): row = [ 1 if row-level > 0 else 0 for row in seq] first = row.index(1) last = len(row) - row[::-1].index(1) - 1 if last - first > 0: total += row[first:last].count(0) return total if '__main__' == __name__: s1 = [2,5,1,2,3,4,7,7,6] print get_volumn(s1) s2 = [ 2, 5, 1, 3, 1, 2, 1, 7, 7, 6] print get_volumn(s2) ``````

### thulka commented Nov 12, 2013

 mkozakov's solution is quite perfect, no allocations, no stack uage. single pass, recursive in python: https://gist.github.com/thulka/7429806 btw, est, your solution wrong: [7,0,5,0,7,0] should be 16, yours returns 14. also you might have hidden loops in the list selectors, so not perfectly single pass. finally, it was about an elegant solution, not just a dumb working one. a simple working one, non single pass would be: for each bin the water = min(leftmax, rightmax) - current height

### chrisdone commented Nov 12, 2013

 My first thought was to use a spreadsheet. Here's that in Haskell: ``````import Control.Lens import qualified Data.Vector as V import Data.Vector ((!),Vector) -- | Calculate the water volume. water :: Vector Int -> Int water = V.sum . V.map (view _3) . loeb . V.imap cell where cell i x xs | i == 0 = edge _1 | i == V.length xs - 1 = edge _2 | otherwise = col i x xs where edge ln = set l (view l (col i x xs)) (x,x,0) where l = cloneLens ln col i x xs = (l,r,min l r - x) where l = neighbor _1 (+) r = neighbor _2 (-) neighbor l o = max x (view l (xs ! (i `o` 1))) `````` It makes a vector where the value of each cell is calculated based on the two neighboring cells. Except edges, who only have one neighbor. It's a direct way of writing how I thought of the problem statement: I'm a cell. My puddle's extent is determined by finding the largest cell to the left of me, and the largest to the right of me. My potential volume is calculated by taking the minimum height of said two cells. My actual volume is calculated by subtracting from that my height. If I'm on the edge (either side), then my height on one side is restricted to me, and my volume is zero. Done. Oh, the `loeb` function is from here: ``````-- | Map over a structure, passing in itself. loeb :: Functor f => f (f b -> b) -> f b loeb x = fmap (\f -> f (loeb x)) x ``````

### neuronsguy commented Nov 12, 2013

 Short, single-pass python solution ```q = [2,5,1,3,1,2,1,7,7,6] def solve(heights): water = *len(heights) total = 0 runoff = 0 for i,height in enumerate(heights): if i == 0: continue above = max(height, heights[i-1] + water[i-1]) - height water[i] = above if water[i] == 0: runoff = 0 else: runoff += water[i] return sum(water)-runoff print q, solve(q)```

### chrisdone commented Nov 13, 2013

 @neuronsguy That's an incomplete solution, it only works in one direction. E.g. `solve([6,7,7,4,3,2,1,5,2])` → `0`

### chrisdone commented Nov 13, 2013

 @DrewFitz Only works properly in one direction. ``````print puddle([2,5,1,2,3,4,7,7,6]) → 10 print puddle([6,7,7,4,3,2,1,5,2]) → 11 ``````

### chrisdone commented Nov 13, 2013

 @HollyGurza Only works properly in one direction. You guys really need to think about the problem in both directions.

### chrisdone commented Nov 13, 2013

 @igor47 Only works in one direction.

### tmoertel commented Nov 14, 2013

 Thanks for the fun problem :-) I ended up solving it recursively, working from the base cases up to a general solution, and then I converted the recursive algorithm into the following Python function taking a height array A. For a problem of size len(A) = N, the function requires O(N) time and O(1) space. For a full derivation of the algorithm, starting from first principles, see the (extensive) discussion in the source code: https://github.com/tmoertel/practice/blob/master/water-holding-capacity/water_holding_capacity.py ```def water_holding_capacity_iterative(A): (i, j) = (0, len(A) - 1) lm = rm = vol = 0 # while segments remain, consume left- or right-end segment while i <= j: if lm < rm: vol += max(0, lm - A[i]) lm = max(lm, A[i]) i += 1 else: vol += max(0, rm - A[j]) rm = max(rm, A[j]) j -= 1 return vol```

### emakarov commented Nov 16, 2013

 python: ``````numbers = [6,7,7,4,3,2,1,5,2] def vol(numbers): prev_max = 0 total = 0 for n in numbers[0:-1]: if n > prev_max: prev_max = n else: total = total + (prev_max-n) total_1 = total total = 0 numbers = numbers[::-1] prev_max = 0 for n in numbers[0:-1]: if n > prev_max: prev_max = n else: total = total + (prev_max-n) total = min(total_1,total) return total ``````

