diff --git a/apps/docs/content/(array)/1-two-sum.mdx b/apps/docs/content/(array)/1-two-sum.mdx
index 6c7753f..20bded5 100644
--- a/apps/docs/content/(array)/1-two-sum.mdx
+++ b/apps/docs/content/(array)/1-two-sum.mdx
@@ -32,6 +32,20 @@ Can you come up with an algorithm that is less than O(n^2) time complexity?
 - `-10^9 <= target <= 10^9`
 - Only one valid answer exists.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["twoSum(nums, target)"]) --> I["seen = {} — value to index"]
+  I --> L{"more (i, num) in nums?"}
+  L -- no --> E(["return []"])
+  L -- yes --> C["complement = target - num"]
+  C --> H{"complement in seen?"}
+  H -- yes --> R(["return [seen[complement], i]"])
+  H -- no --> W["seen[num] = i"]
+  W --> L
+```
+
 ## Solution
 
 ```py
@@ -45,3 +59,7 @@ class Solution:
             seen[num] = i
         return []
 ```
+
+## Explanation
+
+<YouTube id="KLlXCFG5TnA" title="Two Sum - Leetcode 1 - HashMap - Python" />
diff --git a/apps/docs/content/(array)/11-container-with-most-water.mdx b/apps/docs/content/(array)/11-container-with-most-water.mdx
index 0b01c2e..8695ca8 100644
--- a/apps/docs/content/(array)/11-container-with-most-water.mdx
+++ b/apps/docs/content/(array)/11-container-with-most-water.mdx
@@ -27,6 +27,22 @@ Notice that you may not slant the container.
 - `2 <= n <= 10^5`
 - `0 <= height[i] <= 10^4`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["maxArea(height)"]) --> I["res = 0, left = 0, right = len - 1"]
+  I --> W{"left < right?"}
+  W -- no --> E(["return res"])
+  W -- yes --> A["area = (right - left) * min(height[left], height[right])"]
+  A --> M["res = max(res, area)"]
+  M --> C{"height[left] < height[right]?"}
+  C -- yes --> L["left += 1 — drop the shorter wall"]
+  C -- no --> R["right -= 1 — drop the shorter wall"]
+  L --> W
+  R --> W
+```
+
 ## Solution
 
 ```py
@@ -46,3 +62,7 @@ class Solution:
 
         return res
 ```
+
+## Explanation
+
+<YouTube id="UuiTKBwPgAo" title="Container with Most Water - Leetcode 11 - Python" />
diff --git a/apps/docs/content/(array)/121-best-time-to-buy-and-sell-stock.mdx b/apps/docs/content/(array)/121-best-time-to-buy-and-sell-stock.mdx
index d9c2243..75c980c 100644
--- a/apps/docs/content/(array)/121-best-time-to-buy-and-sell-stock.mdx
+++ b/apps/docs/content/(array)/121-best-time-to-buy-and-sell-stock.mdx
@@ -23,6 +23,24 @@ sidebar:
 - `1 <= prices.length <= 10^5`
 - `0 <= prices[i] <= 10^4`
 
+## Approach
+
+::::warning
+The diagram traces the code below literally. That code does not run: `curr` and `ans` are never initialised, so the first `curr > left` test raises `NameError`, and `left` holds a price rather than an index, so `prices[left]` indexes by value.
+::::
+
+```mermaid
+flowchart TD
+  S(["maxProfit(prices)"]) --> I["left = min(prices)"]
+  I --> F{"more right in range(len(prices))?"}
+  F -- no --> E(["return ans"])
+  F -- yes --> W{"curr > left?"}
+  W -- yes --> P["curr -= prices[left]; left += 1"]
+  P --> W
+  W -- no --> M["ans = max(ans, curr)"]
+  M --> F
+```
+
 ## Solution
 
 ```py
@@ -36,3 +54,7 @@ class Solution:
             ans = max(ans, curr)
         return ans
 ```
+
+## Explanation
+
+<YouTube id="1pkOgXD63yU" title="Sliding Window: Best Time to Buy and Sell Stock - Leetcode 121 - Python" />
diff --git a/apps/docs/content/(array)/1365-how-many-numbers-are-smaller-than-the-current-number.mdx b/apps/docs/content/(array)/1365-how-many-numbers-are-smaller-than-the-current-number.mdx
index e250cd2..7d83301 100644
--- a/apps/docs/content/(array)/1365-how-many-numbers-are-smaller-than-the-current-number.mdx
+++ b/apps/docs/content/(array)/1365-how-many-numbers-are-smaller-than-the-current-number.mdx
@@ -26,6 +26,20 @@ sidebar:
 - `2 <= nums.length <= 500`
 - `0 <= nums[i] <= 100`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["smallerNumbersThanCurrent(nums)"]) --> F["freq: count every value in nums"]
+  F --> O["sorted = keys of freq, ascending"]
+  O --> I["count = 0, smaller = {}"]
+  I --> L{"more num in sorted?"}
+  L -- yes --> A["smaller[num] = count — everything already passed is smaller"]
+  A --> B["count += freq[num]"]
+  B --> L
+  L -- no --> M(["return nums.map(n => smaller[n])"])
+```
+
 ## Solution
 
 ```js
diff --git a/apps/docs/content/(array)/1413-minimum-value-to-get-positive-step-by-step-sum.mdx b/apps/docs/content/(array)/1413-minimum-value-to-get-positive-step-by-step-sum.mdx
index 5d4bf8b..fde0a7a 100644
--- a/apps/docs/content/(array)/1413-minimum-value-to-get-positive-step-by-step-sum.mdx
+++ b/apps/docs/content/(array)/1413-minimum-value-to-get-positive-step-by-step-sum.mdx
@@ -37,6 +37,18 @@ startValue = 4 | startValue = 5 | nums
 - `1 <= nums.length <= 100`
 - `-100 <= nums[i] <= 100`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["minStartValue(nums)"]) --> I["prefix = [nums[0]]"]
+  I --> L{"more i in 1..n-1?"}
+  L -- yes --> P["prefix.push(prefix[i-1] + nums[i])"]
+  P --> L
+  L -- no --> M["min = smallest running total"]
+  M --> R(["return max(1, 1 - min) — lift the dip to 1"])
+```
+
 ## Solution
 
