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Solve island connectivity and string multiplication

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.

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/

Minimum Additional Paths to Connect Islands

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

Examples

Example 1

Input: n = 4, connections = [[0,1],[2,3]]
Output: 1

Explanation: There are two components: {0,1} and {2,3}. One additional path between any island in different components connects all islands.

Example 2

Input: n = 5, connections = [[0,1],[1,2]]
Output: 2

Explanation: Components are {0,1,2}, {3}, and {4}. Minimum additional paths needed is 3 - 1 = 2.

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.

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)