chore: clean up content phase organization

apps/docs/content/(01-linear-structures)/binary-search/153-find-minimum-in-rotated-sorted-array.mdx

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+---
+title: '153. Find Minimum in Rotated Sorted Array'
+description: 'Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:'
+sidebar:
+  label: 'Find Minimum in Rotated Sorted Array'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+::::warning
+Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
+::::
+
+### Example 1:
+- Input: `nums = [3,4,5,1,2]`
+- Output: `1`
+- Explanation: The original array was `[1,2,3,4,5]` rotated `3` times.
+
+### Example 2:
+- Input: `nums = [4,5,6,7,0,1,2]`
+- Output: `0`
+- Explanation: The original array was `[0,1,2,4,5,6,7]` and it was rotated `4` times.
+
+### Example 3:
+- Input: `nums = [11,13,15,17]`
+- Output: `11`
+- Explanation: The original array was `[11,13,15,17]` and it was rotated `4` times.
+
+### Constraints:
+
+- `n == nums.length`
+- `1 <= n <= 5000`
+- `-5000 <= nums[i] <= 5000`
+- All the integers of `nums` are unique.
+- `nums` is sorted and rotated between 1 and `n` times.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["findMin(nums)"]) --> I["l = 0, r = n-1, lowest_index = -1"]
+  I --> W{"l <= r?"}
+  W -- no --> E(["return nums[lowest_index]"])
+  W -- yes --> M["m = (l + r) // 2"]
+  M --> C{"nums[m] <= nums[-1]?"}
+  C -- yes --> A["in the right sorted run: record lowest_index = m, r = m - 1"]
+  A --> W
+  C -- no --> B["still in the left sorted run: l = m + 1"]
+  B --> W
+```
+
+## Solution
+
+```py
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        l, r = 0, len(nums) - 1
+        lowest_index = -1
+
+        while l <= r:
+            m = (l + r) // 2
+            if nums[m] <= nums[-1]:
+                lowest_index = m
+                r = m - 1
+            else:
+                l = m + 1
+
+        return nums[lowest_index]
+```
+
+## Explanation
+
+<YouTube id="nIVW4P8b1VA" title="Find Minimum in Rotated Sorted Array - Binary Search - Leetcode 153 - Python" />

apps/docs/content/(01-linear-structures)/binary-search/33-search-in-rotated-sorted-array.mdx

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+---
+title: '33. Search in Rotated Sorted Array'
+description: There is an integer array nums sorted in ascending order (with distinct values)
+sidebar:
+  label: 'Search in Rotated Sorted Array'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+::::warning
+You must write an algorithm with O(log n) runtime complexity.
+::::
+
+### Example 1:
+- Input: `nums = [4,5,6,7,0,1,2], target = 0`
+- Output: `4`
+
+### Example 2:
+- Input: `nums = [4,5,6,7,0,1,2], target = 3`
+- Output: `-1`
+
+### Example 3:
+- Input: `nums = [1], target = 0`
+- Output: `-1`
+
+### Constraints:
+
+- `1 <= nums.length <= 5000`
+- `-10^4 <= nums[i] <= 10^4`
+- All values of `nums` are unique.
+- `nums` is an ascending array that is possibly rotated.
+- `-10^4 <= target <= 10^4`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["search(nums, target)"]) --> I["l = 0, r = n - 1"]
+  I --> W{"l <= r?"}
+  W -- no --> E(["return -1"])
+  W -- yes --> M["mid = floor((l + r) / 2)"]
+  M --> F{"nums[mid] == target?"}
+  F -- yes --> R(["return mid"])
+  F -- no --> H{"nums[l] <= nums[mid]? — left half sorted"}
+  H -- yes --> A{"target outside [nums[l], nums[mid]]?"}
+  A -- yes --> A1["l = mid + 1 — search the right half"]
+  A -- no --> A2["r = mid - 1"]
+  H -- no --> B{"target outside [nums[mid], nums[r]]?"}
+  B -- yes --> B1["r = mid - 1 — search the left half"]
+  B -- no --> B2["l = mid + 1"]
+  A1 --> W
+  A2 --> W
+  B1 --> W
+  B2 --> W
+```
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+var search = function(nums, target) {
+  let l = 0, r = nums.length - 1;   
+
+  while (l <= r){
+    let mid = Math.floor((l + r)/2);
+    if (target === nums[mid]) {
+      return mid
+    }
+
+    // Left sorted portion
+    if (nums[l] <= nums[mid]){
+      if (target > nums[mid] || target < nums[l]){
+        l = mid + 1;
+      }else{
+        r = mid - 1;
+      }
+    }
+    // Right sorted portion
+    else{
+      if(target < nums[mid] || target > nums[r]){
+        r = mid - 1; 
+      }else{
+        l = mid + 1;
+      }
+    }
+  }
+  return -1;
+};
+```
+
+## Explanation
+
+<YouTube id="U8XENwh8Oy8" title="Search in rotated sorted array - Leetcode 33 - Python" />

apps/docs/content/(01-linear-structures)/binary-search/35-search-insert-position.mdx

