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Commit b2cbcd0

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feat: add ts solution to lc problem: No.1761
No.1761.Minimum Degree of a Connected Trio in a Graph
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‎solution/1700-1799/1761.Minimum Degree of a Connected Trio in a Graph/README.md‎

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**方法一:暴力枚举**
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我们先将所有边存入邻接矩阵 $g$ 中,再将每个节点的度数存入数组 $deg$ 中。
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我们先将所有边存入邻接矩阵 $g$ 中,再将每个节点的度数存入数组 $deg$ 中。初始化答案 $ans=+\infty$。
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然后枚举所有的三元组 $(i, j, k),ドル其中 $i \lt j \lt k,ドル如果 $g[i][j] = g[j][k] = g[i][k] = 1,ドル则说明这三个节点构成了一个连通三元组,此时更新答案为 $deg[i] + deg[j] + deg[k] - 6$。返回最小的符合条件的答案即可。
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然后枚举所有的三元组 $(i, j, k),ドル其中 $i \lt j \lt k,ドル如果 $g[i][j] = g[j][k] = g[i][k] = 1,ドル则说明这三个节点构成了一个连通三元组,此时更新答案为 $ans = \min(ans, deg[i] + deg[j] + deg[k] - 6)$。
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枚举完所有的三元组后,如果答案仍然为 $+\infty,ドル说明图中不存在连通三元组,返回 $-1,ドル否则返回答案。
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时间复杂度 $O(n^3),ドル空间复杂度 $O(n^2)$。其中 $n$ 为节点数。
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}
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```
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### **TypeScript**
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```ts
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function minTrioDegree(n: number, edges: number[][]): number {
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const g = Array.from({ length: n }, () => Array(n).fill(false));
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const deg: number[] = Array(n).fill(0);
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for (let [u, v] of edges) {
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u--;
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v--;
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g[u][v] = g[v][u] = true;
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++deg[u];
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++deg[v];
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}
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let ans = Infinity;
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for (let i = 0; i < n; ++i) {
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for (let j = i + 1; j < n; ++j) {
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if (g[i][j]) {
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for (let k = j + 1; k < n; ++k) {
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if (g[i][k] && g[j][k]) {
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ans = Math.min(ans, deg[i] + deg[j] + deg[k] - 6);
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}
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}
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}
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}
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}
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return ans === Infinity ? -1 : ans;
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}
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```
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### **...**
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```

‎solution/1700-1799/1761.Minimum Degree of a Connected Trio in a Graph/README_EN.md‎

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}
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```
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### **TypeScript**
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```ts
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function minTrioDegree(n: number, edges: number[][]): number {
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const g = Array.from({ length: n }, () => Array(n).fill(false));
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const deg: number[] = Array(n).fill(0);
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for (let [u, v] of edges) {
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u--;
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v--;
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g[u][v] = g[v][u] = true;
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++deg[u];
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++deg[v];
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}
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let ans = Infinity;
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for (let i = 0; i < n; ++i) {
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for (let j = i + 1; j < n; ++j) {
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if (g[i][j]) {
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for (let k = j + 1; k < n; ++k) {
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if (g[i][k] && g[j][k]) {
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ans = Math.min(ans, deg[i] + deg[j] + deg[k] - 6);
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}
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}
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}
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}
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}
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return ans === Infinity ? -1 : ans;
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}
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```
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### **...**
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```
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function minTrioDegree(n: number, edges: number[][]): number {
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const g = Array.from({ length: n }, () => Array(n).fill(false));
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const deg: number[] = Array(n).fill(0);
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for (let [u, v] of edges) {
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u--;
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v--;
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g[u][v] = g[v][u] = true;
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++deg[u];
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++deg[v];
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}
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let ans = Infinity;
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for (let i = 0; i < n; ++i) {
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for (let j = i + 1; j < n; ++j) {
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if (g[i][j]) {
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for (let k = j + 1; k < n; ++k) {
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if (g[i][k] && g[j][k]) {
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ans = Math.min(ans, deg[i] + deg[j] + deg[k] - 6);
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}
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}
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}
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}
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}
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return ans === Infinity ? -1 : ans;
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}

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