Feature/markdownlint

.github/workflows/markdown-lint.yml

@@ -24,7 +24,8 @@ jobs:
         with:
           node-version: 24.x
       - name: Install dependencies
-        run: npm install -g markdownlint-cli
+        run: >
+          npm install -g --ignore-scripts markdownlint-cli@0.47.0
 
       - name: Lint
         run: >

Dockerfile

@@ -27,8 +27,8 @@ WORKDIR ${WORKDIR}
 
 RUN  apk add --update --no-cache make nodejs npm \
   && apk add --update --no-cache yamllint \
-  && npm install -g --ignore-scripts markdownlint-cli \
-  && npm install -g --ignore-scripts markdownlint-cli prettier
+  && npm install -g --ignore-scripts markdownlint-cli@0.47.0 \
+  && npm install -g --ignore-scripts prettier@3.7.4
 
 # [!TIP] Use a bind-mount to "/app" to override following "copys"
 # for lint and test against "current" sources in this stage

docs/hackerrank/interview_preparation_kit/dictionaries_and_hashmaps/ctci-ransom-note.md

@@ -49,7 +49,8 @@ The third line contains `n` space-separated strings, each `node[i]`.
 
 - $ 1 \leq m, n \leq 30000 $
 - $ 1 \leq $ length of `magazine[i]` and `note[i]` $ \leq 5 $
-- Each word consists of English alphabetic letters (i.e., `a` to `z` and `A` to `Z`).
+- Each word consists of English alphabetic letters (i.e., `a` to `z` and `A`
+  to `Z`).
 
 ## Sample Input 0
 

docs/hackerrank/interview_preparation_kit/graphs/roads_and_libraries-solution-notes.md

@@ -16,7 +16,8 @@ implemented in C++ that are a little more difficult to understand.
 
 Even so, in the solution found in the article (*), it will not work directly by
 copying and pasting the code since the author (apparently on purpose) wrote
-wrong the calculation formula by exchanging the cost variables of libraries and paths.
+wrong the calculation formula by exchanging the cost variables of libraries and
+paths.
 
 ## Solution sources
 

docs/hackerrank/interview_preparation_kit/greedy_algorithms/angry-children.md

@@ -140,4 +140,5 @@ max(1,2,3,4) - min(1,2,3,4) = 4 - 1 = 3
 
 ## Explanation 2
 
-Here `k = 2`. `arr' = [2, 2]` or `arr' = [1, 1]` give the minimum unfairness of `0`.
+Here `k = 2`. `arr' = [2, 2]` or `arr' = [1, 1]` give the minimum unfairness
+of `0`.

docs/hackerrank/interview_preparation_kit/greedy_algorithms/greedy-florist.md

@@ -74,7 +74,8 @@ the original price of each flower.
 
 ## Explanation 0
 
-There are `n = 3` flowers with costs `c = [2, 5, ,6]` and `k = 3` people in the group.
+There are `n = 3` flowers with costs `c = [2, 5, ,6]` and `k = 3` people in the
+group.
 If each person buys one flower,
 the total cost of prices paid is `2 + 5 + 6 = 13` dollars.
 Thus, we print `13` as our answer.

docs/hackerrank/interview_preparation_kit/linked_lists/insert-a-node-into-a-sorted-doubly-linked-list.md

@@ -21,10 +21,11 @@ Complete the sortedInsert function in the editor below.
 
 sortedInsert has two parameters:
 
-- DoublyLinkedListNode pointer head: a reference to the head of a doubly-linked list
+- DoublyLinkedListNode pointer head: a reference to the head of a doubly-linked
+  list
 
 - int data: An integer denoting the value of the `data`  field for the
-DoublyLinkedListNode you must insert into the list.
+  DoublyLinkedListNode you must insert into the list.
 
 ## Returns
 

docs/hackerrank/interview_preparation_kit/trees/binary-search-tree-lowest-common-ancestor.md

@@ -35,7 +35,8 @@ Node `3` is the lowest node which has nodes  and  as descendants.
 ## Function Description
 
 Complete the function lca in the editor below.
-It should return a pointer to the lowest common ancestor node of the two values given.
+It should return a pointer to the lowest common ancestor node of the two values
+given.
 
 lca has the following parameters:
 

docs/hackerrank/projecteuler/euler002.md

@@ -3,7 +3,8 @@
 - Difficulty: #easy
 - Category: #ProjectEuler+
 
-Each new term in the Fibonacci sequence is generated by adding the previous two terms.
+Each new term in the Fibonacci sequence is generated by adding the previous two
+terms.
 By starting with $ 1 $ and $ 2 $, the first $ 10 $ terms will be:
 
 $$ 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 $$

docs/hackerrank/projecteuler/euler003-solution-notes.md

@@ -33,7 +33,8 @@ The first solution, using the algorithm taught in school, is:
 > Using some test entries, quickly broke the solution at all. So, don't use it.
 > This note is just to record the failed idea.
 
-Since by going through and proving the divisibility of a number $ i $ up to $ n $
+Since by going through and proving the divisibility of a number $ i $ up to
+$ n $
 there are also "remainder" numbers that are also divisible by their opposite,
 let's call it $ j $.
 

docs/projecteuler/problem0002.md

@@ -1,6 +1,7 @@
 # [Even Fibonacci numbers](https://projecteuler.net/problem=2)
 
-Each new term in the Fibonacci sequence is generated by adding the previous two terms.
+Each new term in the Fibonacci sequence is generated by adding the previous
+two terms.
 By starting with 1 and 2, the first 10 terms will be:
 
 $ 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... $

docs/projecteuler/problem0020.md

@@ -3,6 +3,7 @@
 $ n! $ means $ n × (n − 1) × ... × 3 × 2 × 1 $
 
 For example, $ 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800 $,
-and the sum of the digits in the number $ 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27 $.
+and the sum of the digits in the number $ 10! $ is
+$ 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27 $.
 
 Find the sum of the digits in the number $ 100! $