Archipelago Explorer

Archipelago Explorer - Problem

An explorer is on island 0 in an archipelago of n islands numbered 0 to n - 1.

You are given a 2D array bridges where bridges[i] = [u, v] means a two-way bridge connects island u and island v.

Return the total number of islands the explorer can visit (including island 0).

Input & Output

Example 1

$ Input: n=7, bridges=[[0,1],[0,2],[1,3],[4,5],[5,6]]

› Output: 4

💡 Note: Islands 0,1,2,3 reachable. 4,5,6 disconnected.

Example 2

$ Input: n=4, bridges=[[0,1],[1,2],[2,3]]

› Output: 4

💡 Note: All reachable via chain.

Example 3

$ Input: n=5, bridges=[]

› Output: 1

💡 Note: No bridges. Only island 0.

Constraints

1 <= n <= 2*10^5
0 <= bridges.length <= 2*10^5
0 <= u, v < n

Visualization

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Asked in

a Amazon 0

Count = size of connected component containing island 0. BFS explores outward. Union-Find tracks component size.

Common Approaches

✓ BFS from Island 0

⏱️ Time: O(V+E) Space: O(V+E)

Build adjacency list. BFS from 0, count all visited.

Union-Find Component Size

⏱️ Time: O(E*a(V)) Space: O(V)

Union all bridge pairs. Return size[find(0)].

BFS from Island 0 — Algorithm Steps

Build adj list
BFS from 0
Count visited
Return count

Visualization

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Step-by-Step Walkthrough

Start

Visit 0

Expand

Visit 1,2,3

Done

Count=4

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#define MAXN 200001
#define MAXE 400002
int head[MAXN], to[MAXE], nxt[MAXE], ecnt;
int visited[MAXN], queue[MAXN];
void addEdge(int u, int v) {
    to[ecnt] = v;
    nxt[ecnt] = head[u];
    head[u] = ecnt++;
    to[ecnt] = u;
    nxt[ecnt] = head[v];
    head[v] = ecnt++;
}
int main() {
    int n, ch, num, neg, u, v, i;
    ecnt = 0;
    memset(head, -1, sizeof(head));
    while ((ch = getchar()) != EOF && !(ch == '-' || (ch >= '0' && ch <= '9')))
        ;
    neg = 0;
    num = 0;
    if (ch == '-') {
        neg = 1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        num = num * 10 + (ch - '0');
        ch = getchar();
    }
    n = neg ? -num : num;
    while ((ch = getchar()) != EOF && ch != '[')
        ;
    while (1) {
        ch = getchar();
        if (ch == EOF || ch == ']')
            break;
        if (ch == '[') {
            while ((ch = getchar()) != EOF && !(ch == '-' || (ch >= '0' && ch <= '9')))
                ;
            neg = 0;
            num = 0;
            if (ch == '-') {
                neg = 1;
                ch = getchar();
            }
            while (ch >= '0' && ch <= '9') {
                num = num * 10 + (ch - '0');
                ch = getchar();
            }
            u = neg ? -num : num;
            while (ch != EOF && !(ch == '-' || (ch >= '0' && ch <= '9')))
                ch = getchar();
            neg = 0;
            num = 0;
            if (ch == '-') {
                neg = 1;
                ch = getchar();
            }
            while (ch >= '0' && ch <= '9') {
                num = num * 10 + (ch - '0');
                ch = getchar();
            }
            v = neg ? -num : num;
            addEdge(u, v);
            while (ch != ']' && ch != EOF)
                ch = getchar();
        }
    }
    memset(visited, 0, sizeof(int) * n);
    int front = 0, back = 0, count = 0;
    queue[back++] = 0;
    visited[0] = 1;
    while (front < back) {
        int nd = queue[front++];
        count++;
        for (i = head[nd]; i != -1; i = nxt[i]) {
            if (!visited[to[i]]) {
                visited[to[i]] = 1;
                queue[back++] = to[i];
            }
        }
    }
    printf("%d\n", count);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(V+E)

Visit each once

✓ Linear Growth

Space Complexity

O(V+E)

Adj + visited

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

BFS from Island 0 — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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