 ```js
diff --git a/apps/docs/content/(array)/1426-counting-elements.mdx b/apps/docs/content/(array)/1426-counting-elements.mdx
index ae081da..1377109 100644
--- a/apps/docs/content/(array)/1426-counting-elements.mdx
+++ b/apps/docs/content/(array)/1426-counting-elements.mdx
@@ -23,6 +23,19 @@ sidebar:
 - `1 <= arr.length <= 1000`
 - `0 <= arr[i] <= 1000`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["countElements(arr)"]) --> I["arrSet = Set(arr), count = 0"]
+  I --> L{"more n in arr?"}
+  L -- no --> E(["return count"])
+  L -- yes --> H{"arrSet has n + 1?"}
+  H -- yes --> C["count++"]
+  C --> L
+  H -- no --> L
+```
+
 ## Solution
 
 ```js
diff --git a/apps/docs/content/(array)/1480-running-sum-of-1d-array.mdx b/apps/docs/content/(array)/1480-running-sum-of-1d-array.mdx
index a3aef67..3906a0d 100644
--- a/apps/docs/content/(array)/1480-running-sum-of-1d-array.mdx
+++ b/apps/docs/content/(array)/1480-running-sum-of-1d-array.mdx
@@ -27,6 +27,17 @@ sidebar:
 - `1 <= nums.length <= 1000`
 - `-10^6 <= nums[i] <= 10^6`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["runningSum(nums)"]) --> I["prefix = [nums[0]]"]
+  I --> L{"more i in 1..n-1?"}
+  L -- yes --> P["prefix.push(prefix[i-1] + nums[i])"]
+  P --> L
+  L -- no --> E(["return prefix"])
+```
+
 ## Solution
 
 ```js
diff --git a/apps/docs/content/(array)/15-3sum.mdx b/apps/docs/content/(array)/15-3sum.mdx
index 9ca8ec4..92e1a94 100644
--- a/apps/docs/content/(array)/15-3sum.mdx
+++ b/apps/docs/content/(array)/15-3sum.mdx
@@ -32,6 +32,28 @@ Notice that the solution set must not contain duplicate triplets.
 - `3 <= nums.length <= 3000`
 - `-10^5 <= nums[i] <= 10^5`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["threeSum(nums)"]) --> O["nums.sort() — duplicates become adjacent"]
+  O --> F{"more i in 0..n-1?"}
+  F -- no --> E(["return result"])
+  F -- yes --> D{"nums[i] == nums[i-1]?"}
+  D -- yes --> F
+  D -- no --> P["left = i+1, right = n-1, target = -nums[i]"]
+  P --> W{"left < right?"}
+  W -- no --> F
+  W -- yes --> C["current = nums[left] + nums[right]"]
+  C --> Q{"current vs target"}
+  Q -- equal --> A["append triplet, skip equal neighbours, left += 1, right -= 1"]
+  A --> W
+  Q -- "current < target" --> L["left += 1 — need a bigger sum"]
+  L --> W
+  Q -- "current > target" --> R["right -= 1 — need a smaller sum"]
+  R --> W
+```
+
 ## Solution
 
 ```py
@@ -76,3 +98,7 @@ class Solution:
 
         return result
 ```
+
+## Explanation
+
+<YouTube id="jzZsG8n2R9A" title="3Sum - Leetcode 15 - Python" />
diff --git a/apps/docs/content/(array)/150-evaluate-reverse-polish-notation.mdx b/apps/docs/content/(array)/150-evaluate-reverse-polish-notation.mdx
index 04a5134..42eb2ec 100644
--- a/apps/docs/content/(array)/150-evaluate-reverse-polish-notation.mdx
+++ b/apps/docs/content/(array)/150-evaluate-reverse-polish-notation.mdx
@@ -28,6 +28,21 @@ sidebar:
 - `1 <= tokens.length <= 10^4`
 - `tokens[i]` is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["evalRPN(tokens)"]) --> I["stack = [], operations = + - * /"]
+  I --> L{"more token c?"}
+  L -- no --> E(["return stack[0]"])
+  L -- yes --> Q{"c is an operator?"}
+  Q -- no --> N["stack.append(int(c))"]
+  N --> L
+  Q -- yes --> P["y = stack.pop(), x = stack.pop() — order matters"]
+  P --> C["stack.append(operations[c](x, y))"]
+  C --> L
+```
+
 ## Solution
 
 ```py
@@ -53,3 +68,7 @@ class Solution:
 
         return stack[0]
 ```
+
+## Explanation
+
+<YouTube id="iu0082c4HDE" title="Evaluate Reverse Polish Notation - Leetcode 150 - Python" />
diff --git a/apps/docs/content/(array)/153-find-minimum-in-rotated-sorted-array.mdx b/apps/docs/content/(array)/153-find-minimum-in-rotated-sorted-array.mdx
index e4bc0b8..f285bb5 100644
--- a/apps/docs/content/(array)/153-find-minimum-in-rotated-sorted-array.mdx
+++ b/apps/docs/content/(array)/153-find-minimum-in-rotated-sorted-array.mdx
@@ -35,6 +35,21 @@ Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in
 - 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
@@ -53,3 +68,7 @@ class Solution:
 
         return nums[lowest_index]
 ```
+
+## Explanation
+
+<YouTube id="nIVW4P8b1VA" title="Find Minimum in Rotated Sorted Array - Binary Search - Leetcode 153 - Python" />
diff --git a/apps/docs/content/(array)/1636-sort-array-by-increasing-frequency.mdx b/apps/docs/content/(array)/1636-sort-array-by-increasing-frequency.mdx
index eafaf3a..20d1edc 100644
--- a/apps/docs/content/(array)/1636-sort-array-by-increasing-frequency.mdx
+++ b/apps/docs/content/(array)/1636-sort-array-by-increasing-frequency.mdx
@@ -27,6 +27,18 @@ sidebar:
 - `1 <= nums.length <= 100`
 - `-100 <= nums[i] <= 100`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["frequencySort(nums)"]) --> F["freq: count every value in nums"]
+  F --> C["sort nums with a two-key comparator"]
+  C --> K1["primary: freq[a] - freq[b] — rarer first"]
+  C --> K2["tie-break: b - a — larger value first"]
+  K1 --> R(["return nums"])
+  K2 --> R
+```
+
 ## Solution
 
