feat: add ts solution to lc problem: No.2392

No.2392.Build a Matrix With Conditions

Author: yanglbme

solution/2300-2399/2389.Longest Subsequence With Limited Sum/README.md

@@ -51,7 +51,7 @@
 
 将 `nums` 排序，对于每个 `queries[i]`，求出 `nums` 中所有元素之和小于等于 `queries[i]` 的子序列的最大长度。
 
-时间复杂度 $O(n\log n + m\log n)$。其中 $n$ 为 `nums` 的长度，$m$ 为 `queries` 的长度。
+时间复杂度 $O((n+m)\times logn)$。其中 $n$ 和 $m$ 分别为 `nums` 和 `queries` 的长度。
 
 <!-- tabs:start -->
 

solution/2300-2399/2392.Build a Matrix With Conditions/README.md

@@ -71,9 +71,9 @@
 
 **方法一：拓扑排序**
 
-利用拓扑排序，找到一个合法的 $row$ 序列和 $col$ 序列，然后根据这两个序列构造出矩阵。
+利用拓扑排序，找到一个合法的 `row` 序列和 `col` 序列，然后根据这两个序列构造出矩阵。
 
-时间复杂度 $O(m+n+k^2)$。其中 $m$ 为 $rowConditions$ 的长度，$n$ 为 $colConditions$ 的长度，$k$ 为题目中的 $k$。
+时间复杂度 $O(m+n+k)$。其中 $m$ 和 $n$ 分别为 `rowConditions` 和 `colConditions` 的长度，而 $k$ 为题目中给定的正整数。
 
 <!-- tabs:start -->
 
@@ -186,7 +186,7 @@ public:
         this->k = k;
         auto row = f(rowConditions);
         auto col = f(colConditions);
-        if (row.empty() || col.empty()) return vector<vector<int>>();
+        if (row.empty() || col.empty()) return {};
         vector<vector<int>> ans(k, vector<int>(k));
         vector<int> m(k + 1);
         for (int i = 0; i < k; ++i) {
@@ -291,7 +291,52 @@ func buildMatrix(k int, rowConditions [][]int, colConditions [][]int) [][]int {
 ### **TypeScript**
 
 ```ts
+function buildMatrix(
+    k: number,
+    rowConditions: number[][],
+    colConditions: number[][],
+): number[][] {
+    function f(cond) {
+        const g = Array.from({ length: k + 1 }, () => []);
+        const indeg = new Array(k + 1).fill(0);
+        for (const [a, b] of cond) {
+            g[a].push(b);
+            ++indeg[b];
+        }
+        const q = [];
+        for (let i = 1; i < indeg.length; ++i) {
+            if (indeg[i] == 0) {
+                q.push(i);
+            }
+        }
+        const res = [];
+        while (q.length) {
+            for (let n = q.length; n; --n) {
+                const i = q.shift();
+                res.push(i);
+                for (const j of g[i]) {
+                    if (--indeg[j] == 0) {
+                        q.push(j);
+                    }
+                }
+            }
+        }
+        return res.length == k ? res : [];
+    }
 
+    const row = f(rowConditions);
+    const col = f(colConditions);
+    if (!row.length || !col.length) return [];
+    const ans = Array.from({ length: k }, () => new Array(k).fill(0));
+    const m = new Array(k + 1).fill(0);
+    for (let i = 0; i < k; ++i) {
+        m[col[i]] = i;
+    }
+    for (let i = 0; i < k; ++i) {
+        ans[i][m[row[i]]] = row[i];
+    }
+    return ans;
+}
 ```
 
 ### **...**

solution/2300-2399/2392.Build a Matrix With Conditions/README_EN.md

@@ -63,6 +63,8 @@ No matrix can satisfy all the conditions, so we return the empty matrix.
 
