Solution for Double or One Thing in linear time.

solutions/double-or-one-thing/double-or-one.py

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+# Constants:
+# There are many ways to do this
+# But I find this easiest to visualize
+ABC = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
+
+# Functions:
+# We want the first function to output a list of tuples like so: 
+# PEEL --> [(P, 1), (E, 2), (L, 1)]
+def preprocessing(w):
+  pre_word = []
+  letter_list = list(w)
+
+  for i, letter in enumerate(letter_list):
+    happens = 1
+    j = i + 1
+
+    # First part of the statement makes sure we don't go out of range
+    # Second part compares the current letter with the following one
+    while j < len(letter_list) and letter == letter_list[j]:
+      happens += 1
+      letter_list.pop(j)
+
+    # Put the values into a tuple and attach them to a list
+    dupe_tuple = (letter, happens)
+    pre_word.append(dupe_tuple)
+
+  return pre_word
+
+
+#This function returns the lexicographically smallest word:
+def write(ww):
+  duped_chars = []
+  # For every letter except the final one
+  for i in range(len(ww) - 1):
+    tup = ww[i]
+
+    #If a letter is earlier in the alphabet...
+    if ABC.index(tup[0]) < ABC.index(ww[i+1][0]): 
+      times = tup[1] 
+      while times > 0:
+        # ...duplicate it
+        duped_chars.append(tup[0]) 
+        duped_chars.append(tup[0])
+        times -= 1
+
+    # Otherwise just append it as many times as it originally appears
+    else: 
+      times = tup[1]
+      while times > 0:
+        duped_chars.append(tup[0])
+        times -= 1
+
+  # The final letter is added on its own, it is never duplicated
+  times = ww[-1][1] 
+  while times > 0:
+    duped_chars.append(ww[-1][0])
+    times -= 1
+
+  return "".join(duped_chars)    
+
+
+tests = int(input())
+case = 1
+
+while tests > 0:
+    #Call functions here:
+    current_word = input()
+    working_word = preprocessing(current_word)
+    res = write(working_word)
+
+    # This is just a saving throw in case a weird input glitched the code
+    # It makes sure the result doesn't have less letters than the original word
+    if len(res) < len(current_word):
+        res = current_word
+
+    print(f"Case #{case}: {res}")
+
+    tests -= 1 
+    case += 1