 ```js
@@ -40,3 +52,7 @@ var frequencySort = function (nums) {
   return nums.sort((a, b) => freq[a] - freq[b] || b - a);
 };
 ```
+
+## Explanation
+
+<YouTube id="Evq1SfUbhBg" title="Sort Array by Increasing Frequency - Leetcode 1636 - Python" />
diff --git a/apps/docs/content/(array)/167-two-sum-ii-input-array-is-sorted.mdx b/apps/docs/content/(array)/167-two-sum-ii-input-array-is-sorted.mdx
index 0853ffd..16922ae 100644
--- a/apps/docs/content/(array)/167-two-sum-ii-input-array-is-sorted.mdx
+++ b/apps/docs/content/(array)/167-two-sum-ii-input-array-is-sorted.mdx
@@ -35,6 +35,22 @@ Your solution must use only constant extra space.
 - `-1000 <= target <= 1000`
 - The tests are generated such that there is exactly one solution.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["twoSum(numbers, target)"]) --> I["i = 0, j = len - 1"]
+  I --> W{"i < j?"}
+  W -- no --> E(["return []"])
+  W -- yes --> C["c = numbers[i] + numbers[j]"]
+  C --> Q{"c vs target"}
+  Q -- equal --> R(["return [i+1, j+1] — 1-indexed"])
+  Q -- "c < target" --> L["i += 1 — grow the sum"]
+  L --> W
+  Q -- "c > target" --> H["j -= 1 — shrink the sum"]
+  H --> W
+```
+
 ## Solution
 
 ```py
@@ -53,3 +69,7 @@ class Solution:
                 j-=1
         return []
 ```
+
+## Explanation
+
+<YouTube id="cQ1Oz4ckceM" title="TWO SUM II - Amazon Coding Interview Question - Leetcode 167 - Python" />
diff --git a/apps/docs/content/(array)/217-contains-duplicate.mdx b/apps/docs/content/(array)/217-contains-duplicate.mdx
index bde42cd..ea8ff7e 100644
--- a/apps/docs/content/(array)/217-contains-duplicate.mdx
+++ b/apps/docs/content/(array)/217-contains-duplicate.mdx
@@ -27,6 +27,19 @@ sidebar:
 - `1 <= nums.length <= 10^5`
 - `-10^9 <= nums[i] <= 10^9`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["containsDuplicate(nums)"]) --> I["freq = {}"]
+  I --> L{"more n in nums?"}
+  L -- no --> E(["return false"])
+  L -- yes --> C["freq[n] = (freq[n] or 0) + 1"]
+  C --> Q{"freq[n] >= 2?"}
+  Q -- yes --> R(["return true"])
+  Q -- no --> L
+```
+
 ## Solution
 
 ```js
@@ -45,3 +58,7 @@ var containsDuplicate = function(nums) {
   return false;
 };
 ```
+
+## Explanation
+
+<YouTube id="3OamzN90kPg" title="Contains Duplicate - Leetcode 217 - Python" />
diff --git a/apps/docs/content/(array)/238-product-of-array-except-self.mdx b/apps/docs/content/(array)/238-product-of-array-except-self.mdx
index 647c1ea..3e622d3 100644
--- a/apps/docs/content/(array)/238-product-of-array-except-self.mdx
+++ b/apps/docs/content/(array)/238-product-of-array-except-self.mdx
@@ -27,6 +27,17 @@ Can you solve the problem in O(1) extra space complexity? (The output array does
 - `-30 <= nums[i] <= 30`
 - The input is generated such that `answer[i]` is guaranteed to fit in a 32-bit integer.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["productExceptSelf(nums)"]) --> I["answer = ones(n), rightArr = ones(n)"]
+  I --> P1["Pass 1 — left to right: answer[i] = nums[i-1] * answer[i-1]"]
+  P1 --> P2["Pass 2 — right to left: rightArr[i] = nums[i+1] * rightArr[i+1]"]
+  P2 --> P3["Pass 3: answer[i] *= rightArr[i]"]
+  P3 --> E(["return answer — no division needed"])
+```
+
 ## Solution
 
 ```js
@@ -51,3 +62,7 @@ var productExceptSelf = function(nums) {
   return answer;
 };
 ```
+
+## Explanation
+
+<YouTube id="bNvIQI2wAjk" title="Product of Array Except Self - Leetcode 238 - Python" />
diff --git a/apps/docs/content/(array)/268-missing-number.mdx b/apps/docs/content/(array)/268-missing-number.mdx
index 36100cf..7309442 100644
--- a/apps/docs/content/(array)/268-missing-number.mdx
+++ b/apps/docs/content/(array)/268-missing-number.mdx
@@ -34,6 +34,18 @@ Could you implement a solution using only O(1) extra space complexity and O(n) r
 - `0 <= nums[i] <= n`
 - All the numbers of `nums` are unique.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["missingNumber(nums)"]) --> I["numSet = Set(nums), expectedCount = n + 1"]
+  I --> L{"more i in 0..expectedCount-1?"}
+  L -- yes --> Q{"numSet has i?"}
+  Q -- no --> R(["return i"])
+  Q -- yes --> L
+  L -- no --> E(["return -1"])
+```
+
 ## Solution
 
 ```js
@@ -53,3 +65,7 @@ var missingNumber = function(nums) {
   return -1;
 };
 ```
+
+## Explanation
+
+<YouTube id="WnPLSRLSANE" title="Missing Number - Blind 75 - Leetcode 268 - Python" />
diff --git a/apps/docs/content/(array)/303-range-sum-query-immutable.mdx b/apps/docs/content/(array)/303-range-sum-query-immutable.mdx
index bbf2916..821498f 100644
--- a/apps/docs/content/(array)/303-range-sum-query-immutable.mdx
+++ b/apps/docs/content/(array)/303-range-sum-query-immutable.mdx
@@ -20,6 +20,21 @@ sidebar:
 - `0 <= left <= right < nums.length`
 - At most `10^4` calls will be made to `sumRange`.
 
+## Approach
+
+```mermaid
+flowchart TD
+  subgraph build["constructor(nums) — O(n) once"]
+    A["prefix = [0]"] --> B{"more n in nums?"}
+    B -- yes --> C["prefix.push(last + n)"]
+    C --> B
+  end
+  build --> Q
+  subgraph Q["sumRange(left, right) — O(1) per call"]
+    D(["return prefix[right + 1] - prefix[left]"])
+  end
+```
+
 ## Solution
 
