feat: refactored docs structure

apps/docs/content/(05-decision-space)/1d-dynamic-programming/(problems)/198_house-robber.mdx

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+---
+title: '198. House Robber'
+description: You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+issue: "https://github.com/prdlk/leetcode/issues/158"
+sidebar:
+  label: 'House Robber'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `nums = [1,2,3,1]`
+- Output: `4`
+- Explanation: Rob house `1` (money = `1`) and then rob house `3` (money = `3`). Total amount you can rob = `1 + 3 = 4`.
+
+### Example 2:
+- Input: `nums = [2,7,9,3,1]`
+- Output: `12`
+- Explanation: Rob house `1` (money = `2`), rob house `3` (money = `9`) and rob house `5` (money = `1`). Total amount you can rob = `2 + 9 + 1 = 12`.
+
+### Constraints:
+
+- `1 <= nums.length <= 100`
+- `0 <= nums[i] <= 400`
+
+## Solution
+
+```py
+class Solution:
+    def rob(self, nums: list[int]) -> int:
+        n = len(nums)
+        if n == 0:
+            return 0
+        if n <= 2:
+            return max(nums)
+
+        dp = [0] * n
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+        for i in range(2, n):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+        return dp[-1]
+```

apps/docs/content/(05-decision-space)/1d-dynamic-programming/(problems)/322_coin-change.mdx

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+---
+title: '322. Coin Change'
+description: You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+  - { name: "Breadth-First Search", slug: "breadth-first-search" }
+  - { name: "Knapsack Problem", slug: "knapsack-problem" }
+  - { name: "Complete Knapsack", slug: "complete-knapsack" }
+issue: "https://github.com/prdlk/leetcode/issues/159"
+sidebar:
+  label: 'Coin Change'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `coins = [1,2,5], amount = 11`
+- Output: `3`
+- Explanation: `11 = 5 + 5 + 1`
+
+### Example 2:
+- Input: `coins = [2], amount = 3`
+- Output: `-1`
+
+### Example 3:
+- Input: `coins = [1], amount = 0`
+- Output: `0`
+
+### Constraints:
+
+- `1 <= coins.length <= 12`
+- `1 <= coins[i] <= 2^31 - 1`
+- `0 <= amount <= 10^4`
+
+## Solution
+
+```py
+from math import inf
+
+
+class Solution:
+    def coinChange(self, coins: list[int], amount: int) -> int:
+        n = len(coins)
+        memo = [[-1] * (amount + 1) for _ in range(n + 1)]
+
+        def min_coins(coins, i, amount, memo):
+            if amount == 0:
+                return 0
+            if i == 0 or amount < 0:
+                return inf
+            if memo[i][amount] != -1:
+                return memo[i][amount]
+
+            skip = min_coins(coins, i - 1, amount, memo)
+            use = min_coins(coins, i, amount - coins[i - 1], memo)
+            if use != inf:
+                use += 1
+
+            memo[i][amount] = min(skip, use)
+            return memo[i][amount]
+
+        result = min_coins(coins, n, amount, memo)
+        return result if result != inf else -1  # pyright: ignore[reportReturnType]
+```

apps/docs/content/(05-decision-space)/1d-dynamic-programming/(problems)/70_climbing-stairs.mdx

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+---
+title: '70. Climbing Stairs'
+description: You are climbing a staircase. It takes n steps to reach the top
+icon: dot
+topics:
+  - { name: "Math", slug: "math" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+  - { name: "Memoization", slug: "memoization" }
+issue: "https://github.com/prdlk/leetcode/issues/157"
+sidebar:
+  label: 'Climbing Stairs'
+  badge: 'Easy'
+---
+
+### Example 1:
+- Input: `n = 2`
+- Output: `2`
+- Explanation: There are two ways to climb to the top. `1`. `1` step + `1` step `2`. `2` steps
+
+### Example 2:
+- Input: `n = 3`
+- Output: `3`
+- Explanation: There are three ways to climb to the top. `1`. `1` step + `1` step + `1` step `2`. `1` step + `2` steps `3`. `2` steps + `1` step
+
+### Constraints:
+
+- `1 <= n <= 45`
+
+## Solution
+
+```py
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        if n == 1:
+            return 1
+        if n == 2:
+            return 2
+
+        first, second = 1, 2
+
+        for _ in range(3, n + 1):
+            third = first + second
+            first, second = second, third
+
+        return second
+```

apps/docs/content/(05-decision-space)/1d-dynamic-programming/index.mdx

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+---
+title: 1-D Dynamic Programming
+description: 'The result for each state is stored in a list, and each new result is computed from the results before it.'
+hidden: true
+---

apps/docs/content/(05-decision-space)/1d-dynamic-programming/meta.ts

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+import { defineMeta } from "blume";
+
+export default defineMeta({ title: "1-D Dynamic Programming", order: 21, icon: "columns-3", display: "page" });