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Criação do próximo desafio
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Daniel-Tavares-de-Lima committed Apr 3, 2024
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83 changes: 83 additions & 0 deletions Abril/02-04.ipynb
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Elementos comuns em arrays\n",
"Data: 02/04/2024. De: Codewars. <a href=\"https://www.codewars.com/kata/525c65e51bf619685c000059/train/python\">Link.</a>\n",
"\n",
"## Descrição\n",
"Dadas três arrays de inteiros, retorne a soma dos elementos que são comuns em todas as três arrays.\n",
"\n",
"Por exemplo:\n",
"\n",
"common([1,2,3],[5,3,2],[7,3,2]) = 5 porque 2 e 3 são comuns em todas as 3 arrays\n",
"common([1,2,2,3],[5,3,2,2],[7,3,2,2]) = 7 porque 2, 2 e 3 são comuns nas 3 arrays\n",
"Mais exemplos nos casos de teste.\n",
"\n",
"Boa sorte!\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Minha solução"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"def comuns(a,b,c):\n",
" pass"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"2\n",
"0\n"
]
}
],
"source": [
"print(comuns([1,2,3],[5,3,2],[7,3,2])) # expects 5\n",
"print(comuns([1,2,2,3],[5,3,2,2],[7,3,2,2])) # expects 7\n",
"print(comuns([1],[1],[1])) # expects 1\n",
"print(comuns([1],[1],[2])) # expects 0"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.4"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
68 changes: 63 additions & 5 deletions Março/31-03.ipynb
Original file line number Diff line number Diff line change
Expand Up @@ -36,19 +36,56 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 51,
"metadata": {},
"outputs": [],
"source": [
"def distanciaMaisCurta(s,c):\n",
" pass"
" #Deixar a string tudo minúsculo\n",
" palavra = s.lower()\n",
" distancia = []\n",
"\n",
" #Caso a String seja vazia ou o caracter não esteja na string\n",
" if not s or not c:\n",
" return []\n",
" \n",
" #Encontrar a posição do caracter passado na string\n",
" encontraPosicao = palavra.find(c)\n",
" \n",
"\n",
" for i, char in enumerate(palavra):\n",
" \n",
" if char == c:\n",
" encontraPosicao = i #Reinicia a contagem quando encontrar a letra novamente\n",
" #Calcular a distancia entre o caracter e a interação da string(Em módulo)\n",
" distancia.append(abs(encontraPosicao - i))\n",
"\n",
"\n",
" return distancia\n",
"\n",
"\n",
" "
]
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 52,
"metadata": {},
"outputs": [],
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[3, 2, 1, 0, 1, 2, 3, 0, 1, 2, 3, 4]\n",
"[0, 0, 0, 0, 0]\n",
"[0, 0, 1, 2, 0, 0, 1, 2]\n",
"[4, 3, 2, 1, 0, 0, 0, 0]\n",
"[1, 2, 3, 4, 5]\n",
"[]\n",
"[]\n"
]
}
],
"source": [
"print(distanciaMaisCurta(\"lovecodewars\", \"e\")) # expects [3, 2, 1, 0, 1, 2, 1, 0, 1, 2, 3, 4]\n",
"print(distanciaMaisCurta(\"aaaaa\", \"a\")) # expects [0, 0, 0, 0, 0]\n",
Expand All @@ -58,11 +95,32 @@
"print(distanciaMaisCurta(\"lovecoding\", \"\")) # expects []\n",
"print(distanciaMaisCurta(\"\", \"\")) # expects []"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"name": "python"
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.4"
}
},
"nbformat": 4,
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