feat: auth & admin

.agents/skills/vercel-react-best-practices/rules/js-min-max-loop.md

@@ -13,14 +13,14 @@ Finding the smallest or largest element only requires a single pass through the
 
 ```typescript
 interface Project {
-  id: string
-  name: string
-  updatedAt: number
+  id: string;
+  name: string;
+  updatedAt: number;
 }
 
 function getLatestProject(projects: Project[]) {
-  const sorted = [...projects].sort((a, b) => b.updatedAt - a.updatedAt)
-  return sorted[0]
+  const sorted = [...projects].sort((a, b) => b.updatedAt - a.updatedAt);
+  return sorted[0];
 }
 ```
 
@@ -30,8 +30,8 @@ Sorts the entire array just to find the maximum value.
 
 ```typescript
 function getOldestAndNewest(projects: Project[]) {
-  const sorted = [...projects].sort((a, b) => a.updatedAt - b.updatedAt)
-  return { oldest: sorted[0], newest: sorted[sorted.length - 1] }
+  const sorted = [...projects].sort((a, b) => a.updatedAt - b.updatedAt);
+  return { oldest: sorted[0], newest: sorted[sorted.length - 1] };
 }
 ```
 
@@ -41,31 +41,31 @@ Still sorts unnecessarily when only min/max are needed.
 
 ```typescript
 function getLatestProject(projects: Project[]) {
-  if (projects.length === 0) return null
-  
-  let latest = projects[0]
-  
+  if (projects.length === 0) return null;
+
+  let latest = projects[0];
+
   for (let i = 1; i < projects.length; i++) {
     if (projects[i].updatedAt > latest.updatedAt) {
-      latest = projects[i]
+      latest = projects[i];
     }
   }
-  
-  return latest
+
+  return latest;
 }
 
 function getOldestAndNewest(projects: Project[]) {
-  if (projects.length === 0) return { oldest: null, newest: null }
-  
-  let oldest = projects[0]
-  let newest = projects[0]
-  
+  if (projects.length === 0) return { oldest: null, newest: null };
+
+  let oldest = projects[0];
+  let newest = projects[0];
+
   for (let i = 1; i < projects.length; i++) {
-    if (projects[i].updatedAt < oldest.updatedAt) oldest = projects[i]
-    if (projects[i].updatedAt > newest.updatedAt) newest = projects[i]
+    if (projects[i].updatedAt < oldest.updatedAt) oldest = projects[i];
+    if (projects[i].updatedAt > newest.updatedAt) newest = projects[i];
   }
-  
-  return { oldest, newest }
+
+  return { oldest, newest };
 }
 ```
 
@@ -74,9 +74,9 @@ Single pass through the array, no copying, no sorting.
 **Alternative (Math.min/Math.max for small arrays):**
 
 ```typescript
-const numbers = [5, 2, 8, 1, 9]
-const min = Math.min(...numbers)
-const max = Math.max(...numbers)
+const numbers = [5, 2, 8, 1, 9];
+const min = Math.min(...numbers);
+const max = Math.max(...numbers);
 ```
 
 This works for small arrays, but can be slower or just throw an error for very large arrays due to spread operator limitations. Maximal array length is approximately 124000 in Chrome 143 and 638000 in Safari 18; exact numbers may vary - see [the fiddle](https://jsfiddle.net/qw1jabsx/4/). Use the loop approach for reliability.