Quick Overview

This question evaluates graph connectivity reasoning (computing minimum edges to unify components) and string-based big-integer arithmetic (multiplying numbers represented as strings), testing competency in graph theory, component analysis, and numeric string manipulation.

Solve island connectivity and string multiplication

Company: Meta

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Technical Screen

##### Question Given N distinct islands (e.g., a, b, c…) and a list of existing undirected paths between them, compute the minimum number of additional paths required so that all islands become connected. LeetCode 43. Multiply Strings – Given two non-negative integers represented as strings, return their product as a string. (Clarify whether the carry can exceed 10.) https://leetcode.com/problems/multiply-strings/description/

Overview: This question evaluates graph connectivity reasoning (computing minimum edges to unify components) and string-based big-integer arithmetic (multiplying numbers represented as strings), testing competency in graph theory, component analysis, and numeric string manipulation.

You are given n islands labeled from 0 to n-1 and a list of existing undirected paths (connections) between them. Each connection is a pair [u, v] with 0 <= u, v < n and u != v. Determine the minimum number of additional paths needed to make all islands belong to a single connected component. You may add a path between any two distinct islands.

Constraints

  • 1 <= n <= 200000
  • 0 <= len(connections) <= 200000
  • connections[i] = [u, v], where 0 <= u, v < n and u != v
  • Islands are distinct and labeled 0 to n-1
  • Multiple or duplicate connections may be present and should be treated as a single connection for connectivity

Hints

  1. The minimum number of additional paths required equals the number of connected components minus one.
  2. Use DFS/BFS or Union-Find (Disjoint Set Union) to count connected components efficiently.
  3. Be careful to handle duplicate connections without affecting the component count.

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Approach

We treat islands as nodes in an undirected graph. Using Union-Find, we merge endpoints of each existing path to form connected components. The minimum number of edges required to connect k components into a single component is k - 1, since each new edge can join two components. Thus, after counting components via Union-Find, we return components - 1.

Time complexity:
O(n + m) where m = len(connections)
Space complexity:
O(n)