二维阵列的水容量

Question

I have to do a little exercise at my university but I am already stuck for a while. The exercise is about calculating the water capacity of a 2D array, the user has to enter the width (w) and the height (h) of the 2D array, and then all the elements of the array, which represent the height at that location. Really simple example:

10 10 10
10 2 10
10 10 10

The output will then be 8, because that is the maximum water that fits in there. Another example is:

 6 4
 1 5 1 5 4 3
 5 1 5 1 2 4
 1 5 1 4 1 5
 3 1 3 6 4 1

Output will be 14.

What also important to mention is: The width and height of the array can not be larger than 1000 and the heights of the element cannot be larger than 10^5.

Now I basically have the solution, but it is not fast enough for larger inputs. What I did is the following: I add the heights to a TreeSet and then every time I poll the last one (the highest) and then I go through the array (not looking at the edges) and use DFS and check for every position if the water can stay in there. If the water doesn't go out of the array than calculate the positions that are under water, if it goes out of the array then poll again and do the same.

I also tried looking at the peaks in the array, by going vertically and horizontally. For the example above you get this:

0 5 0 5 4 0
5 0 5 0 0 4
0 5 0 4 0 5
3 1 3 6 4 0

What I did with this was give the peaks a color let say (black) and then for all the white colors take the minimum peak value with DFS again and then take that minimum to calculate the water capacity. But this doesn't work, because for example:

7 7 7 7 7
7 4 4 4 7
7 2 3 1 7
7 4 4 4 7
7 7 7 7 7

Now 3 is a peak, but the water level is 7 everywhere. So this won't work.

But because my solution is not fast enough, I am looking for a more efficient one. This is the part of the code where the magic happens:

    while (p.size() != 0 || numberOfNodesVisited!= (w-2)*(h-2)) {
        max = p.pollLast();
        for (int i=1; i < h-1; i++) {
            for (int j=1; j < w-1; j++) {
                if (color[i][j] == 0) {
                    DFSVisit(profile, i, j);
                    if (!waterIsOut) {
                        sum+= solveSubProblem(heights, max);
                        numberOfNodesVisited += heights.size();
                        for(int x = 0; x < color.length; x++) {
                            color2[x] = color[x].clone();
                        }
                    } else {
                        for(int x = 0; x < color2.length; x++) {
                            color[x] = color2[x].clone();
                        }
                        waterIsOut = false;
                    }
                    heights.clear();
                }
            }
        }
   }

Note I am resetting the paths and the colors every time, I think this is the part that has to be improved.

And my DFS: I have three colors 2 (black) it is visited, 1 (gray) if it is an edge and 0 (white) if is not visited and not an edge.

 public void DFSVisit(int[][] profile, int i, int j) {
    color[i][j] = 2; // black
    heights.add(profile[i][j]);
    if (!waterIsOut && heights.size() < 500) { 
        if (color[i+1][j] == 0 && max > profile[i+1][j]) { // up
            DFSVisit(profile, i+1, j);
        } else if (color[i+1][j] == 1 && max > profile[i+1][j]) {
            waterIsOut = true;
        }
        if (color[i-1][j] == 0 && max > profile[i-1][j]) { // down
            DFSVisit(profile, i-1, j);
        } else if (color[i-1][j] == 1 && max > profile[i-1][j]) {
            waterIsOut = true;
        }
        if (color[i][j+1] == 0 && max > profile[i][j+1]) { // right
            DFSVisit(profile, i, j+1);
        } else if (color[i][j+1] == 1  && max > profile[i][j+1]) {
            waterIsOut = true;
        }
        if (color[i][j-1] == 0  && max > profile[i][j-1]) { //left
            DFSVisit(profile, i, j-1);
        } else if (color[i][j-1] == 1  && max > profile[i][j-1]) {
            waterIsOut = true;
        }
    }
}

UPDATE @dufresnb referred to talentbuddy.co where the same exercise is given at https://www.talentbuddy.co/challenge/526efd7f4af0110af3836603 . However I tested al lot of solutions and a few of them actually make it through my first four test cases, most of them however already fail on the easy ones. Talent buddy did a bad job on making test cases: in fact they only have two. If you want to see the solutions they have just register and enter this code (language C): it is enough to pass their test cases

#include <stdio.h>

void rain(int m, int *heights, int heights_length) {
    //What tests do we have here?
    if (m==6)
        printf("5");
    else if (m==3)
        printf("4");
    //Looks like we need some more tests.
}

UPDATE @tobias_k solution is a working solution, however just like my solution it is not efficient enough to pass the larger input test cases, does anyone have an idea for an more efficient implementation?

Any ideas and help will be much appreciated.

Answer 1

Here's my take on the problem. The idea is as follows: You repeatedly flood-fill the array using increasing "sea levels". The level a node is first flooded will be the same level that the water would stay pooled over that node when the "flood" retreats.

• for each height starting from the lowest to the highest level:
  • put the outer nodes into a set, called fringe
  • while there are more nodes in the fringe set, pop a node from the set
    • if this node was first reached in this iteration and its height is lesser or equal to the current flood height, memorize the current flood height for tha tnode
    • add all its neighbours that have not yet been flooded and have a height lesser or equal to the current flood height to the fringe

As it stands, this will have compexity O(nmz) for an nxm array with maximum elevation z , but with some optimization we can get it down to O(nm) . For this, instead of using just one fringe, and each time working our way from the outside all the way inwards, we use multiple fringe sets, one for each elevation level, and put the nodes that we reach in the fringe corresponding to their own height (or the current fringe, if they are lower). This way, each node in the array is added to and removed from a fringe exactly once . And that's as fast as it possibly gets.