 ```js
@@ -50,3 +65,7 @@ class NumArray {
  * var param_1 = obj.sumRange(left,right)
  */
 ```
+
+## Explanation
+
+<YouTube id="2pndAmo_sMA" title="Range Sum Query Immutable - Leetcode 303 - Python" />
diff --git a/apps/docs/content/(array)/33-search-in-rotated-sorted-array.mdx b/apps/docs/content/(array)/33-search-in-rotated-sorted-array.mdx
index 185460a..81515a0 100644
--- a/apps/docs/content/(array)/33-search-in-rotated-sorted-array.mdx
+++ b/apps/docs/content/(array)/33-search-in-rotated-sorted-array.mdx
@@ -32,6 +32,29 @@ You must write an algorithm with O(log n) runtime complexity.
 - `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
@@ -69,3 +92,7 @@ var search = function(nums, target) {
   return -1;
 };
 ```
+
+## Explanation
+
+<YouTube id="U8XENwh8Oy8" title="Search in rotated sorted array - Leetcode 33 - Python" />
diff --git a/apps/docs/content/(array)/347-top-k-frequent-elements.mdx b/apps/docs/content/(array)/347-top-k-frequent-elements.mdx
index cf4d31c..bf9b30c 100644
--- a/apps/docs/content/(array)/347-top-k-frequent-elements.mdx
+++ b/apps/docs/content/(array)/347-top-k-frequent-elements.mdx
@@ -31,6 +31,16 @@ Your algorithm's time complexity must be better than O(n log n), where n is the
 - `k` is in the range `[1, the number of unique elements in the array]`.
 - It is guaranteed that the answer is unique.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["topKFrequent(nums, k)"]) --> F["freq: count every value in nums"]
+  F --> K["keys of freq, cast to Number"]
+  K --> O["sort by freq[b] - freq[a] — most frequent first"]
+  O --> R(["return slice(0, k)"])
+```
+
 ## Solution
 
 ```js
@@ -49,3 +59,7 @@ var topKFrequent = function (nums, k) {
     .slice(0, k);
 };
 ```
+
+## Explanation
+
+<YouTube id="YPTqKIgVk-k" title="Top K Frequent Elements - Bucket Sort - Leetcode 347 - Python" />
diff --git a/apps/docs/content/(array)/35-search-insert-position.mdx b/apps/docs/content/(array)/35-search-insert-position.mdx
index 889e15a..ed458c2 100644
--- a/apps/docs/content/(array)/35-search-insert-position.mdx
+++ b/apps/docs/content/(array)/35-search-insert-position.mdx
@@ -31,6 +31,22 @@ You must write an algorithm with O(log n) runtime complexity.
 - `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
@@ -49,3 +65,7 @@ class Solution:
 
         return (l + r) // 2 + 1
 ```
+
+## Explanation
+
+<YouTube id="K-RYzDZkzCI" title="Search Insert Position - Binary Search - Leetcode 35 - Python" />
diff --git a/apps/docs/content/(array)/49-group-anagrams.mdx b/apps/docs/content/(array)/49-group-anagrams.mdx
index 75c52bc..a6730ca 100644
--- a/apps/docs/content/(array)/49-group-anagrams.mdx
+++ b/apps/docs/content/(array)/49-group-anagrams.mdx
@@ -27,6 +27,21 @@ sidebar:
 - `0 <= strs[i].length <= 100`
 - `strs[i]` consists of lowercase English letters.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["groupAnagrams(strs)"]) --> K["sorted[i] = letters of strs[i], sorted — the anagram key"]
+  K --> I["anagrams = {}"]
+  I --> L{"more i in 0..n-1?"}
+  L -- no --> E(["return Object.values(anagrams)"])
+  L -- yes --> Q{"anagrams has sorted[i]?"}
+  Q -- no --> N["anagrams[sorted[i]] = [strs[i]]"]
+  Q -- yes --> P["anagrams[sorted[i]].push(strs[i])"]
+  N --> L
+  P --> L
+```
+
 ## Solution
 
 ```js
@@ -49,3 +64,7 @@ var groupAnagrams = function(strs) {
   return Object.values(anagrams);
 };
 ```
+
+## Explanation
+
+<YouTube id="vzdNOK2oB2E" title="Group Anagrams - Categorize Strings by Count - Leetcode 49" />
diff --git a/apps/docs/content/(array)/560-subarray-sum-equals-k.mdx b/apps/docs/content/(array)/560-subarray-sum-equals-k.mdx
index 2678d07..5e67546 100644
--- a/apps/docs/content/(array)/560-subarray-sum-equals-k.mdx
+++ b/apps/docs/content/(array)/560-subarray-sum-equals-k.mdx
@@ -22,6 +22,21 @@ sidebar:
 - `-1000 <= nums[i] <= 1000`
 - `-10^7 <= k <= 10^7`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["subarraySum(nums, k)"]) --> I["map = {0: 1} — the empty prefix; sum = 0, count = 0"]
+  I --> L{"more n in nums?"}
+  L -- no --> E(["return count"])
+  L -- yes --> A["sum += n"]
+  A --> Q{"map has sum - k?"}
+  Q -- yes --> C["count += map.get(sum - k) — every earlier prefix that closes a window"]
+  Q -- no --> W["map.set(sum, count of sum + 1)"]
+  C --> W
+  W --> L
+```
+
 ## Solution
 
