Playground: Reorganize String

Clase 37 de 41 • Curso de Algoritmos Avanzados: Estructuras de Datos Lineales

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John Eduard Meneses González

student•

My solution!!

Carlos Hernandez

student•

Solución utilizando el patrón 2 apuntadores:

def reorganizeString(text: str) -> str:
  left = 0
  right = 1
  detected_repeated_chars = False
  text = list(text)

  while left < len(text) and right < len(text):
      left_char, right_char = text[left], text[right]
      if left_char != right_char:
          if detected_repeated_chars:
              text[left + 1], text[right] = right_char, left_char
              detected_repeated_chars = False
          left += 1
          right = left + 1
      else:
          detected_repeated_chars = True
          right += 1
  if detected_repeated_chars: return ''

  return "".join(text)

response = reorganizeString("aacab")
print(response)

response = reorganizeString("aaab")
print(response)

Miguel Angel Reyes Moreno

student•

propuesta:

def reorganizeString(text: str) -> str: frequency_characters = {} s_len_half = (len(text) + 1) // 2 for char in text: frequency_characters[char] = frequency_characters.get(char, 0) + 1 if frequency_characters[char] > s_len_half: return "" max_heap = [] for char, freq in frequency_characters.items(): heapq.heappush(max_heap, (-freq, char)) result = [] prev_char = None prev_freq = 0 while max_heap: freq, char = heapq.heappop(max_heap) result.append(char) if prev_freq < 0: heapq.heappush(max_heap, (prev_freq, prev_char)) prev_char = char prev_freq = freq + 1 return "".join(result)

Juan Pablo Reina Gutierrez

student•

Mi solución:

Rafael Estiven Uribe Álvarez

student•

Mi posible solución: