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Gonzalo Diaz
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[lint] markdownlint fixes
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README.md

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installed in your SO.
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- [Install and run with make](#install-and-run-using-make) require runtime tools
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and "make" installed in your SO.
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- [Install and run in Docker](#install-and-running-with-docker-) require Docker and
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docker-compose installed.
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- [Install and run in Docker](#install-and-running-with-docker-) require Docker
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and docker-compose installed.
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- (⭐️)
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[Install and in Docker with make](#install-and-running-with-docker--using-make)
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require docker-compose and make installed.

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

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- $ 1 \leq m, n \leq 30000 $
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- $ 1 \leq $ length of `magazine[i]` and `note[i]` $ \leq 5 $
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- Each word consists of English alphabetic letters (i.e., `a` to `z` and `A` to `Z`).
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- Each word consists of English alphabetic letters (i.e., `a` to `z` and `A` to
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`Z`).
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## Sample Input 0
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docs/hackerrank/interview_preparation_kit/greedy_algorithms/angry-children.md

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## Explanation 2
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Here `k = 2`. `arr' = [2, 2]` or `arr' = [1, 1]` give the minimum unfairness of `0`.
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Here `k = 2`. `arr' = [2, 2]` or `arr' = [1, 1]` give the minimum unfairness
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of `0`.

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

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

docs/projecteuler/problem0002.md

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

docs/projecteuler/problem0020.md

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

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