 ```js
@@ -49,3 +64,7 @@ var subarraySum = function(nums, k) {
   return count
 };
 ```
+
+## Explanation
+
+<YouTube id="fFVZt-6sgyo" title="Subarray Sum Equals K - Prefix Sums - Leetcode 560 - Python" />
diff --git a/apps/docs/content/(array)/704-binary-search.mdx b/apps/docs/content/(array)/704-binary-search.mdx
index f92972b..513d1c7 100644
--- a/apps/docs/content/(array)/704-binary-search.mdx
+++ b/apps/docs/content/(array)/704-binary-search.mdx
@@ -29,6 +29,22 @@ You must write an algorithm with O(log n) runtime complexity.
 - 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
@@ -54,3 +70,7 @@ var search = function(nums, target) {
     return -1;
 };
 ```
+
+## Explanation
+
+<YouTube id="s4DPM8ct1pI" title="Binary Search - Leetcode 704 - Python" />
diff --git a/apps/docs/content/(array)/739-daily-temperatures.mdx b/apps/docs/content/(array)/739-daily-temperatures.mdx
index 56ccefb..2b8e63a 100644
--- a/apps/docs/content/(array)/739-daily-temperatures.mdx
+++ b/apps/docs/content/(array)/739-daily-temperatures.mdx
@@ -25,6 +25,20 @@ sidebar:
 - `1 <= temperatures.length <= 10^5`
 - `30 <= temperatures[i] <= 100`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["dailyTemperatures(temperatures)"]) --> I["ans = zeros(n), stack = [] — indices of days still waiting"]
+  I --> L{"more (i, temp)?"}
+  L -- no --> E(["return ans"])
+  L -- yes --> W{"stack non-empty and temperatures[stack[-1]] < temp?"}
+  W -- yes --> P["prev = stack.pop(); ans[prev] = i - prev — today resolves that day"]
+  P --> W
+  W -- no --> A["stack.append(i)"]
+  A --> L
+```
+
 ## Solution
 
 ```py
@@ -39,3 +53,7 @@ class Solution:
             stack.append(i)
         return ans
 ```
+
+## Explanation
+
+<YouTube id="cTBiBSnjO3c" title="Daily Temperatures - Monotonic Stack - Leetcode 739 - Python" />
diff --git a/apps/docs/content/(array)/74-search-a-2d-matrix.mdx b/apps/docs/content/(array)/74-search-a-2d-matrix.mdx
index 3af2355..d53794c 100644
--- a/apps/docs/content/(array)/74-search-a-2d-matrix.mdx
+++ b/apps/docs/content/(array)/74-search-a-2d-matrix.mdx
@@ -27,6 +27,32 @@ You must write a solution in O(log(m * n)) time complexity.
 - `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
@@ -59,3 +85,7 @@ class Solution:
                 return True
         return False
 ```
+
+## Explanation
+
+<YouTube id="Ber2pi2C0j0" title="Search a 2D Matrix - Leetcode 74 - Python" />
diff --git a/apps/docs/content/(array)/853-car-fleet.mdx b/apps/docs/content/(array)/853-car-fleet.mdx
index f05f906..89068d1 100644
--- a/apps/docs/content/(array)/853-car-fleet.mdx
+++ b/apps/docs/content/(array)/853-car-fleet.mdx
@@ -32,6 +32,22 @@ sidebar:
 - All the values of `position` are unique.
 - `0 < speed[i] <= 10^6`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["carFleet(target, position, speed)"]) --> O["zip and sort by position, descending — nearest the target first"]
+  O --> I["stack = [] — arrival time of each fleet's leader"]
+  I --> L{"more (pos, spd)?"}
+  L -- no --> E(["return len(stack) — one entry per fleet"])
+  L -- yes --> T["time = (target - pos) / spd"]
+  T --> Q{"stack empty or time > stack[-1]?"}
+  Q -- yes --> A["stack.append(time) — slower, so it starts a new fleet"]
+  Q -- no --> B["discard — it catches the car ahead and joins that fleet"]
+  A --> L
+  B --> L
+```
+
 ## Solution
 
 ```py
@@ -45,3 +61,7 @@ class Solution:
 
         return len(stack)
 ```
+
+## Explanation
+
+<YouTube id="Pr6T-3yB9RM" title="Car Fleet - Leetcode 853 - Python" />
diff --git a/apps/docs/content/(array)/875-koko-eating-bananas.mdx b/apps/docs/content/(array)/875-koko-eating-bananas.mdx
index 7682605..721f566 100644
--- a/apps/docs/content/(array)/875-koko-eating-bananas.mdx
+++ b/apps/docs/content/(array)/875-koko-eating-bananas.mdx
@@ -26,6 +26,22 @@ sidebar:
 - `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
@@ -60,3 +76,7 @@ var minEatingSpeed = function(piles, h) {
   return l;
 };
 ```
+
+## Explanation
+
+<YouTube id="U2SozAs9RzA" title="Koko Eating Bananas - Binary Search - Leetcode 875 - Python" />
diff --git a/apps/docs/content/(array)/977-squares-of-a-sorted-array.mdx b/apps/docs/content/(array)/977-squares-of-a-sorted-array.mdx
index a79dd32..a94ca09 100644
--- a/apps/docs/content/(array)/977-squares-of-a-sorted-array.mdx
+++ b/apps/docs/content/(array)/977-squares-of-a-sorted-array.mdx
@@ -27,6 +27,21 @@ Squaring each element and sorting the new array is very trivial, could you find
 - `-10^4 <= nums[i] <= 10^4`
 - `nums` is sorted in non-decreasing order.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["sortedSquares(nums)"]) --> I["ans = Array(n), left = 0, right = n - 1"]
+  I --> L{"more i from n-1 down to 0?"}
+  L -- no --> E(["return ans"])
+  L -- yes --> C{"abs(nums[left]) < abs(nums[right])?"}
+  C -- yes --> R["square = nums[right]; right--"]
+  C -- no --> F["square = nums[left]; left++"]
+  R --> W["ans[i] = square * square — fill from the back, largest first"]
+  F --> W
+  W --> L
+```
+
 ## Solution
 
 ```js
@@ -53,3 +68,7 @@ var sortedSquares = function(nums) {
   return ans;
 };
 ```
+
+## Explanation
+
+<YouTube id="FPCZsG_AkUg" title="Squares of a Sorted Array - Leetcode 977 - Python" />
diff --git a/apps/docs/content/(hash-table)/141-linked-list-cycle.mdx b/apps/docs/content/(hash-table)/141-linked-list-cycle.mdx
index 996012b..f1413d9 100644
--- a/apps/docs/content/(hash-table)/141-linked-list-cycle.mdx
+++ b/apps/docs/content/(hash-table)/141-linked-list-cycle.mdx
@@ -33,6 +33,20 @@ Can you solve it using O(1) (i.e. constant) memory?
 - `-10^5 <= Node.val <= 10^5`
 - pos is -1 or a valid index in the linked-list.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["hasCycle(head)"]) --> I["fast = slow = head"]
+  I --> W{"fast and fast.next?"}
+  W -- no --> E(["return False — ran off the end, no cycle"])
+  W -- yes --> A["fast = fast.next.next — two steps"]
+  A --> B["slow = slow.next — one step"]
+  B --> Q{"fast is slow?"}
+  Q -- yes --> R(["return True — the gap closed, so it loops"])
+  Q -- no --> W
+```
+
 ## Solution
 
 ```py
@@ -54,3 +68,7 @@ class Solution:
                 return True
         return False
 ```
+
+## Explanation
+
+<YouTube id="gBTe7lFR3vc" title="Linked List Cycle - Floyd's Tortoise and Hare - Leetcode 141 - Python" />
diff --git a/apps/docs/content/(hash-table)/146-lru-cache.mdx b/apps/docs/content/(hash-table)/146-lru-cache.mdx
index 60b72e0..53f4791 100644
--- a/apps/docs/content/(hash-table)/146-lru-cache.mdx
+++ b/apps/docs/content/(hash-table)/146-lru-cache.mdx
@@ -20,6 +20,31 @@ sidebar:
 - `0 <= value <= 10^5`
 - At most 2 * 10^5 calls will be made to get and put.
 
