feat(content): add new problem pages, reorganize docs into (problems) subfolders, and update meta display to page

2026-11-03 13:27:36 +00:00 · 2026-10-05 13:07:56 -04:00
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+---
+title: '1004. Max Consecutive Ones III'
+description: Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Binary Search", slug: "binary-search" }
+  - { name: "Sliding Window", slug: "sliding-window" }
+  - { name: "Prefix Sum", slug: "prefix-sum" }
+issue: "https://github.com/prdlk/leetcode/issues/56"
+sidebar:
+  label: 'Max Consecutive Ones III'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2`
+- Output: `6`
+- Explanation: `[1,1,1,0,0,1,1,1,1,1,1]` Bolded numbers were flipped from `0` to `1`. The longest subarray is underlined.
+
+### Example 2:
+- Input: `nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3`
+- Output: `10`
+- Explanation: `[0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]` Bolded numbers were flipped from `0` to `1`. The longest subarray is underlined.
+
+### Constraints:
+
+- `1 <= nums.length <= 10^5`
+- `nums[i]` is either 0 or 1.
+- `0 <= k <= nums.length`
+
+## Solution
+
+```py
+class Solution:
+    def longestOnes(self, nums: list[int], k: int) -> int:
+        left = 0
+        cur = 0
+        ans = 0
+
+        for right in range(len(nums)):
+            if nums[right] == 0:
+                cur += 1
+            while cur > k:
+                if nums[left] == 0:
+                    cur -= 1
+                left += 1
+
+            ans = max(ans, right - left + 1)
+
+        return ans
+```
@@ -0,0 +1,60 @@
+---
+title: '121. Best Time to Buy and Sell Stock'
+description: You are given an array prices where prices[i] is the price of a given stock on the i^th day
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+video: "1pkOgXD63yU"
+issue: "https://github.com/prdlk/leetcode/issues/50"
+sidebar:
+  label: 'Best Time to Buy and Sell Stock'
+  badge: 'Easy'
+---
+
+### Example 1:
+- Input: `prices = [7,1,5,3,6,4]`
+- Output: `5`
+- Explanation: Buy on day `2` (price = `1`) and sell on day `5` (price = `6`), profit = `6-1 = 5`. Note that buying on day `2` and selling on day `1` is not allowed because you must buy before you sell.
+
+### Example 2:
+- Input: `prices = [7,6,4,3,1]`
+- Output: `0`
+- Explanation: In this case, no transactions are done and the max profit = `0`.
+
+### Constraints:
+
+- `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
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        left = min(prices)
+        for right in range(len(prices)):
+            while curr > left:
+                curr -= prices[left]
+                left += 1
+            ans = max(ans, curr)
+        return ans
+```
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+---
+title: '209. Minimum Size Subarray Sum'
+description: Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose sum is greater than or equal to target. If there is no such subarray, return 0 instead
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Binary Search", slug: "binary-search" }
+  - { name: "Sliding Window", slug: "sliding-window" }
+  - { name: "Prefix Sum", slug: "prefix-sum" }
+issue: "https://github.com/prdlk/leetcode/issues/54"
+sidebar:
+  label: 'Minimum Size Subarray Sum'
+  badge: 'Medium'
+---
+
+::::warning
+If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).
+::::
+
+### Example 1:
+- Input: `target = 7, nums = [2,3,1,2,4,3]`
+- Output: `2`
+- Explanation: The subarray `[4,3]` has the minimal length under the problem constraint.
+
+### Example 2:
+- Input: `target = 4, nums = [1,4,4]`
+- Output: `1`
+
+### Example 3:
+- Input: `target = 11, nums = [1,1,1,1,1,1,1,1]`
+- Output: `0`
+
+### Constraints:
+
+- `1 <= target <= 10^9`
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^4`
+
+## Solution
+
+```py
+class Solution:
+    def minSubArrayLen(self, target: int, nums: list[int]) -> int:
+        window_sum = 0
+        length = len(nums) + 1
+        left = 0
+
+        for right in range(len(nums)):
+            window_sum += nums[right]
+            while window_sum >= target:
+                length = min(length, right - left + 1)
+                window_sum -= nums[left]
+                left += 1
+        if length > len(nums):
+            return 0
+        return length
+```
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+---
+title: '3. Longest Substring Without Repeating Characters'
+description: Given a string s, find the length of the longest substring without duplicate characters
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+  - { name: "Sliding Window", slug: "sliding-window" }
+video: "wiGpQwVHdE0"
+issue: "https://github.com/prdlk/leetcode/issues/51"
+sidebar:
+  label: 'Longest Substring Without Repeating Characters'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `s = "abcabcbb"`
+- Output: `3`
+- Explanation: The answer is "abc", with the length of `3`. Note that "bca" and "cab" are also correct answers.
