Feature/markdownlint

.github/workflows/markdown-lint.yml

@@ -32,7 +32,12 @@ jobs:
           node-version: ${{ matrix.node-version }}
 
       - name: Install dependencies
-        run: npm install -g markdownlint-cli
+        run: >
+          npm install -g --ignore-scripts markdownlint-cli@0.47.0
+
+      - name: Test dependencies
+        run: |
+          markdownlint --version
 
       - name: Lint
         run: >

Dockerfile

@@ -17,7 +17,7 @@ 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@0.47.0
 
 # [!TIP] Use a bind-mount to "/app" to override following "copys"
 # for lint and test against "current" sources in this stage

docs/hackerrank/implementation/countApplesAndOranges.md

@@ -19,8 +19,9 @@ In the diagram below:
   fell $ d $ units to the tree's left, and a positive value of $ d $ means it falls
   $ d $ units to the tree's right.
 
-Given the value of $ d $ for $ m $ apples and $ n $ oranges, determine how many apples
-and oranges will fall on Sam's house (i.e., in the inclusive range $ [s, t] $)?
+Given the value of $ d $ for $ m $ apples and $ n $ oranges, determine how many
+apples and oranges will fall on Sam's house (i.e., in the inclusive range
+$ [s, t] $)?
 
 For example, Sam's house is between $ s = 7 $ and $ t = 10 $. The apple tree is
 located at $ a = 4 $ and the orange at $ b = 12 $. There are $ m = 3 $ apples

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/dictionaries_and_hashmaps/sherlock_and_anagrams-solution-notes.md

@@ -27,4 +27,5 @@ in scientific notation, losing precision.
 This loss of precision can result in an erroneous result
 in the final calculation of permutations.
 
-To avoid this problem, it is necessary to introduce large number handling using BigInt.
+To avoid this problem, it is necessary to introduce large number handling using
+BigInt.

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

@@ -141,4 +141,5 @@ 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/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

@@ -33,9 +33,9 @@ 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 $
-there are also "remainder" numbers that are also divisible by their opposite,
-let's call it $ j $.
+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 $.
 
 At first it seemed attractive to test numbers $ i $ up to half of $ n $ then
 test whether $ i $ or $ j $ are prime. 2 problems arise:

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! $