+## Approach
+
+```mermaid
+flowchart TD
+  subgraph state["state — dict for O(1) lookup, doubly linked list for O(1) reorder"]
+    H["head — least recent"] --- N["... nodes ..."] --- T["tail — most recent"]
+  end
+  subgraph get["get(key)"]
+    G1{"key in cache?"} -- no --> G2(["return -1"])
+    G1 -- yes --> G3["_remove_node then _add_node — move it beside the tail"]
+    G3 --> G4(["return node.value"])
+  end
+  subgraph put["put(key, value)"]
+    P1{"key in cache?"} -- yes --> P2["update value, _remove_node then _add_node"]
+    P1 -- no --> P3["new Node, cache[key] = node, _add_node"]
+    P3 --> P4{"len(cache) > capacity?"}
+    P4 -- yes --> P5["evict head.next — the least recent — and del cache[lru.key]"]
+    P4 -- no --> P6(["done"])
+    P2 --> P6
+    P5 --> P6
+  end
+  state --> get
+  state --> put
+```
+
 ## Solution
 
 ```py
@@ -84,3 +109,7 @@ class LRUCache:
 # param_1 = obj.get(key)
 # obj.put(key,value)
 ```
+
+## Explanation
+
+<YouTube id="7ABFKPK2hD4" title="LRU Cache - Twitch Interview Question - Leetcode 146" />
diff --git a/apps/docs/content/(hash-table)/1832-check-if-the-sentence-is-pangram.mdx b/apps/docs/content/(hash-table)/1832-check-if-the-sentence-is-pangram.mdx
index e669c06..954e1c8 100644
--- a/apps/docs/content/(hash-table)/1832-check-if-the-sentence-is-pangram.mdx
+++ b/apps/docs/content/(hash-table)/1832-check-if-the-sentence-is-pangram.mdx
@@ -22,6 +22,19 @@ sidebar:
 - `1 <= sentence.length <= 1000`
 - `sentence` consists of lowercase English letters.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["checkIfPangram(sentence)"]) --> I["freq = {}"]
+  I --> L{"more char c?"}
+  L -- yes --> C["freq[c] = (freq[c] or 0) + 1"]
+  C --> L
+  L -- no --> Q{"distinct keys in freq == 26?"}
+  Q -- yes --> R(["return true"])
+  Q -- no --> F(["return false"])
+```
+
 ## Solution
 
 ```js
diff --git a/apps/docs/content/(hash-table)/242-valid-anagram.mdx b/apps/docs/content/(hash-table)/242-valid-anagram.mdx
index dff1203..89ba2fa 100644
--- a/apps/docs/content/(hash-table)/242-valid-anagram.mdx
+++ b/apps/docs/content/(hash-table)/242-valid-anagram.mdx
@@ -25,6 +25,19 @@ What if the inputs contain Unicode characters? How would you adapt your solution
 - `1 <= s.length, t.length <= 5 * 10^4`
 - `s` and `t` consist of lowercase English letters.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["isAnagram(s, t)"]) --> I["freq: count every char of s"]
+  I --> L{"more char c in t?"}
+  L -- no --> E(["return true"])
+  L -- yes --> Q{"freq[c] missing or 0?"}
+  Q -- yes --> R(["return false — t has a char s cannot cover"])
+  Q -- no --> D["freq[c] -= 1"]
+  D --> L
+```
+
 ## Solution
 
 ```js
@@ -46,3 +59,7 @@ var isAnagram = function(s, t) {
   return true;
 };
 ```
+
+## Explanation
+
+<YouTube id="9UtInBqnCgA" title="Valid Anagram - Leetcode 242 - Python" />
diff --git a/apps/docs/content/(hash-table)/3-longest-substring-without-repeating-characters.mdx b/apps/docs/content/(hash-table)/3-longest-substring-without-repeating-characters.mdx
index c5fbe45..72d73d2 100644
--- a/apps/docs/content/(hash-table)/3-longest-substring-without-repeating-characters.mdx
+++ b/apps/docs/content/(hash-table)/3-longest-substring-without-repeating-characters.mdx
@@ -28,6 +28,21 @@ sidebar:
 - `0 <= s.length <= 10^5`
 - `s` consists of English letters, digits, symbols and spaces.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["lengthOfLongestSubstring(s)"]) --> I["left = 0, ans = 0, window = set()"]
+  I --> L{"more (right, c) in s?"}
+  L -- no --> E(["return ans"])
+  L -- yes --> W{"c already in window?"}
+  W -- yes --> P["window.remove(s[left]); left += 1 — shrink until c is free"]
+  P --> W
+  W -- no --> A["window.add(c)"]
+  A --> M["ans = max(ans, right - left + 1)"]
+  M --> L
+```
+
 ## Solution
 
 ```py
@@ -45,3 +60,7 @@ class Solution:
 
         return ans
 ```
+
+## Explanation
+
+<YouTube id="wiGpQwVHdE0" title="Longest Substring Without Repeating Characters - Leetcode 3 - Python" />
diff --git a/apps/docs/content/(hash-table)/387-first-unique-character-in-a-string.mdx b/apps/docs/content/(hash-table)/387-first-unique-character-in-a-string.mdx
index 065a16c..d5f831e 100644
--- a/apps/docs/content/(hash-table)/387-first-unique-character-in-a-string.mdx
+++ b/apps/docs/content/(hash-table)/387-first-unique-character-in-a-string.mdx
@@ -26,6 +26,18 @@ sidebar:
 - `1 <= s.length <= 10^5`
 - `s` consists of only lowercase English letters.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["firstUniqChar(s)"]) --> F["Pass 1 — freq: count every char of s"]
+  F --> L{"Pass 2 — more i in 0..len-1?"}
+  L -- no --> E(["return -1"])
+  L -- yes --> Q{"freq[s[i]] == 1?"}
+  Q -- yes --> R(["return i — leftmost wins because the scan is in order"])
+  Q -- no --> L
+```
+
 ## Solution
 
 ```js
@@ -45,3 +57,7 @@ var firstUniqChar = function (s) {
   return -1;
 };
 ```
+
+## Explanation
+
+<YouTube id="rBENYgWy3xU" title="First Unique Character in a String - Leetcode 387 - Python" />
diff --git a/apps/docs/content/(hash-table)/424-longest-repeating-character-replacement.mdx b/apps/docs/content/(hash-table)/424-longest-repeating-character-replacement.mdx
index c807620..f4deb7d 100644
--- a/apps/docs/content/(hash-table)/424-longest-repeating-character-replacement.mdx
+++ b/apps/docs/content/(hash-table)/424-longest-repeating-character-replacement.mdx
@@ -24,6 +24,21 @@ sidebar:
 - `s` consists of only uppercase English letters.
 - `0 <= k <= s.length`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["characterReplacement(s, k)"]) --> I["count = {}, res = 0, l = 0, maxF = 0"]
+  I --> L{"more (r, c) in s?"}
+  L -- no --> E(["return res"])
+  L -- yes --> A["count[c] += 1; maxF = max(maxF, count[c])"]
+  A --> W{"(r - l + 1) - maxF > k? — too many chars to replace"}
+  W -- yes --> P["count[s[l]] -= 1; l += 1 — shrink from the left"]
+  P --> W
+  W -- no --> M["res = max(res, r - l + 1)"]
+  M --> L
+```
+
 ## Solution
 
