{"id":2353,"date":"2026-05-27T18:20:26","date_gmt":"2026-05-27T18:20:26","guid":{"rendered":"https:\/\/tucumandevelopers.com\/index.php\/2026\/05\/27\/stacks-en-entrevistas-tecnicas-3-problemas-resueltos-paso-a-paso\/"},"modified":"2026-05-27T18:20:26","modified_gmt":"2026-05-27T18:20:26","slug":"stacks-en-entrevistas-tecnicas-3-problemas-resueltos-paso-a-paso","status":"publish","type":"post","link":"https:\/\/tucumandevelopers.com\/index.php\/2026\/05\/27\/stacks-en-entrevistas-tecnicas-3-problemas-resueltos-paso-a-paso\/","title":{"rendered":"Stacks en entrevistas t\u00e9cnicas: 3 problemas resueltos paso a paso"},"content":{"rendered":"
\n
<\/div>\n<\/details>\n

<\/a> \u00bfC\u00f3mo lo pensamos? <\/h3>\n

“El \u00faltimo que abri\u00f3 es el primero que debe cerrarse.” Justo eso es lo que un stack hace bien.<\/p>\n

Recorremos el string. Cada apertura va al stack. Cada cierre debe coincidir con el tope. Si al final el stack queda vac\u00edo, todo cerr\u00f3 bien.<\/p>\n

<\/a><\/p>\n

<\/a> Soluci\u00f3n <\/h3>\n
\n
function<\/span> validParentheses<\/span>(<\/span>s<\/span>)<\/span> {<\/span> const<\/span> stack<\/span> =<\/span> [];<\/span> const<\/span> pairs<\/span> =<\/span> {<\/span> \"<\/span>)<\/span>\"<\/span>:<\/span> \"<\/span>(<\/span>\"<\/span>,<\/span> \"<\/span>}<\/span>\"<\/span>:<\/span> \"<\/span>{<\/span>\"<\/span>,<\/span> \"<\/span>]<\/span>\"<\/span>:<\/span> \"<\/span>[<\/span>\"<\/span> };<\/span> for <\/span>(<\/span>const<\/span> char<\/span> of<\/span> s<\/span>)<\/span> {<\/span> if <\/span>(<\/span>char<\/span> ===<\/span> \"<\/span>(<\/span>\"<\/span> ||<\/span> char<\/span> ===<\/span> \"<\/span>{<\/span>\"<\/span> ||<\/span> char<\/span> ===<\/span> \"<\/span>[<\/span>\"<\/span>)<\/span> {<\/span> stack<\/span>.<\/span>push<\/span>(<\/span>char<\/span>);<\/span> }<\/span> else<\/span> if <\/span>(<\/span>stack<\/span>.<\/span>pop<\/span>()<\/span> !==<\/span> pairs<\/span>[<\/span>char<\/span>])<\/span> {<\/span> return<\/span> false<\/span>;<\/span> }<\/span> }<\/span> return<\/span> stack<\/span>.<\/span>length<\/span> ===<\/span> 0<\/span>;<\/span> }<\/span> <\/code><\/pre>\n
\n<\/p><\/div>\n<\/p><\/div>\n
Lo importante:<\/p>\n
    \n
  1.  pairs<\/code> mapea cada cierre con su apertura.<\/li>\n
  2. Aperturas van al stack. Cierres hacen pop<\/code> y validan.<\/li>\n
  3. Si el stack est\u00e1 vac\u00edo al hacer pop<\/code>, devuelve undefined<\/code> y la comparaci\u00f3n falla. C\u00f3modo, as\u00ed no necesitamos un chequeo extra.<\/li>\n
  4. Al final, el stack debe estar vac\u00edo.<\/li>\n<\/ol>\n
    \n
    Complejidad: O(n) tiempo y O(n) espacio.<\/p>\n<\/blockquote>\n
      <\/a> Problema 2: Reverse String (easy) <\/h2>\n
    Adaptaci\u00f3n de “Reverse String” de LeetCode<\/a>. Vamos a invertir una palabra usando stack.<\/p>\n
    Problema: dado un string s<\/code>, devuelve el string invertido.<\/p>\n
    \nEjemplos<\/summary>\n
     <\/p>\n
    \n
    Input: \"stack\" \u2192 \"kcats\" Input: \"hello\" \u2192 \"olleh\" <\/code><\/pre>\n
    \n<\/p><\/div>\n<\/p><\/div>\n<\/details>\n
      <\/a> \u00bfC\u00f3mo lo pensamos? <\/h3>\n
    “Invertir” es la pista. Metes los elementos en orden y los sacas en orden inverso. Eso es exactamente lo que hace un stack.<\/p>\n
    En la vida real usar\u00edas s.split(\"\").reverse().join(\"\")<\/code> y listo. Aqu\u00ed lo hacemos con stack para ver el patr\u00f3n en acci\u00f3n.<\/p>\n
    <\/a><\/p>\n
      <\/a> Soluci\u00f3n <\/h3>\n
    \n
    var<\/span> reverseString<\/span> =<\/span> function <\/span>(<\/span>s<\/span>)<\/span> {<\/span> const<\/span> stack<\/span> =<\/span> [...<\/span>s<\/span>];<\/span> \/\/ crea un stack con los caracteres<\/span> let<\/span> reversed<\/span> =<\/span> \"\"<\/span>;<\/span> while <\/span>(<\/span>stack<\/span>.<\/span>length<\/span> ><\/span> 0<\/span>)<\/span> {<\/span> reversed<\/span> +=<\/span> stack<\/span>.<\/span>pop<\/span>();<\/span> }<\/span> return<\/span> reversed<\/span>;<\/span> };<\/span> <\/code><\/pre>\n
    \n<\/p><\/div>\n<\/p><\/div>\n
    Metemos todos los caracteres al stack y los vamos sacando uno por uno. Como pop<\/code> devuelve el \u00faltimo que entr\u00f3, los caracteres salen al rev\u00e9s.<\/p>\n
    \n
    Complejidad: O(n) tiempo y O(n) espacio.<\/p>\n<\/blockquote>\n
      <\/a> Problema 3: Simplify Path (medium) <\/h2>\n
    “Simplify Path” de LeetCode<\/a>.<\/p>\n
    Problema: dada una ruta absoluta de Unix, convi\u00e9rtela a su forma can\u00f3nica.<\/p>\n
    Reglas:<\/p>\n