Add Claude skills for React best practices and web design

- react-best-practices: Performance optimization patterns (client-side only)
- web-design-guidelines: UI review against Web Interface Guidelines

Co-authored-by: Ona <no-reply@ona.com>

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

@@ -0,0 +1,82 @@
+---
+title: Use Loop for Min/Max Instead of Sort
+impact: LOW
+impactDescription: O(n) instead of O(n log n)
+tags: javascript, arrays, performance, sorting, algorithms
+---
+
+## Use Loop for Min/Max Instead of Sort
+
+Finding the smallest or largest element only requires a single pass through the array. Sorting is wasteful and slower.
+
+**Incorrect (O(n log n) - sort to find latest):**
+
+```typescript
+interface Project {
+  id: string
+  name: string
+  updatedAt: number
+}
+
+function getLatestProject(projects: Project[]) {
+  const sorted = [...projects].sort((a, b) => b.updatedAt - a.updatedAt)
+  return sorted[0]
+}
+```
+
+Sorts the entire array just to find the maximum value.
+
+**Incorrect (O(n log n) - sort for oldest and newest):**
+
+```typescript
+function getOldestAndNewest(projects: Project[]) {
+  const sorted = [...projects].sort((a, b) => a.updatedAt - b.updatedAt)
+  return { oldest: sorted[0], newest: sorted[sorted.length - 1] }
+}
+```
+
+Still sorts unnecessarily when only min/max are needed.
+
+**Correct (O(n) - single loop):**
+
+```typescript
+function getLatestProject(projects: Project[]) {
+  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]
+    }
+  }
+  
+  return latest
+}
+
+function getOldestAndNewest(projects: Project[]) {
+  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]
+  }
+  
+  return { oldest, newest }
+}
+```
+
+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)
+```
+
+This works for small arrays but can be slower for very large arrays due to spread operator limitations. Use the loop approach for reliability.