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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0547_number_of_provinces;

// #Medium #Depth_First_Search #Breadth_First_Search #Graph #Union_Find
// #Algorithm_II_Day_6_Breadth_First_Search_Depth_First_Search
// #Graph_Theory_I_Day_8_Standard_Traversal #Level_2_Day_19_Union_Find
// #2022_08_02_Time_2_ms_(69.51%)_Space_54.2_MB_(42.16%)

import java.util.Arrays;

/**
 * 547 - Number of Provinces\.
 *
 * Medium
 *
 * There are `n` cities. Some of them are connected, while some are not. If city `a` is connected directly with city `b`, and city `b` is connected directly with city `c`, then city `a` is connected indirectly with city `c`.
 *
 * A **province** is a group of directly or indirectly connected cities and no other cities outside of the group.
 *
 * You are given an `n x n` matrix `isConnected` where `isConnected[i][j] = 1` if the i^th city and the j^th city are directly connected, and `isConnected[i][j] = 0` otherwise.
 *
 * Return _the total number of **provinces**_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/12/24/graph1.jpg)
 *
 * **Input:** isConnected = \[\[1,1,0],[1,1,0],[0,0,1]]
 *
 * **Output:** 2
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/12/24/graph2.jpg)
 *
 * **Input:** isConnected = \[\[1,0,0],[0,1,0],[0,0,1]]
 *
 * **Output:** 3
 *
 * **Constraints:**
 *
 * *   `1 <= n <= 200`
 * *   `n == isConnected.length`
 * *   `n == isConnected[i].length`
 * *   `isConnected[i][j]` is `1` or `0`.
 * *   `isConnected[i][i] == 1`
 * *   `isConnected[i][j] == isConnected[j][i]`
**/
public class Solution {
    public int findCircleNum(int[][] arr) {
        int[] parent = new int[arr.length];
        Arrays.fill(parent, -1);
        int ans = 0;
        for (int i = 0; i < arr.length - 1; i++) {
            for (int j = i + 1; j < arr[i].length; j++) {
                if (arr[i][j] == 1) {
                    ans += union(i, j, parent);
                }
            }
        }
        return arr.length - ans;
    }

    private int union(int a, int b, int[] arr) {
        int ga = find(a, arr);
        int gb = find(b, arr);
        if (ga != gb) {
            arr[gb] = ga;
            return 1;
        }
        return 0;
    }

    private int find(int a, int[] arr) {
        if (arr[a] == -1) {
            return a;
        }
        arr[a] = find(arr[a], arr);
        return arr[a];
    }
}