test: 827 solution

py

Author: QuBenhao

daily-problems.json

@@ -1,4 +1,4 @@
 {
-    "daily": "3227",
+    "daily": "827",
     "plans": ["3674", "problems", "3675", "problems", "3676", "problems"]
 }

golang/solution_test.go

@@ -1,10 +1,10 @@
 package golang
 
 import (
-	problem "leetCode/problems/problems_3227"
+	problem "leetCode/problems/problems_827"
 	"testing"
 )
 
 func TestSolution(t *testing.T) {
-	TestEach(t, "3227", "problems", problem.Solve)
+	TestEach(t, "827", "problems", problem.Solve)
 }

problems/problems_827/Solution.cpp

@@ -0,0 +1,28 @@
+//go:build ignore
+#include "cpp/common/Solution.h"
+
+
+using namespace std;
+using json = nlohmann::json;
+
+class Solution {
+public:
+    int largestIsland(vector<vector<int>>& grid) {
+        
+    }
+};
+
+json leetcode::qubh::Solve(string input_json_values) {
+	vector<string> inputArray;
+	size_t pos = input_json_values.find('\n');
+	while (pos != string::npos) {
+		inputArray.push_back(input_json_values.substr(0, pos));
+		input_json_values = input_json_values.substr(pos + 1);
+		pos = input_json_values.find('\n');
+	}
+	inputArray.push_back(input_json_values);
+
+	Solution solution;
+	vector<vector<int>> grid = json::parse(inputArray.at(0));
+	return solution.largestIsland(grid);
+}

problems/problems_827/Solution.java

@@ -0,0 +1,18 @@
+package problems.problems_827;
+
+import com.alibaba.fastjson.JSON;
+import java.util.*;
+import qubhjava.BaseSolution;
+
+
+public class Solution extends BaseSolution {
+    public int largestIsland(int[][] grid) {
+        
+    }
+
+    @Override
+    public Object solve(String[] inputJsonValues) {
+        int[][] grid = jsonArrayToInt2DArray(inputJsonValues[0]);
+        return JSON.toJSON(largestIsland(grid));
+    }
+}

problems/problems_827/problem.md

@@ -0,0 +1,41 @@
+# 827. Making A Large Island [Rating: 1933.96]
+
+<p>You are given an <code>n x n</code> binary matrix <code>grid</code>. You are allowed to change <strong>at most one</strong> <code>0</code> to be <code>1</code>.</p>
+
+<p>Return <em>the size of the largest <strong>island</strong> in</em> <code>grid</code> <em>after applying this operation</em>.</p>
+
+<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>s.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[1,0],[0,1]]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[1,1],[1,0]]
+<strong>Output:</strong> 4
+<strong>Explanation: </strong>Change the 0 to 1 and make the island bigger, only one island with area = 4.</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[1,1],[1,1]]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> Can&#39;t change any 0 to 1, only one island with area = 4.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == grid.length</code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>1 &lt;= n &lt;= 500</code></li>
+	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
+</ul>

problems/problems_827/problem_zh.md

@@ -0,0 +1,42 @@
+# 827. 最大人工岛 [难度分: 1933.96]
+
+<p>给你一个大小为 <code>n x n</code> 二进制矩阵 <code>grid</code> 。<strong>最多</strong> 只能将一格&nbsp;<code>0</code> 变成&nbsp;<code>1</code> 。</p>
+
+<p>返回执行此操作后，<code>grid</code> 中最大的岛屿面积是多少？</p>
+
+<p><strong>岛屿</strong> 由一组上、下、左、右四个方向相连的&nbsp;<code>1</code> 形成。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1:</strong></p>
+
+<pre>
+<strong>输入: </strong>grid = [[1, 0], [0, 1]]
+<strong>输出:</strong> 3
+<strong>解释:</strong> 将一格0变成1，最终连通两个小岛得到面积为 3 的岛屿。
+</pre>
+
+<p><strong class="example">示例 2:</strong></p>
+
+<pre>
+<strong>输入: </strong>grid =<strong> </strong>[[1, 1], [1, 0]]
+<strong>输出:</strong> 4
+<strong>解释:</strong> 将一格0变成1，岛屿的面积扩大为 4。</pre>
+
+<p><strong class="example">示例 3:</strong></p>
+
+<pre>
+<strong>输入: </strong>grid = [[1, 1], [1, 1]]
+<strong>输出:</strong> 4
+<strong>解释:</strong> 没有0可以让我们变成1，面积依然为 4。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n == grid.length</code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>1 &lt;= n &lt;= 500</code></li>
+	<li><code>grid[i][j]</code> 为 <code>0</code> 或 <code>1</code></li>
+</ul>