Here's some code. I've done it in Python, but you should be able to transfer this to Java -- just pretend it's executable pseudo-code. You can add a counter to see that the body of the while loop is indeed executed 24 times, and the result, for this example, is 14.

# setup and preparations
a = """1 5 1 5 4 3
       5 1 5 1 2 4
       1 5 1 4 1 5
       3 1 3 6 4 1"""
array = [[int(x) for x in line.strip().split()] 
         for line in a.strip().splitlines()]
cols, rows = len(array[0]), len(array)
border = set([(i, 0     ) for i in range(rows)] + 
             [(i, cols-1) for i in range(rows)] + 
             [(0, i     ) for i in range(cols)] + 
             [(rows-1, i) for i in range(cols)])
lowest  = min(array[x][y] for (x, y) in border) # lowest on border
highest = max(map(max, array))                  # highest overall

# distribute fringe nodes to separate fringes, one for each height level
import collections
fringes = collections.defaultdict(set) # maps points to sets
for (x, y) in border:
    fringes[array[x][y]].add((x, y))

# 2d-array how high the water can stand above each cell
fill_height = [[None for _ in range(cols)] for _ in range(rows)]
# for each consecutive height, flood-fill from current fringe inwards
for height in range(lowest, highest + 1):
    while fringes[height]: # while this set is non-empty...
        # remove next cell from current fringe and set fill-height
        (x, y) = fringes[height].pop()
        fill_height[x][y] = height
        # put not-yet-flooded neighbors into fringe for their elevation
        for x2, y2 in [(x-1, y), (x, y-1), (x+1, y), (x, y+1)]:
            if 0 <= x2 < rows and 0 <= y2 < cols and fill_height[x2][y2] is None:
                # get fringe for that height, auto-initialize with new set if not present
                fringes[max(height, array[x2][y2])].add((x2, y2))

# sum of water level minus ground level for all the cells
volume = sum(fill_height[x][y] - array[x][y] for x in range(cols) for y in range(rows))
print "VOLUME", volume

To read your larger test cases from files, replace the a = """...""" at the top with this:

with open("test") as f:
    a = f.read()

The file should contain just the raw array as in your question, without dimension information, separated with spaces and line breaks.

Answer 2

talentbuddy.co has this problem as one of their coding tasks. It's called rain, if you make an account you can view other peoples solutions.

#include <iostream>
#include <vector>

bool check(int* myHeights, int x, int m, bool* checked,int size)
{
    checked[x]=true;
    if(myHeights[x-1]==myHeights[x] && (x-1)%m!=0 && !checked[x-1])
    {
        if(!check(myHeights,x-1,m,checked,size))return false;
    }
    else if((x-1)%m==0 && myHeights[x-1]<=myHeights[x])
    {
        return false;
    }
    if(myHeights[x+1]==myHeights[x] && (x+1)%m!=m-1 && !checked[x+1])
    {
        if(!check(myHeights,x+1,m,checked,size))return false;
    }
    else if((x+1)%m==m-1 && myHeights[x+1]<=myHeights[x])
    {
        return false;
    }
    if(myHeights[x-m]==myHeights[x] && (x-m)>m && !checked[x-m])
    {
        if(!check(myHeights,x-m,m,checked,size))return false;
    }
    else if((x-m)<m && myHeights[x-m]<=myHeights[x])
    {
        return false;
    }
    if(myHeights[x+m]==myHeights[x] && (x+m)<size-m && !checked[x+m])
    {
        if(!check(myHeights,x+m,m,checked,size))return false;
    }
    else if((x+m)>size-m && myHeights[x+m]<=myHeights[x])
    {
        return false;
    }
    return true;
}

void rain(int m, const std::vector<int> &heights) 
{
    int total=0;
    int max=1;
    if(m<=2 || heights.size()/m<=2)
    {
        std::cout << total << std::endl;
        return;
    }
    else
    {
        int myHeights[heights.size()];
        for(int x=0;x<heights.size();++x)
        {
            myHeights[x]=heights[x];
        }
        bool done=false;
        while(!done)
        {
            done=true;
            for(int x=m+1;x<heights.size()-m;++x)
            {
                if(x<=m || x%m==0 || x%m==m-1)
                {
                    continue;
                }

                int lower=0;
                if(myHeights[x]<myHeights[x-1])++lower;
                if(myHeights[x]<myHeights[x+1])++lower;
                if(myHeights[x]<myHeights[x-m])++lower;
                if(myHeights[x]<myHeights[x+m])++lower;

                if(lower==4)
                {
                    ++total;
                    ++myHeights[x];
                    done=false;
                }
                else if(lower>=2)
                {
                    bool checked[heights.size()];
                    for(int y=0;y<heights.size();++y)
                    {
                        checked[y]=false;
                    }
                    if(check(myHeights,x,m,checked,heights.size()))
                    {
                        ++total;
                        ++myHeights[x];
                        done=false;
                    }
                }
            }
        }
    }
    std::cout << total << std::endl;
    return;
}