 ## Solutions
 
+Topological Sort.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -170,7 +172,7 @@ public:
         this->k = k;
         auto row = f(rowConditions);
         auto col = f(colConditions);
-        if (row.empty() || col.empty()) return vector<vector<int>>();
+        if (row.empty() || col.empty()) return {};
         vector<vector<int>> ans(k, vector<int>(k));
         vector<int> m(k + 1);
         for (int i = 0; i < k; ++i) {
@@ -275,7 +277,52 @@ func buildMatrix(k int, rowConditions [][]int, colConditions [][]int) [][]int {
 ### **TypeScript**
 
 ```ts
+function buildMatrix(
+    k: number,
+    rowConditions: number[][],
+    colConditions: number[][],
+): number[][] {
+    function f(cond) {
+        const g = Array.from({ length: k + 1 }, () => []);
+        const indeg = new Array(k + 1).fill(0);
+        for (const [a, b] of cond) {
+            g[a].push(b);
+            ++indeg[b];
+        }
+        const q = [];
+        for (let i = 1; i < indeg.length; ++i) {
+            if (indeg[i] == 0) {
+                q.push(i);
+            }
+        }
+        const res = [];
+        while (q.length) {
+            for (let n = q.length; n; --n) {
+                const i = q.shift();
+                res.push(i);
+                for (const j of g[i]) {
+                    if (--indeg[j] == 0) {
+                        q.push(j);
+                    }
+                }
+            }
+        }
+        return res.length == k ? res : [];
+    }
 
+    const row = f(rowConditions);
+    const col = f(colConditions);
+    if (!row.length || !col.length) return [];
+    const ans = Array.from({ length: k }, () => new Array(k).fill(0));
+    const m = new Array(k + 1).fill(0);
+    for (let i = 0; i < k; ++i) {
+        m[col[i]] = i;
+    }
+    for (let i = 0; i < k; ++i) {
+        ans[i][m[row[i]]] = row[i];
+    }
+    return ans;
+}
 ```
 
 ### **...**

solution/2300-2399/2392.Build a Matrix With Conditions/Solution.cpp

@@ -6,7 +6,7 @@ class Solution {
         this->k = k;
         auto row = f(rowConditions);
         auto col = f(colConditions);
-        if (row.empty() || col.empty()) return vector<vector<int>>();
+        if (row.empty() || col.empty()) return {};
         vector<vector<int>> ans(k, vector<int>(k));
         vector<int> m(k + 1);
         for (int i = 0; i < k; ++i) {

solution/2300-2399/2392.Build a Matrix With Conditions/Solution.ts

@@ -0,0 +1,46 @@
+function buildMatrix(
+    k: number,
+    rowConditions: number[][],
+    colConditions: number[][],
+): number[][] {
+    function f(cond) {
+        const g = Array.from({ length: k + 1 }, () => []);
+        const indeg = new Array(k + 1).fill(0);
+        for (const [a, b] of cond) {
+            g[a].push(b);
+            ++indeg[b];
+        }
+        const q = [];
+        for (let i = 1; i < indeg.length; ++i) {
+            if (indeg[i] == 0) {
+                q.push(i);
+            }
+        }
+        const res = [];
+        while (q.length) {
+            for (let n = q.length; n; --n) {
+                const i = q.shift();
+                res.push(i);
+                for (const j of g[i]) {
+                    if (--indeg[j] == 0) {
+                        q.push(j);
+                    }
+                }
+            }
+        }
+        return res.length == k ? res : [];
+    }
+
+    const row = f(rowConditions);
+    const col = f(colConditions);
+    if (!row.length || !col.length) return [];
+    const ans = Array.from({ length: k }, () => new Array(k).fill(0));
+    const m = new Array(k + 1).fill(0);
+    for (let i = 0; i < k; ++i) {
+        m[col[i]] = i;
+    }
+    for (let i = 0; i < k; ++i) {
+        ans[i][m[row[i]]] = row[i];
+    }
+    return ans;
+}