Iterating through a List with Elements which contains two Values and sum up only the connected ones (Java)

Question

I need help with the following problem. When I have a list of elements which contains two values, that represents the start and endpoint of a bridge. These Values are representing a "bridge". So for example [0,1] means a bridge connects island0 with island1.

How can I iterate through that list starting with 0 and only sum up the values which are connected and ignoring the other ones?

Example 1 - Every Island is indirectly connected

[0,1]
[1,2]
[2,3]
[3,4]

Solution should be: 0+1+2+3+4 = 10

Example 2

[0,1]
[1,2]
[3,4] !!! NOT CONNECTED

Solution should be: 0+1+2 = 3

Thanks!!!

Answer 1

try this simple solution

import java.util.ArrayList;
import java.util.List;

public class TestBridge {

    public static final class Bridge {
        int start;

        int end;

        Bridge(int start, int end) {
            this.start = start;
            this.end = end;
        }
    }

    private static int getSum(List<Bridge> bridges) {
        int sum = 0;
        int prevEnd = -1;
        for (Bridge bridge : bridges) {
            int actualStart = bridge.start;
            int actualEnd = bridge.end;
            if (actualStart == 0 || prevEnd == actualStart) {
                prevEnd = actualEnd;
                sum += actualEnd;
            }
        }
        return sum;
    }

    private static List<Bridge> getFirstExample() {
        List<Bridge> bridges = new ArrayList<>();
        bridges.add(new Bridge(0, 1));
        bridges.add(new Bridge(1, 2));
        bridges.add(new Bridge(2, 3));
        bridges.add(new Bridge(3, 4));
        return bridges;
    }

    private static List<Bridge> getSecondExample() {
        List<Bridge> bridges = new ArrayList<>();
        bridges.add(new Bridge(0, 1));
        bridges.add(new Bridge(1, 2));
        bridges.add(new Bridge(3, 4));
        return bridges;
    }

    public static void main(String[] args) {
        System.out.println(getSum(getFirstExample()));
        System.out.println(getSum(getSecondExample()));
    }
}

Answer 2

Hagen??

With some contraints it could be easy, without it can become more complicated. With constraints I mean the first bridge should be the source of truth. Just think about those cases:

[0,1]
[1,2]
[3,4]
[4,5]

(two parts internally connected with the same length)

[0,1]
[1,2]
[3,4]
[4,5]

(The first Bridge is the excluded one)

What happens if Bridges 1+2 are connected and 3+4 are connected but not both parts together? I mean this...

[0,1]
[2,3]
[3,4]
[4,5]

Assuming the first bridge as the source of thruth you can try this:

public class IslandConnector {
    private final List<Bridge> includedBridges = new ArrayList<>();
    private List<Bridge> excludedBridges;

    public IslandConnector(List<Bridge> bridges) {
        this.excludedBridges = new ArrayList<>();
        for(Bridge bridge : bridges) {
            if(includedBridges.isEmpty()) {
                includedBridges.add(bridge);
            }
            else {
                if(!tryIncludeBridge(bridge)) {
                    excludedBridges.add(bridge);
                }
            }           
        }
    }

    private boolean tryIncludeBridge(Bridge bridge) {
        for(Bridge includedBridge: includedBridges) {
            if(bridge.hasSameS(includedBridge)) {
                includeBridge(bridge);
                return true;
            }
        }

        return false;
    }

    private void includeBridge(Bridge bridge) {
        includedBridges.add(bridge);

        for(Bridge excludedBridge: excludedBridges) {
            if(bridge.hasSameS(excludedBridge)) {
                excludedBridges.remove(excludedBridge);
                includeBridge(excludedBridge);
            }
        }
    }

    public List<Bridge> getIncludedBridges() {
        return includedBridges;
    }

    public static void main(String... args) {
        System.out.println(new IslandConnector(Arrays.asList(
            new Bridge(0, 1),
            new Bridge(1, 2),
            new Bridge(2, 3),
            new Bridge(3, 4)
        )).getIncludedBridges());