### emakarov commented Nov 16, 2013

 or one-way loop in python ``````def vol2(numbers): prev_max_left = 0 prev_max_right = 0 total_left = 0 total_right = 0 for n in xrange(0,len(numbers)-1): if numbers[n] > prev_max_left: prev_max_left = numbers[n] elif numbers[n] < prev_max_left: total_left = total_left + (prev_max_left - numbers[n]) if numbers[-1-n] > prev_max_right: prev_max_right = numbers[-1-n] elif numbers[-1-n] < prev_max_right: total_right = total_right + (prev_max_right - numbers[-1-n]) return min(total_left,total_right) ``````

### divipp commented Nov 20, 2013

 Another Haskell solution, O(1) space when switched to foldl' and strict tuples: ``````water is = w1 + w2 + t1 - length is * m1 where (t1,m1,w1) = foldl f (0,0,0) is (t2,m2,w2) = foldl f (0,0,0) \$ reverse is f (t,m,w) i = (t+i, max m i, w + max 0 (m-i)) ``````

### andrewnguyen42 commented Dec 2, 2013

 Elegant one liners are cool! Sadly mine is not one of those. I start by calculating what the maximum amount of water that could be held is. Then, I go level by level and remove blocks that would have fallen off. Think of it as if you dipped a lego structure into a bathtub and pulled it out: every hole is filled at first, but as you pull it out of the water, each level loses some water. https://gist.github.com/andrewnguyen42/7745194

### vvsh commented Dec 5, 2013

 rightMax should be equals land.length - 1

### allumbra commented Dec 19, 2013

 Short JS solution: https://gist.github.com/allumbra/8035125 1 pass. Keeps track of last time wall height was seen.

### joneschris commented Dec 20, 2013

 Another Python solution - https://gist.github.com/joneschris/8049901

### markcrandall commented Dec 20, 2013

 Here is a solution using C# and one lambda ```using System; using System.Linq; namespace TwitterInterviewAnswer { class Program { static void Main(string[] args) { //the data int[] wallHeights = {2, 5, 1, 2, 3, 4, 7, 7, 6}; //the lambda expression Func volume = walls => walls.Skip(1).Take(walls.Count() - 2) .Select((s, i) => new {s, i}) .Select(v => new[] { walls.Take(v.i + 1) .Max(), walls .Skip(v.i + 2) .Max() } .Min() - v.s) .Where(w => w > 0) .Sum(); //the output Console.WriteLine("The volume is: {0}", volume.Invoke(wallHeights)); Console.ReadKey(); } } }```
Owner Author

### mkozakov commented Jan 3, 2014

 @jbojcic your code won't work for something like [7,3,2,4]. You will fill in water that should spill out

### jbojcic commented Jan 19, 2014

 @mkozakov it won't fill up water that should spill out but it will spill out everything, which is also wrong