+
+### Example 2:
+- Input: `s = "bbbbb"`
+- Output: `1`
+- Explanation: The answer is "b", with the length of `1`.
+
+### Example 3:
+- Input: `s = "pwwkew"`
+- Output: `3`
+- Explanation: The answer is "wke", with the length of `3`. Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
+
+### Constraints:
+
+- `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
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        left = 0
+        ans = 0
+        window = set()
+        for right, c in enumerate(s):
+            while c in window:
+                window.remove(s[left])
+                left += 1
+            window.add(c)
+            ans = max(ans, right - left + 1)
+
+        return ans
+```
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+---
+title: '424. Longest Repeating Character Replacement'
+description: You are given a string s and an integer k. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+  - { name: "Sliding Window", slug: "sliding-window" }
+video: "gqXU1UyA8pk"
+issue: "https://github.com/prdlk/leetcode/issues/52"
+sidebar:
+  label: 'Longest Repeating Character Replacement'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `s = "ABAB", k = 2`
+- Output: `4`
+- Explanation: Replace the two 'A's with two 'B's or vice versa.
+
+### Example 2:
+- Input: `s = "AABABBA", k = 1`
+- Output: `4`
+- Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA". The substring "BBBB" has the longest repeating letters, which is `4`. There may exists other ways to achieve this answer too.
+
+### Constraints:
+
+- `1 <= s.length <= 10^5`
+- `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
+class Solution:
+    def characterReplacement(self, s: str, k: int) -> int:
+        count = {}
+        res = 0
+
+        l = 0
+        maxF = 0
+        for r, c in enumerate(s):
+            count[c] = 1 + count.get(c, 0)
+            maxF = max(maxF, count[c])
+
+            while (r - l + 1) - maxF > k:
+                count[s[l]] -= 1
+                l += 1
+
+            res = max(res, r - l + 1)
+        return res
+```
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+---
+title: '567. Permutation in String'
+description: Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "Two Pointers", slug: "two-pointers" }
+  - { name: "String", slug: "string" }
+  - { name: "Sliding Window", slug: "sliding-window" }
+video: "UbyhOgBN834"
+issue: "https://github.com/prdlk/leetcode/issues/53"
+sidebar:
+  label: 'Permutation in String'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `s1 = "ab", s2 = "eidbaooo"`
+- Output: `true`
+- Explanation: `s2` contains one permutation of `s1` ("ba").
+
+### Example 2:
+- Input: `s1 = "ab", s2 = "eidboaoo"`
+- Output: `false`
+
+### Constraints:
+
+- `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
+from collections import Counter
+
+
+class Solution:
+    def checkInclusion(self, s1: str, s2: str) -> bool:
+        k = len(s1)
+        if k > len(s2):
+            return False
+
+        need = Counter(s1)
+        window = Counter()
+
+        for right, c in enumerate(s2):
+            window[c] += 1
+            if right >= k:
+                left_char = s2[right - k]
+                window[left_char] -= 1
+                if window[left_char] == 0:
+                    del window[left_char]
+            if window == need:
+                return True
+
+        return False
+```
@@ -0,0 +1,46 @@
+---
+title: '643. Maximum Average Subarray I'
+description: You are given an integer array nums consisting of n elements, and an integer k
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Sliding Window", slug: "sliding-window" }
+issue: "https://github.com/prdlk/leetcode/issues/55"
+sidebar:
+  label: 'Maximum Average Subarray I'
+  badge: 'Easy'
+---
+
+### Example 1:
+- Input: `nums = [1,12,-5,-6,50,3], k = 4`
+- Output: `12.75000`
+- Explanation: Maximum average is `(12 - 5 - 6 + 50`) / `4 = 51 / 4 = 12.75`
+
+### Example 2:
+- Input: `nums = [5], k = 1`
+- Output: `5.00000`
+
+### Constraints:
+
+- `n == nums.length`
+- `1 <= k <= n <= 10^5`
+- `-10^4 <= nums[i] <= 10^4`
+
+## Solution
+
+```py
+class Solution:
+    def findMaxAverage(self, nums: list[int], k: int) -> float:
+        window_sum = 0
+        for i in range(k):
+            window_sum += nums[i]
+        largest = window_sum
+
+        for right in range(k, len(nums)):
+            left = right - k
+            window_sum -= nums[left]
+            window_sum += nums[right]
+            largest = max(largest, window_sum)
+
+        return largest / k
+```