Twitter Interview Solution

gistfile1.java

/* 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;
	}
}

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

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

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

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

@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.

@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.

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.

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.

Quick PHP version: https://gist.github.com/orukusaki/bb189d9f69ad09e2cd5a

Quick PHP version: https://gist.github.com/orukusaki/bb189d9f69ad09e2cd5a

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

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

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

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

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

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

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

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

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

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

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;
}

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;
}

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 (<height) xs
    area        = (length elems * height) - (sum elems)

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 (<height) xs
    area        = (length elems * height) - (sum elems)

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

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

Short Java
http://it.laplit.com/index.php/humanoid-blog/128-article-7
https://gist.github.com/AndroidHumanoid/7241202

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;
}

Short Java
http://it.laplit.com/index.php/humanoid-blog/128-article-7
https://gist.github.com/AndroidHumanoid/7241202

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;
}

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

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

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

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

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

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

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

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

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

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

My take: https://gist.github.com/leikind/7247086

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

My take: https://gist.github.com/leikind/7247086

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

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

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

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

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

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.

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.

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[0]),
b = input.indexOf(sort[1], (sort[0]==sort[1]) ? a+1 : 0),
last, first = ( a > b ) ? (last=a,b) : (last=b,a);
for (var j = first+1; j < last; j++ ) v+=sort[1]-input[j];
return v + calc(input.filter(function(z,y){return (y < first-1) || (y >= last) }))
}

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[0]),
b = input.indexOf(sort[1], (sort[0]==sort[1]) ? a+1 : 0),
last, first = ( a > b ) ? (last=a,b) : (last=b,a);
for (var j = first+1; j < last; j++ ) v+=sort[1]-input[j];
return v + calc(input.filter(function(z,y){return (y < first-1) || (y >= last) }))
}

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;

}

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;

}

The Groovy version:

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

The Groovy version:

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

@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.

@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.

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:

1. It has to be a free-space square (as opposed to a solid block) to begin with.
2. 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

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:

1. It has to be a free-space square (as opposed to a solid block) to begin with.
2. 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

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

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

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.

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.

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

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

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

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

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

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

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

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

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

                           #                                                    
                          ###                                                   
               #~~~~~~~~######~~~~~~~#                                          
   #~~~~~~~~~~~###~~~~~~#######~~~~~##~#~~~~~~~##~~~~~~~~~~~~~~~~~~#            
   #~~~~~~~~~~####~~~~##########~~~~#####~~##~~##~~~~~~~~~~#~~~~~~##            
#~###~~~~~~~#######~~~##########~~~###############~~~~~~~~~#~~~~#####~#~##      
#~###~~~############~#############~###############~~~~~~~~####~~###########     
######~#############~##############################~#~#~~~#################     
######~#################################################~###################    
########################################################~###################~~##

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

                           #                                                    
                          ###                                                   
               #~~~~~~~~######~~~~~~~#                                          
   #~~~~~~~~~~~###~~~~~~#######~~~~~##~#~~~~~~~##~~~~~~~~~~~~~~~~~~#            
   #~~~~~~~~~~####~~~~##########~~~~#####~~##~~##~~~~~~~~~~#~~~~~~##            
#~###~~~~~~~#######~~~##########~~~###############~~~~~~~~~#~~~~#####~#~##      
#~###~~~############~#############~###############~~~~~~~~####~~###########     
######~#############~##############################~#~#~~~#################     
######~#################################################~###################    
########################################################~###################~~##

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):

https://gist.github.com/AndroidHumanoid/7241202
http://it.laplit.com/index.php/ct-menu-item-8/128-article-7

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;
}

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):

https://gist.github.com/AndroidHumanoid/7241202
http://it.laplit.com/index.php/ct-menu-item-8/128-article-7

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;
}

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

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

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)

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)

Mot solutions are actually wrong...

Mot solutions are actually wrong...

Hi all,

Another solution in Python using numpy (and matplotlib):

https://gist.github.com/yannick1974/7334581

Yannick

Hi all,

Another solution in Python using numpy (and matplotlib):

https://gist.github.com/yannick1974/7334581

Yannick

So does this work for:

4,0,0,4,0,4,0,0,4,2

?

So does this work for:

4,0,0,4,0,4,0,0,4,2

?

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.

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.

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)

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)

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

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

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

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

Short, single-pass python solution

q = [2,5,1,3,1,2,1,7,7,6]
def solve(heights):
    water = [0]*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)

Short, single-pass python solution

q = [2,5,1,3,1,2,1,7,7,6]
def solve(heights):
    water = [0]*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)

@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

@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

@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

@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

@HollyGurza Only works properly in one direction.

You guys really need to think about the problem in both directions.

@HollyGurza Only works properly in one direction.

You guys really need to think about the problem in both directions.

@igor47 Only works in one direction.

@igor47 Only works in one direction.

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

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

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

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

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)

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)

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))

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))

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

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

rightMax should be equals land.length - 1

rightMax should be equals land.length - 1

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

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

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

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

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<int[], int> 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();
        }
    }
}

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<int[], int> 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();
        }
    }
}

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

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

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

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

# 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)

# 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)

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;
    }
}

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;
    }
}

@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 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.

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

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

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

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

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

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

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

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

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

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

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

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