### sumerc commented Feb 18, 2014

 ```# Actually the original implementation is as following which is done in n-pass: def water_vol_orig(a): lv = rv = -1 result = lmax = rmax = l = 0 r = len(a)-1 # no pass on array while (l < r): lv = a[l] if lv > lmax: lmax = lv rv = a[r] if rv > rmax: rmax = rv if lmax >= rmax: result += rmax - rv r -= 1 else: result += lmax - lv l += 1 return result # This implementation avoids the lookup for the max item in the array to finish in n-pass. # However, it is doing 3 comparisons inside the inner loop and in dynamic languages like Python/Java # indexed access to an array is expensive, using enumerators is far more effective. # So more efficient way of doing this in Python may be: #1) find the max item #2) approach to max item from left and right. def water_vol_fast(a): def _imax(x): # find max in single pass, faster than max(..,itemgetter(..)) max_idx = max_val = 0 for i, v in enumerate(a): if v > max_val: max_val = v max_idx = i return max_idx def _volp(x): max = result = 0 for e in x: if e > max: max = e else: result += max - e return result imax = _imax(a) # n-pass return _volp(a[:imax]) + _volp(reversed(a[imax:])) # n pass # This version will have 2N comparisons(1N for finding max and 1N for calculating the volume) and no index # based access to any array. # If these are profiled, one can see the difference: # water_vol_orig takes 0.405603 CPU Secs, and water_vol will take 0.265202 which is nearly %80 faster. # You can test above by: # for test: import random a = [int(1000*random.random()) for i in range(10000)] assert water_vol_orig(a) == water_vol_fast(a)```

### Mak-Sym commented Feb 23, 2014

 Hm... tried to run original solution against the data with 3 maximums and it didn't work... ``````public class TestTest { public static void main (String[] args) throws java.lang.Exception { int[] testData1 = {1, 2, 1, 3, 2, 1, 0, 2, 4, 3, 1, 5, 3, 1, 4, 1, 2, 1, 0}; int expectedResult1 = 17; int[] testData2 = {4, 1, 2, 1, 3, 2, 1, 0, 2, 4, 3, 1, 5, 3, 1, 4, 1, 2, 1, 0}; int expectedResult2 = 40; int[] testData3 = {1, 3, 4, 1, 2, 1, 3, 2, 1, 0, 2, 4, 3, 1, 5, 3, 1, 4, 1, 2, 1, 0}; int expectedResult3 = 40; int[] testData4 = {2, 1, 3, 4, 1, 2, 1, 3, 2, 1, 0, 2, 4, 3, 1, 5, 3, 1, 4, 1, 2, 1, 0}; int expectedResult4 = 41; int[] testData5 = {2, 1, 3, 4, 1, 2, 1, 3, 2, 1, 0, 2, 4, 3, 1, 5, 3, 1, 4, 1, 2, 1, 0, 5, 2, 1, 3}; int expectedResult5 = 61; test(expectedResult1, testData1); test(expectedResult2, testData2); test(expectedResult3, testData3); test(expectedResult4, testData4); test(expectedResult5, testData5); //[ERROR] Test failure! Expected: 61 but was: 62 } public static void test(int expected, int[] array){ int result = calculateVolume(array); if(result != expected) { System.out.println("[ERROR] Test failure! Expected: " + expected + " but was: " + result); } else { System.out.println("[INFO] Test successful!"); } } public static int calculateVolume(int[] land) { int leftMax = 0; int rightMax = 0; int left = 0; int right = land.length - 1; int volume = 0; while(left < right) { if(land[left] > leftMax) { leftMax = land[left]; } if(land[right] > rightMax) { rightMax = land[right]; } if(leftMax >= rightMax) { volume += rightMax - land[right]; right--; } else { volume += leftMax - land[left]; left++; } } return volume; } } ``````

### sumerc commented Feb 24, 2014

 @Mak-Sym Original implementation returns 51 and your expectation should not be 61. 51 is the correct volume for input set: {2, 1, 3, 4, 1, 2, 1, 3, 2, 1, 0, 2, 4, 3, 1, 5, 3, 1, 4, 1, 2, 1, 0, 5, 2, 1, 3}. Please do not forget the fact that water will spill from left/right if there is no surrounding block.

### Mak-Sym commented Feb 25, 2014

 Thanks! There is typo in my post - sorry about that I meant this one: ``````int[] testData5 = {2, 1, 3, 5, 1, 2, 1, 3, 2, 1, 0, 2, 4, 3, 1, 5, 3, 1, 4, 1, 2, 1, 0, 5, 2, 1, 3}; //should be 61, but was 62 ``````

### Mak-Sym commented Feb 25, 2014

 Whoops - found bug in my calculations - it should be 62 (not sure how I missed this during numerous manual verifications and my v1 impl :) )

### Pranz commented May 8, 2014

 unspoiled solution; I think mine works the same way as yours. https://gist.github.com/Pranz/45eff77d201ef1e2dc61

### bourneagain commented Sep 10, 2014

 This seems to work ... https://gist.github.com/bourneagain/0cb90851106756746933

### isaiah-perumalla commented Aug 13, 2015

 here is an alternate but simple nlog(n) solution https://gist.github.com/isaiah-perumalla/105b72e6b99f69b599ec

### harrytallbelt commented May 18, 2016 • edited

 just an F# solution for aestetic pleasure ```let oneWayCapacity a = let mutable max = 0 a |> List.map (fun e -> if e > max then max <- e; 0 else max - e) let totalCapacity a = let aL = a |> oneWayCapacity let aR = a |> List.rev |> oneWayCapacity |> List.rev List.map2 min aL aR |> List.sum let test a = a |> totalCapacity |> printfn "totalCapacity %A = %i" a let () = [1; 2; 1; 3; 5; 3; 1; 2; 4; 1] |> test [2; 5; 10; 7; 3; 8; 5; 2] |> test [2; 5; 1; 2; 3; 4; 7; 7; 6] |> test``` https://gist.github.com/harry-tallbelt/cc236e710c9edf858853c118bc38c9ad