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Long Luo's Life Notes

夏至日测地球:利用太阳影子计算地球半径 太阳温度是怎么计算出来的? 《大象的时间,老鼠的时间》读书笔记:生命节奏背后的数学规律 小港流到哪里去? 如何用一根棍子测出地球有多大?复刻埃拉托色尼的春分实验 2007江苏高考数学第20题解析:一道通向黄金分割数的数列压轴题 Google经典面试题: 鸡蛋应该怎么扔? 2010年江苏高考数学压轴题解析:巧用余弦定理与数学归纳法 2011年清华大学自主招生数学题解析:一道经典数列题的解法与思路 2011年清华大学自主招生数学题解析:一道经典数列题的解法与思路 2006年江西高考理科数学压轴题解析:递推、放缩与不等式结构 2006年江西高考理科数学压轴题解析:递推、放缩与不等式结构 一道初中数学极值题的多种解法:柯西不等式、几何法、函数法详解 扔几个骰子,怎么算出期望?——拼多多校招笔试算法题的数学故事 拼多多校招笔试算法题:一行公式搞定“多多的魔术盒子” 斯特林公式(Stirling's Formula):我一个阶乘表达式,怎么就和圆扯上关系了呢? 我爱做题:2010年江西高考理科数学压轴题 热机的效率上限在哪里?解析卡诺循环(Carnot Cycle) 为什么 2024 年会有 366 天? 数学之美:几何视角下的高斯积分(Gaussian Integral) 从最小二乘法到正态分布:高斯是如何找到失踪的谷神星的? 正态分布(Normal Distribution)公式为什么长这样? 高速公路编号背后的数学密码 2024阿里巴巴全球数学竞赛预选赛试题及解答 库函数 (libm) 是如何计算三角函数值的? payne hanek 归约算法 音乐背后的数学 素描背后的物理 cody waite 浮点数 Remez Algorithm 参数归约算法(Argument Range Reduction):如何在浮点数环境下计算超大数字的三角函数值? 素描背后的数学 发生在计算机内存里的进化:解密遗传算法(Genetic Algorithm) CORDIC算法:一种高效计算三角函数值的方法 墨卡托的魔术:地图是如何欺骗你的眼睛的? PID 算法到底在干什么?工程师最常用的控制方法 解密卡尔曼滤波(Kalman Filter)算法:深入解析卡尔曼滤波算法原理与在线可视化实例 从记忆到洞察:轻松掌握泰勒展开式(Taylor Series)的记忆技巧 哪个更大呢? $2^{100!}$ 还是 $2^{100}!$ ? Google经典编程竞赛题:计算 $(3 + \sqrt{5})^n$ 的小数点前三位数 手写数字识别:解码机器学习的背后的数学原理 The Answers of MRI Tutorial Videos gdb 操作指南 Linux 网络命令指南 贝塞尔曲线(Bezier Curve):优雅背后的数学原理 LeetCode 380. Insert Delete GetRandom O(1) Data Structures: Thought Process from HashMap to HashMap + Array LeetCode 2475. 数组中不等三元组的数目 2种 O(n) 时间复杂度算法 LeetCode 295. Find Median from Data Stream Two Heaps with the Follow Ups LeetCode 295. Find Median from Data Stream Two Heaps with the Follow Ups LeetCode 1668. 最大重复子字符串 不用API,比KMP更易理解简洁优雅的暴力解法 LeetCode 334. Increasing Triplet Subsequence Why Greedy Works? LeetCode 迷宫问题(The Maze)
LeetCode 947. Most Stones Removed with Same Row or Column It is Literally a Graph: DFS and Union Find
2022-11-14 · via Long Luo's Life Notes

By Long Luo

This article is the solution It is Literally a Graph: DFS and Union Find of Problem 947. Most Stones Removed with Same Row or Column .

Intuition

We can find that this is a graph theory problem with analysis.

Imagine the stone on the 2D coordinate plane as the vertex of the graph, If the x-coord or the y-coord of two stones are the same, there is an edge between them.

This can be show as follows:

947. Most Stones Removed with Same Row or Column 1

According to the rule that stones can be removed, we should remove those points that are in the same row or column with other points as late as possible. That is, the more points in the same row or column with point A, the later point A should be removed. In this way, we can delete as many points as possible through point A.

It can be found that all vertices in a connected graph can be deleted to only one vertex.

947. Most Stones Removed with Same Row or Column 2

Since these vertices are in a connected graph, all vertices of the connected graph can be traversed by DFS or BFS.

Therefore: the maximum number of stones that can be removed = the number of all stones - the number of connected components.

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class Solution {
public int removeStones(int[][] stones) {
int n = stones.length;
if (n <= 1) {
return 0;
}

List<Integer>[] graph = new List[n];
for (int i = 0; i < n; i++) {
graph[i] = new ArrayList<>();
}

for (int i = 0; i < n; i++) {
int[] u = stones[i];
for (int j = 0; j < n; j++) {
if (i == j) {
continue;
}

int[] v = stones[j];
if (u[0] == v[0] || u[1] == v[1]) {
graph[i].add(j);
}
}
}

boolean[] visited = new boolean[n];
int ans = 0;

for (int i = 0; i < n; i++) {
if (visited[i]) {
continue;
}

dfs(graph, visited, i);
ans++;
}

return n - ans;
}

private static void dfs(List<Integer>[] graph, boolean[] visited, int start) {

visited[start] = true;

List<Integer> neighbors = graph[start];
for (int x : neighbors) {
if (visited[x]) {
continue;
}

dfs(graph, visited, x);
}
}
}

Analysis

  • Time Complexity: \(O(n^2)\).
  • Space Complexity: \(O(n)\).

Union Find

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class Solution {
public int removeStones(int[][] stones) {
if (stones == null || stones.length <= 1) {
return 0;
}

int n = stones.length;

UnionFind uf = new UnionFind();
for (int[] edge : stones) {
uf.union(edge[0] + 10001, edge[1]);
}

return n - uf.getCount();
}

class UnionFind {
Map<Integer, Integer> parents;
int count;

public UnionFind() {
parents = new HashMap<>();
count = 0;
}

public int getCount() {
return count;
}

public int find(int x) {
if (!parents.containsKey(x)) {
parents.put(x, x);
count++;
}

if (x != parents.get(x)) {
parents.put(x, find(parents.get(x)));
}

return parents.get(x);
}

public void union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX == rootY) {
return;
}

parents.put(rootX, rootY);
count--;
}
}
}

Analysis

  • Time Complexity: \(O(n \log n)\)
  • Space Complexity: \(O(n)\)

All suggestions are welcome. If you have any query or suggestion please comment below. Please upvote👍 if you like💗 it. Thank you:-)

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