Submission - Edward-TL

solutions/Edward-TL/solution.py

@@ -0,0 +1,48 @@
+from typing import List
+
+
+class Solution:
+
+
+    def duplicate_zeros(self, arr: List[int]):
+        # AQUÍ VA TU SOLUCIÓN
+        zeros = 0
+        original_length = len(arr)  #No '-1' beacues is used only at 'For's, and they use < not <= 
+        new_length = original_length
+
+        if original_length > 0:    #Just in case i have 1 element on the array
+            element = 0
+            while element in range(0, new_length):      #I only need to check the last value that would fit on the new array
+                if arr[element] == 0:
+                    zeros += 1
+                    new_length = original_length - zeros    #And here's how i know if it would fit
+
+                element += 1
+        else:
+            if arr[0] == 0:
+                    zeros += 1
+
+        check_position = original_length - zeros - 1 #Because of arr[0]
+        modified_position = original_length - 1 #Because of arr[0]
+
+        if zeros == 1:      #Imaging that the only 0 is the last, it won't change
+            check_value = arr[modified_position]
+
+            if check_value == 0:
+                zeros = 0
+
+
+        while zeros > 0 and check_position >= 0:
+            check_value = arr[check_position]       #I know I could use the arr, but this was easier for debugging
+
+            if check_value != 0:
+                arr[modified_position] = check_value
+
+            if check_value == 0:
+                arr[modified_position] = check_value #It's 0 but in the new position
+                modified_position -= 1               #Move in position
+                arr[modified_position] = check_value #It's 0 but in the new position
+                zeros -= 1
+
+            check_position -= 1
+            modified_position -= 1