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+---
+title: '35. Search Insert Position'
+description: Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order
+sidebar:
+  label: 'Search Insert Position'
+  badge: 'Easy'
+---
+
+<Badge variant="accent">Array</Badge>
+
+::::warning
+You must write an algorithm with O(log n) runtime complexity.
+::::
+
+### Example 1:
+- Input: `nums = [1,3,5,6], target = 5`
+- Output: `2`
+
+### Example 2:
+- Input: `nums = [1,3,5,6], target = 2`
+- Output: `1`
+
+### Example 3:
+- Input: `nums = [1,3,5,6], target = 7`
+- Output: `4`
+
+### Constraints:
+
+- `1 <= nums.length <= 10^4`
+- `-10^4 <= nums[i] <= 10^4`
+- `nums` contains distinct values sorted in ascending order.
+- `-10^4 <= target <= 10^4`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["searchInsert(nums, target)"]) --> I["l = 0, r = n - 1"]
+  I --> W{"l <= r?"}
+  W -- yes --> M["m = (l + r) // 2"]
+  M --> Q{"nums[m] vs target"}
+  Q -- equal --> R(["return m"])
+  Q -- "nums[m] < target" --> L["l = m + 1"]
+  L --> W
+  Q -- "nums[m] > target" --> H["r = m - 1"]
+  H --> W
+  W -- no --> E(["return (l + r) // 2 + 1 — the insert slot"])
+```
+
+## Solution
+
+```py
+class Solution:
+    def searchInsert(self, nums: List[int], target: int) -> int:
+        l, r = 0, len(nums) - 1
+
+        while l <= r:
+            m = (l + r) // 2
+            if nums[m] == target:
+                return m
+            elif nums[m] < target:
+                l = m + 1
+            elif nums[m] > target:
+                r = m - 1
+
+        return (l + r) // 2 + 1
+```
+
+## Explanation
+
+<YouTube id="K-RYzDZkzCI" title="Search Insert Position - Binary Search - Leetcode 35 - Python" />

apps/docs/content/(01-linear-structures)/binary-search/704-binary-search.mdx

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+---
+title: '704. Binary Search'
+description: Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1
+sidebar:
+  label: 'Binary Search'
+  badge: 'Easy'
+---
+
+<Badge variant="accent">Array</Badge>
+
+::::warning
+You must write an algorithm with O(log n) runtime complexity.
+::::
+
+### Example 1:
+- Input: `nums = [-1,0,3,5,9,12], target = 9`
+- Output: `4`
+- Explanation: `9` exists in `nums` and its index is `4`
+
+### Example 2:
+- Input: `nums = [-1,0,3,5,9,12], target = 2`
+- Output: `-1`
+- Explanation: `2` does not exist in `nums` so return `-1`
+
+### Constraints:
+
+- `1 <= nums.length <= 10^4`
+- `-10^4 < nums[i], target < 10^4`
+- All the integers in `nums` are unique.
+- `nums` is sorted in ascending order.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["search(nums, target)"]) --> I["left = 0, right = n - 1"]
+  I --> W{"left <= right?"}
+  W -- no --> E(["return -1"])
+  W -- yes --> M["mid = left + floor((right - left) / 2)"]
+  M --> Q{"nums[mid] vs target"}
+  Q -- equal --> R(["return mid"])
+  Q -- "nums[mid] < target" --> L["left = mid + 1"]
+  L --> W
+  Q -- "nums[mid] > target" --> H["right = mid - 1"]
+  H --> W
+```
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+var search = function(nums, target) {
+    let left = 0;
+    let right = nums.length - 1;
+
+    while(left <= right) {
+        const mid = left + Math.floor((right - left) / 2);
+        if (nums[mid] === target) {
+            return mid;
+        } else if(nums[mid] < target){
+            left = mid + 1;
+        } else if (nums[mid] > target) {
+            right = mid - 1;
+        }
+    }
+    return -1;
+};
+```
+
+## Explanation
+
+<YouTube id="s4DPM8ct1pI" title="Binary Search - Leetcode 704 - Python" />

