Solving Twitter Puddle with Zipper

Question

This is a follow-up to my previous question. Consider the following puzzle

I would like to generate a waterLevel array, so that the i-th item is the water level at the i-th point and then sum them up to solve the puzzle.

waterLevel[i] =
  max(0, min(max of left neighbors, max of right neighbors) - height[i])

I would probably try to code it with Zipper

 waterLevels = heights.toZipper.cobind {z => 
   max(0, min(max(z.left), max(z.right)) - z.focus
 }.toList

Does it make sense ?

This would definitely work, but you'll need to handle the edges and make sure you don't end up with a negative amount of water. — Travis Brown, Jun 03 '14 at 12:44
Yes, you are right. I will update the question. I wonder if the zipper solution is `O(N)` and does not scan the array N times — Michael, Jun 03 '14 at 13:38
:( Thanks. That's what I was afraid of. So it does not look like a good example of zipper usage. — Michael, Jun 03 '14 at 14:03
Well, zippers let you describe the computation very elegantly. But no, you probably wouldn't actually want to use them here. — Travis Brown, Jun 03 '14 at 14:25

score 1 · Answer 1 · answered Oct 26 '14 at 18:04

My solution with java, it comes with tests with expected solution:

package com.company;
import java.util.*;

enum Orientation {DOWN, UP};

class Walls{
    public static void main(String []args){

        HashMap<String, Integer> tests = new HashMap<String,Integer>();
        tests.put("2 5 1 2 3 4 7 7 6", 10);
        tests.put("2 2 5 1 3 1 2 1 7 7 6", 17);
        tests.put("2 7 2 7 4 7 1 7 3 7", 18);
        tests.put("4 6 7 7 4 3 2 1 5 2", 10);
        tests.put("5 2 5 1 2 3 4 7 7 6 2 7 1 2 3 4 5 5 4", 26);

        Iterator it = tests.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry pairs = (Map.Entry)it.next();
            it.remove();

            String[] strings = ((String)pairs.getKey()).split(" ");
            int[] walls = new int[strings.length];
            for (int i = 0; i < walls.length; i++){
                walls[i] = Integer.parseInt(strings[i].trim());
            }
            System.out.println(pairs.getKey()+" result="+accumulatedWater(walls)+" expected= " +pairs.getValue());

        }
    }

    static int accumulatedWater(int []wall){
        int MAX = 0;
        int start = 0;

        for(int i=0;i < wall.length;i++){  //let's go to the first peak
            if(wall[i] >= MAX){
                MAX = wall[i];
                start = i;
            }else{
                break;
            }
        }
        int []accumulate_max = new int[MAX+1];  // sums up to certain height
        int []accumulate_max_step = new int[MAX+1]; // steps up to certain height

        Orientation going = Orientation.DOWN;
        int prev = MAX;
        int local_sum=0;
        int total_sum=0;

        int PREVPEAK = MAX;

        for(int i=start+1; i< wall.length; i++){

            if( i == wall.length -1 ||
                wall[i] < prev  && going == Orientation.UP ){
                going = Orientation.DOWN;
                if(wall[i-1] >= MAX){
                    total_sum +=  accumulate_max_step[MAX-1] * MAX - accumulate_max[MAX-1];
                    MAX = wall[i-1];
                    PREVPEAK = MAX;
                    accumulate_max = new int[MAX+1];
                    accumulate_max_step = new int[MAX+1];
                    local_sum = 0;
                }else{
                    int indexNewPeak = (i == wall.length -1 && wall[i]> wall[i-1]) ? i : i-1;
                    int water =  accumulate_max_step[wall[indexNewPeak]-1] * wall[indexNewPeak] - accumulate_max[wall[indexNewPeak]-1];
                    if(wall[indexNewPeak] > PREVPEAK){
                        local_sum = water;
                        PREVPEAK = wall[indexNewPeak];
                    }else{
                        local_sum += water;
                    }

                }
            }else if(wall[i]>prev){
                going = Orientation.UP;
            }

            for(int j=wall[i];j <= MAX;j++){
                accumulate_max[j]      += wall[i];
                accumulate_max_step[j] += 1;
            }

            prev = wall[i];
        }
        return total_sum + local_sum;
    }
}

Solving Twitter Puddle with Zipper

1 Answers1