problems/problems_827/solution.go

@@ -0,0 +1,22 @@
+package problem827
+
+import (
+	"encoding/json"
+	"log"
+	"strings"
+)
+
+func largestIsland(grid [][]int) int {
+    
+}
+
+func Solve(inputJsonValues string) any {
+	inputValues := strings.Split(inputJsonValues, "\n")
+	var grid [][]int
+
+	if err := json.Unmarshal([]byte(inputValues[0]), &grid); err != nil {
+		log.Fatal(err)
+	}
+
+	return largestIsland(grid)
+}

problems/problems_827/solution.py

@@ -0,0 +1,78 @@
+import copy
+
+import solution
+from typing import *
+
+
+class Solution(solution.Solution):
+    def solve(self, test_input=None):
+        return self.largestIsland(test_input)
+
+    def largestIsland(self, grid: List[List[int]]) -> int:
+        DIRS = [(-1, 0), (1, 0), (0, -1), (0, 1)]
+        def point_to_idx(x, y):
+            return x * n + y
+
+
+
+        n = len(grid)
+        uf = UnionFind(n * n)
+        for i, row in enumerate(grid):
+            for j, val in enumerate(row):
+                if val == 1:
+                    p = point_to_idx(i, j)
+                    for dx, dy in DIRS:
+                        if 0 <= (nx := i + dx) < n and 0 <= (ny := j + dy) < n and grid[nx][ny] == 1:
+                            uf.union(p, point_to_idx(nx, ny))
+        ans = max(uf.count)
+        for i, row in enumerate(grid):
+            for j, val in enumerate(row):
+                if val == 0:
+                    tot = 1
+                    explored = set()
+                    for dx, dy in DIRS:
+                        if 0 <= (nx := i + dx) < n and 0 <= (ny := j + dy) < n and grid[nx][ny] == 1:
+                            root = uf.find(point_to_idx(nx, ny))
+                            if root in explored:
+                                continue
+                            explored.add(root)
+                            tot += uf.count[root]
+                    ans = max(ans, tot)
+
+        return ans
+
+class UnionFind:
+    def __init__(self, size):
+        self.parent = list(range(size))
+        self.rank = [1] * size
+        self.count = [1] * size
+        self.size = size
+        self.cc = size
+
+    def find(self, x):
+        while self.parent[x] != x:
+            self.parent[x] = self.parent[self.parent[x]]  # 路径压缩
+            x = self.parent[x]
+        return x
+
+    def union(self, x, y):
+        root_x = self.find(x)
+        root_y = self.find(y)
+
+        if root_x == root_y:
+            return False  # 已经在同一集合
+
+        # 按秩合并
+        if self.rank[root_x] > self.rank[root_y]:
+            self.parent[root_y] = root_x
+            self.count[root_x] += self.count[root_y]
+        else:
+            self.parent[root_x] = root_y
+            if self.rank[root_x] == self.rank[root_y]:
+                self.rank[root_y] += 1
+            self.count[root_y] += self.count[root_x]
+        self.cc -= 1
+        return True
+
+    def is_connected(self, x, y):
+        return self.find(x) == self.find(y)