apps/docs/content/(01-linear-structures)/binary-search/74-search-a-2d-matrix.mdx

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+---
+title: '74. Search a 2D Matrix'
+description: 'You are given an m x n integer matrix matrix with the following two properties:'
+sidebar:
+  label: 'Search a 2D Matrix'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+::::warning
+You must write a solution in O(log(m * n)) time complexity.
+::::
+
+### Example 1:
+- Input: `matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3`
+- Output: `true`
+
+### Example 2:
+- Input: `matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13`
+- Output: `false`
+
+### Constraints:
+
+- `m == matrix.length`
+- `n == matrix[i].length`
+- `1 <= m, n <= 100`
+- `-10^4 <= matrix[i][j], target <= 10^4`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["searchMatrix(matrix, target)"]) --> I["top = 0, bot = ROWS - 1"]
+  I --> W1{"top <= bot?"}
+  W1 -- yes --> R1["row = (top + bot) // 2"]
+  R1 --> C1{"target vs that row's range"}
+  C1 -- "above matrix[row][-1]" --> U["top = row + 1"]
+  C1 -- "below matrix[row][0]" --> V["bot = row - 1"]
+  C1 -- inside --> B(["break — row found"])
+  U --> W1
+  V --> W1
+  W1 -- no --> F(["return False — no row can hold it"])
+  B --> P["row = (top + bot) // 2; l = 0, r = COLS - 1"]
+  P --> W2{"l <= r?"}
+  W2 -- no --> G(["return False"])
+  W2 -- yes --> M["m = (l + r) // 2"]
+  M --> C2{"target vs matrix[row][m]"}
+  C2 -- greater --> X["l = m + 1"]
+  C2 -- smaller --> Y["r = m - 1"]
+  C2 -- equal --> T(["return True"])
+  X --> W2
+  Y --> W2
+```
+
+## Solution
+
+```py
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        ROWS, COLS = len(matrix), len(matrix[0])
+
+        top, bot = 0, ROWS - 1
+        while top <= bot:
+            row = (top + bot) // 2
+            if target > matrix[row][-1]:
+                top = row + 1
+            elif target < matrix[row][0]:
+                bot = row - 1
+            else:
+                break
+
+        if not (top <= bot):
+            return False
+
+        row = (top + bot) // 2
+        l, r = 0, COLS - 1
+        while l <= r:
+            m = (l + r) // 2
+            if target > matrix[row][m]:
+                l = m + 1
+            elif target < matrix[row][m]:
+                r = m - 1
+            else:
+                return True
+        return False
+```
+
+## Explanation
+
+<YouTube id="Ber2pi2C0j0" title="Search a 2D Matrix - Leetcode 74 - Python" />

apps/docs/content/(01-linear-structures)/binary-search/875-koko-eating-bananas.mdx

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+---
+title: '875. Koko Eating Bananas'
+description: Koko loves to eat bananas. There are n piles of bananas, the i^th pile has piles[i] bananas. The guards have gone and will come back in h hours
+sidebar:
+  label: 'Koko Eating Bananas'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+### Example 1:
+- Input: `piles = [3,6,7,11], h = 8`
+- Output: `4`
+
+### Example 2:
+- Input: `piles = [30,11,23,4,20], h = 5`
+- Output: `30`
+
+### Example 3:
+- Input: `piles = [30,11,23,4,20], h = 6`
+- Output: `23`
+
+### Constraints:
+
+- `1 <= piles.length <= 10^4`
+- `piles.length <= h <= 10^9`
+- `1 <= piles[i] <= 10^9`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["minEatingSpeed(piles, h)"]) --> I["l = 1, r = max(piles)"]
+  I --> W{"l < r?"}
+  W -- no --> E(["return l — the smallest workable speed"])
+  W -- yes --> M["mid = floor((l + r) / 2)"]
+  M --> K["kWorks(mid): hours = sum of ceil(p / mid)"]
+  K --> Q{"hours <= h?"}
+  Q -- yes --> A["r = mid — keep mid, try slower"]
+  Q -- no --> B["l = mid + 1 — too slow, speed up"]
+  A --> W
+  B --> W
+```
+
+## Solution
+
+```js
+/**
+ * @param {number[]} piles
+ * @param {number} h
+ * @return {number}
+ */
+var minEatingSpeed = function(piles, h) {
+  // Function to determine if K works for the H
+   function kWorks(k){
+    let hours = 0;
+    for (let p of piles){
+      hours += Math.ceil(p/k)
+    }
+    return hours <= h;
+  }
+
+  // Setup Binary Search
+  let l = 1;
+  let r = Math.max(...piles);
+
+  // Find the K value which works for H and consumes all bananas
+  while (l < r){
+    const mid = Math.floor((l + r) / 2);
+    if (kWorks(mid)){
+      r = mid;
+    }else{
+      l = mid + 1;
+    }
+  }
+  return l;
+};
+```
+
+## Explanation
+
+<YouTube id="U2SozAs9RzA" title="Koko Eating Bananas - Binary Search - Leetcode 875 - Python" />

apps/docs/content/(01-linear-structures)/binary-search/index.mdx

@@ -0,0 +1,5 @@
+---
+title: Binary Search
+description: 'Half of a sorted range is removed at each step until the target is found.'
+hidden: true
+---

apps/docs/content/(01-linear-structures)/binary-search/meta.ts

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+import { defineMeta } from "blume";
+
+export default defineMeta({ title: "Binary Search", order: 5, icon: "search", display: "group" });