 ```py
@@ -45,3 +60,7 @@ class Solution:
             res = max(res, r - l + 1)
         return res
 ```
+
+## Explanation
+
+<YouTube id="gqXU1UyA8pk" title="Longest Repeating Character Replacement - Leetcode 424 - Python" />
diff --git a/apps/docs/content/(hash-table)/451-sort-characters-by-frequency.mdx b/apps/docs/content/(hash-table)/451-sort-characters-by-frequency.mdx
index a0415a1..a5d3d30 100644
--- a/apps/docs/content/(hash-table)/451-sort-characters-by-frequency.mdx
+++ b/apps/docs/content/(hash-table)/451-sort-characters-by-frequency.mdx
@@ -28,6 +28,18 @@ sidebar:
 - `1 <= s.length <= 5 * 10^5`
 - `s` consists of uppercase and lowercase English letters and digits.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["frequencySort(s)"]) --> F["freq: count every char of s"]
+  F --> C["split s and sort with a two-key comparator"]
+  C --> K1["primary: freq[b] - freq[a] — most frequent first"]
+  C --> K2["tie-break: a.localeCompare(b) — alphabetical"]
+  K1 --> R(["join and return"])
+  K2 --> R
+```
+
 ## Solution
 
 ```js
@@ -47,3 +59,7 @@ var frequencySort = function (s) {
     .join("");
 };
 ```
+
+## Explanation
+
+<YouTube id="OXdXc9HTrIg" title="Sort Characters By Frequency - Leetcode 451 - Python" />
diff --git a/apps/docs/content/(hash-table)/567-permutation-in-string.mdx b/apps/docs/content/(hash-table)/567-permutation-in-string.mdx
index 3483848..f5e4734 100644
--- a/apps/docs/content/(hash-table)/567-permutation-in-string.mdx
+++ b/apps/docs/content/(hash-table)/567-permutation-in-string.mdx
@@ -22,6 +22,24 @@ sidebar:
 - `1 <= s1.length, s2.length <= 10^4`
 - `s1` and `s2` consist of lowercase English letters.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["checkInclusion(s1, s2)"]) --> G{"k = len(s1) > len(s2)?"}
+  G -- yes --> X(["return False"])
+  G -- no --> I["need = Counter(s1), window = Counter()"]
+  I --> L{"more (right, c) in s2?"}
+  L -- no --> E(["return False"])
+  L -- yes --> A["window[c] += 1"]
+  A --> B{"right >= k? — window is now longer than k"}
+  B -- yes --> C["drop s2[right - k] from window, deleting the key at 0"]
+  B -- no --> Q
+  C --> Q{"window == need?"}
+  Q -- yes --> R(["return True"])
+  Q -- no --> L
+```
+
 ## Solution
 
 ```py
@@ -49,3 +67,7 @@ class Solution:
 
         return False
 ```
+
+## Explanation
+
+<YouTube id="UbyhOgBN834" title="Permutation in String - Leetcode 567 - Python" />
diff --git a/apps/docs/content/(linked-list)/206-reverse-linked-list.mdx b/apps/docs/content/(linked-list)/206-reverse-linked-list.mdx
index 76f547d..537c3ae 100644
--- a/apps/docs/content/(linked-list)/206-reverse-linked-list.mdx
+++ b/apps/docs/content/(linked-list)/206-reverse-linked-list.mdx
@@ -29,6 +29,19 @@ A linked list can be reversed either iteratively or recursively. Could you imple
 - The number of nodes in the list is the range [0, 5000].
 - `-5000 <= Node.val <= 5000`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["reverseList(head)"]) --> I["prev = None, curr = head"]
+  I --> W{"curr?"}
+  W -- no --> E(["return prev — the old tail is the new head"])
+  W -- yes --> A["next_ = curr.next — save it before overwriting"]
+  A --> B["curr.next = prev — flip the link"]
+  B --> C["prev = curr; curr = next_ — step forward"]
+  C --> W
+```
+
 ## Solution
 
 ```py
@@ -48,3 +61,7 @@ class Solution:
             curr = next_
         return prev
 ```
+
+## Explanation
+
+<YouTube id="G0_I-ZF0S38" title="Reverse Linked List - Iterative AND Recursive - Leetcode 206 - Python" />
diff --git a/apps/docs/content/(linked-list)/21-merge-two-sorted-lists.mdx b/apps/docs/content/(linked-list)/21-merge-two-sorted-lists.mdx
index 24f2fbb..d74695e 100644
--- a/apps/docs/content/(linked-list)/21-merge-two-sorted-lists.mdx
+++ b/apps/docs/content/(linked-list)/21-merge-two-sorted-lists.mdx
@@ -26,6 +26,23 @@ sidebar:
 - `-100 <= Node.val <= 100`
 - Both list1 and list2 are sorted in non-decreasing order.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["mergeTwoLists(list1, list2)"]) --> G1{"list1 empty?"}
+  G1 -- yes --> X1(["return list2"])
+  G1 -- no --> G2{"list2 empty?"}
+  G2 -- yes --> X2(["return list1"])
+  G2 -- no --> H["head = the smaller first node; advance that list"]
+  H --> C["current = head"]
+  C --> W{"list1 and list2 both non-empty?"}
+  W -- yes --> P["current.next = the smaller node; advance that list; current = current.next"]
+  P --> W
+  W -- no --> T["current.next = list1 or list2 — append the leftover tail"]
+  T --> E(["return head"])
+```
+
 ## Solution
 