        System.out.println(new IslandConnector(Arrays.asList(
            new Bridge(0, 1),
            new Bridge(1, 2),
            new Bridge(3, 4)
        )).getIncludedBridges());

        System.out.println(new IslandConnector(Arrays.asList(
            new Bridge(0, 1),
            new Bridge(2, 3),
            new Bridge(3, 4)
        )).getIncludedBridges());
    }
}

printing

[[0,1], [1,2], [2,3], [3,4]]
[[0,1], [1,2]]
[[0,1]]

Answer 3

Try this solution with HashMap

public class BridgeSolution {

    public static void main(String[] args) {

        Map<Integer, Integer> bridgeMap = new HashMap<>();

        List<Integer[]> bridges = new ArrayList<>();
        bridges.add(new Integer[]{0, 1});
        bridges.add(new Integer[]{1, 2});
        bridges.add(new Integer[]{2, 3});
        bridges.add(new Integer[]{3, 4});


        for (int i = 0; i < bridges.size(); i++) {
            bridgeMap.put(bridges.get(i)[0], bridges.get(i)[1]);
        }

        Integer start = bridges.get(0)[0];

        Integer value = null;
        Integer sum = start;
        while ((value = bridgeMap.get(start)) != null) {
            sum += value;
            start = value;
        }

        System.out.println(sum);
    }
}

Answer 4

Having the following class:

public class Graph<V> {
    private static class Edge<VV> {
        private final VV from;
        private final VV to;

        private Edge(VV from, VV to) {
            this.from = from;
            this.to = to;
        }
    }

    private final Set<V> vertices = new HashSet<>();
    private final Collection<Edge<V>> edges = new ArrayList<>();

    public void addEdge(V from, V to) {
        vertices.add(from);
        vertices.add(to);
        edges.add(new Edge(from, to));
    }


    public Set<V> closure(V start) {
        Set<V> result = new HashSet<>();
        closure(start, result);
        return result;
    }

    private void closure(V start, Set<V> seen) {
        if (vertices.contains(start) && !seen.contains(start)) {
            seen.add(start);
            for (Edge<V> e: edges) {
                if (start.equals(e.from)) {
                    closure(e.to, seen);
                }
            }
        }
    }
}

You can do this:

    Graph<Integer> graph = new Graph<>();
    graph.addEdge(0,1);
    graph.addEdge(1,2);
    graph.addEdge(3,4);
    int sum = 0;
    for (int v: graph.closure(0)) {
        System.out.println("Vertex: " + v);
        sum += v;
    }
    System.out.println("Sum: " + sum);

With the following output:

Vertex: 0
Vertex: 1
Vertex: 2
Sum: 3

Answer 5

Try the below. Change the Bridge object to suit your requirements

public class Bridge {
    int start;
    int end;

    public Bridge(int start, int end) {
        this.start = start;
        this.end = end;
    }

    public int getStart() {
        return start;
    }

    public int getEnd() {
        return end;
    }
}

Calculate the sum

import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

public class BridgeSum {
    public static void main(String... args) throws IOException {
        Bridge B1 = new Bridge(0, 1);
        Bridge B2 = new Bridge(1, 2);
        Bridge B3 = new Bridge(2, 4);
        Bridge B4 = new Bridge(4, 8);
        Bridge B5 = new Bridge(8, 9);

        List<Bridge> list = new ArrayList<>();
        list.add(B5);
        list.add(B2);
        list.add(B3);
        list.add(B1);
        list.add(B4);

        list.sort((a, b) -> a.getStart() > b.getStart() ? 0 : -1);

        int sum = 0;
        for (int i = 0; i < list.size(); i++) {
            if (i == 0) {
                sum = sum + list.get(i).getStart() + list.get(i).getEnd();
            } else {
                if (list.get(i).getStart() == list.get(i - 1).getEnd()) {
                    sum = sum + list.get(i).getEnd();
                } else {
                    break;
                }
            }
        }

        System.out.println(sum);
    }
}