Connected Road to Destination- Google Top Interview Questions


Problem Statement :


You are given integers sx, sy, ex, ey and two-dimensional list of integers roads. 

You are currently located at coordinate (sx, sy) and want to move to destination (ex, ey). 

Each element in roads contains (x, y) which is a road that will be added at that coordinate. 

Roads are added one by one in order. 

You can only move to adjacent (up, down, left, right) coordinates if there is a road in that coordinate or if it's the destination coordinate. For example, at (x, y) we can move to (x + 1, y) if (x + 1, y) is a road or the destination.

Return the minimum number of roads in order that must be added before there is a path consisting of roads that allows us to get to (ex, ey) from (sx, sy). If there is no solution, return -1.

Constraints

0 ≤ n ≤ 100,000 where n is the length of roads

Example 1

Input
sx = 0

sy = 0

ex = 1

ey = 2

roads = [

    [9, 9],

    [0, 1],

    [0, 2],

    [0, 3],

    [3, 3]

]

Output

3

Explanation

We need to add the first three roads which allows us to go from (0, 0), (0, 1), (0, 2), (1, 2). Note that we 
must take (9, 9) since roads must be added in order.



Solution :



title-img




                        Solution in C++ :

pair<int, int> directions[] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

int solve(int sx, int sy, int ex, int ey, vector<vector<int>>& roads) {
    map<pair<int, int>, int> Roads;
    const int n = (int)roads.size();
    for (int i = 0; i < n; i++) Roads[{roads[i][0], roads[i][1]}] = i + 1;

    Roads[{ex, ey}] = 0;

    priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<>> q;
    q.push({0, sx, sy});

    map<pair<int, int>, int> cost;
    cost[{sx, sy}] = 0;

    while (!q.empty()) {
        auto [curr_index, x, y] = q.top();
        q.pop();

        if (x == ex && y == ey) return curr_index;
        for (auto [dx, dy] : directions) {
            int nx = x + dx;
            int ny = y + dy;
            if (Roads.count({nx, ny})) {
                if (!cost.count({nx, ny}) || cost[{nx, ny}] > max(Roads[{nx, ny}], curr_index)) {
                    cost[{nx, ny}] = max(Roads[{nx, ny}], curr_index);
                    q.push({cost[{nx, ny}], nx, ny});
                }
            }
        }
    }

    return -1;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int sx, int sy, int ex, int ey, int[][] roads) {
        if (sx == ex && sy == ey) {
            return 0;
        }

        // We sort by the number of roads the current destination needs
        Queue<int[]> q = new PriorityQueue<>((a, b) -> a[2] - b[2]);

        // Use to hash/store each road
        long time = (long) (1e9);

        // The has for target
        long target = ex * time + ey;

        Map<Long, Integer> road = new HashMap<>();
        Set<Long> visit = new HashSet<>();

        // Store each road into the map
        int count = 1;
        for (int[] row : roads) {
            road.put(row[0] * time + row[1], count++);
        }

        // [x , y , number of roads]

        q.offer(new int[] {sx, sy, 0});

        while (!q.isEmpty()) {
            int[] cur = q.poll();
            long num = cur[0] * time + cur[1];

            // If visited, skip
            if (visit.contains(num)) {
                continue;
            }
            visit.add(num);

            // each hashcode for each move (up,down,left,right)
            long num1 = (cur[0] - 1) * time + cur[1];
            long num2 = (cur[0] + 1) * time + cur[1];
            long num3 = (cur[0]) * time + cur[1] - 1;
            long num4 = (cur[0]) * time + cur[1] + 1;

            // if any of them is equal to the target, then we return the number of roads
            if (num1 == target || num2 == target || num3 == target || num4 == target) {
                return cur[2];
            }

            // If the road map contains up/down/left/right move, put it into the pq with the number
            // of roads it need to build.

            if (road.containsKey(num1)) {
                q.offer(new int[] {cur[0] - 1, cur[1], Math.max(cur[2], road.get(num1))});
            }
            if (road.containsKey(num2)) {
                q.offer(new int[] {cur[0] + 1, cur[1], Math.max(cur[2], road.get(num2))});
            }
            if (road.containsKey(num3)) {
                q.offer(new int[] {cur[0], cur[1] - 1, Math.max(cur[2], road.get(num3))});
            }
            if (road.containsKey(num4)) {
                q.offer(new int[] {cur[0], cur[1] + 1, Math.max(cur[2], road.get(num4))});
            }
        }
        return -1;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, sx, sy, ex, ey, roads):
        parent = {}
        seen = set()

        # Find
        def ufind(x, y):
            if (x, y) not in parent:
                parent[(x, y)] = (x, y)
                return (x, y)
            if parent[(x, y)] == (x, y):
                return (x, y)

            d = parent[(x, y)]
            p = ufind(d[0], d[1])
            parent[(x, y)] = p
            return p

        # Union
        def uunion(x1, y1, x2, y2):
            p1 = ufind(x1, y1)
            p2 = ufind(x2, y2)

            parent[p2] = parent[p1]

        directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]
        for dx, dy in directions:
            nx, ny = sx + dx, sy + dy
            if (nx, ny) == (ex, ey):
                return 0

        seen.add((sx, sy))
        seen.add((ex, ey))
        for index, (x, y) in enumerate(roads):
            # Look at neighbors
            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                if (nx, ny) in seen:
                    uunion(x, y, nx, ny)

            seen.add((x, y))
            if ufind(ex, ey) == ufind(sx, sy):
                return index + 1
        return -1
                    


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