 ```py
@@ -64,3 +81,7 @@ class Solution:
         current.next = list1 or list2
         return head
 ```
+
+## Explanation
+
+<YouTube id="XIdigk956u0" title="Merge Two Sorted Lists - Leetcode 21 - Python" />
diff --git a/apps/docs/content/(stack)/155-min-stack.mdx b/apps/docs/content/(stack)/155-min-stack.mdx
index e284439..364b963 100644
--- a/apps/docs/content/(stack)/155-min-stack.mdx
+++ b/apps/docs/content/(stack)/155-min-stack.mdx
@@ -23,6 +23,28 @@ You must implement a solution with O(1) time complexity for each function.
 - Methods pop, top and getMin operations will always be called on non-empty stacks.
 - At most 3 * 10^4 calls will be made to push, pop, top, and getMin.
 
+## Approach
+
+```mermaid
+flowchart TD
+  subgraph state["two parallel stacks — same depth, always"]
+    A["stack — the values"]
+    B["minStack — the minimum as of that depth"]
+  end
+  state --> P
+  subgraph P["push(value)"]
+    P1["stack.append(value)"] --> P2["minStack.append(min(value, minStack[-1]))"]
+  end
+  state --> O
+  subgraph O["pop()"]
+    O1["pop both stacks together"]
+  end
+  state --> R
+  subgraph R["top() / getMin() — O(1)"]
+    R1(["stack[-1] / minStack[-1]"])
+  end
+```
+
 ## Solution
 
 ```py
@@ -54,3 +76,7 @@ class MinStack:
 # param_3 = obj.top()
 # param_4 = obj.getMin()
 ```
+
+## Explanation
+
+<YouTube id="qkLl7nAwDPo" title="Design Min Stack - Amazon Interview Question - Leetcode 155 - Python" />
diff --git a/apps/docs/content/(string)/20-valid-parentheses.mdx b/apps/docs/content/(string)/20-valid-parentheses.mdx
index e26830b..d67f136 100644
--- a/apps/docs/content/(string)/20-valid-parentheses.mdx
+++ b/apps/docs/content/(string)/20-valid-parentheses.mdx
@@ -33,6 +33,24 @@ sidebar:
 - `1 <= s.length <= 10^4`
 - `s` consists of parentheses only '()[]{}'.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["isValid(s)"]) --> I["stack = [], pairs = closer to opener"]
+  I --> L{"more char c?"}
+  L -- no --> Z{"stack empty?"}
+  Z -- yes --> T(["return True"])
+  Z -- no --> U(["return False — unclosed openers left"])
+  L -- yes --> Q{"c is a closer?"}
+  Q -- no --> A["stack.append(c) — it is an opener"]
+  A --> L
+  Q -- yes --> M{"stack non-empty and pairs[c] == stack[-1]?"}
+  M -- yes --> P["stack.pop() — matched"]
+  P --> L
+  M -- no --> F(["return False — mismatch"])
+```
+
 ## Solution
 
 ```py
@@ -52,3 +70,7 @@ class Solution:
 
         return True if not stack else False
 ```
+
+## Explanation
+
+<YouTube id="WTzjTskDFMg" title="Valid Parentheses - Stack - Leetcode 20 - Python" />
diff --git a/apps/docs/content/(two-pointers)/125-valid-palindrome.mdx b/apps/docs/content/(two-pointers)/125-valid-palindrome.mdx
index 027d09f..1237832 100644
--- a/apps/docs/content/(two-pointers)/125-valid-palindrome.mdx
+++ b/apps/docs/content/(two-pointers)/125-valid-palindrome.mdx
@@ -28,6 +28,20 @@ sidebar:
 - `1 <= s.length <= 2 * 10^5`
 - `s` consists only of printable ASCII characters.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["isPalindrome(s)"]) --> N["normal = s stripped of non-alphanumerics, lowercased"]
+  N --> I["i = 0, j = normal.length - 1"]
+  I --> W{"i < j?"}
+  W -- no --> E(["return true — the pointers met in the middle"])
+  W -- yes --> Q{"normal[i] != normal[j]?"}
+  Q -- yes --> F(["return false"])
+  Q -- no --> A["i++; j-- — step both inward"]
+  A --> W
+```
+
 ## Solution
 
 ```js
@@ -49,3 +63,7 @@ var isPalindrome = function(s) {
   return true;
 };
 ```
+
+## Explanation
+
+<YouTube id="jJXJ16kPFWg" title="Valid Palindrome - Leetcode 125 - Python" />
diff --git a/apps/docs/content/(two-pointers)/344-reverse-string.mdx b/apps/docs/content/(two-pointers)/344-reverse-string.mdx
index 9e94efe..6da31a9 100644
--- a/apps/docs/content/(two-pointers)/344-reverse-string.mdx
+++ b/apps/docs/content/(two-pointers)/344-reverse-string.mdx
@@ -25,6 +25,18 @@ You must do this by modifying the input array in-place with O(1) extra memory.
 - `1 <= s.length <= 10^5`
 - `s[i]` is a printable ascii character.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["reverseString(s)"]) --> I["i = 0, j = s.length - 1"]
+  I --> W{"i < j?"}
+  W -- no --> E(["done — s was reversed in place, nothing returned"])
+  W -- yes --> A["swap s[i] and s[j]"]
+  A --> B["i++; j--"]
+  B --> W
+```
+
 ## Solution
 
 ```js
@@ -43,3 +55,7 @@ var reverseString = function(s) {
     }   
 };
 ```
+
+## Explanation
+
+<YouTube id="_d0T_2Lk2qA" title="Reverse String - 3 Ways - Leetcode 344 - Python" />
diff --git a/apps/docs/content/(two-pointers)/392-is-subsequence.mdx b/apps/docs/content/(two-pointers)/392-is-subsequence.mdx
index 36c8b59..66d4087 100644
--- a/apps/docs/content/(two-pointers)/392-is-subsequence.mdx
+++ b/apps/docs/content/(two-pointers)/392-is-subsequence.mdx
@@ -26,6 +26,22 @@ Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 10^9, and y
 - `0 <= t.length <= 10^4`
 - `s` and `t` consist only of lowercase English letters.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["isSubsequence(s, t)"]) --> G{"len(s) > len(t)?"}
+  G -- yes --> X(["return False"])
+  G -- no --> I["i = 0 into s, j = 0 into t"]
+  I --> W{"i < len(s) and j < len(t)?"}
+  W -- no --> E(["return i == len(s) — every char of s was matched"])
+  W -- yes --> Q{"s[i] == t[j]?"}
+  Q -- yes --> A["i += 1 — consume the match"]
+  Q -- no --> B["j += 1 — always advance t"]
+  A --> B
+  B --> W
+```
+
 ## Solution
 
 ```py
@@ -42,3 +58,7 @@ class Solution:
 
         return i == len(s)
 ```
+
+## Explanation
+
+<YouTube id="99RVfqklbCE" title="Is Subsequence - Leetcode 392" />