How to Insert Into A List With Slicing In Python

Used to solve Leet Code’s Duplicate Zeroes Puzzle

The aim of this page📝 is to utilize Python slicing to insert a data structure into a list at a given index. Slicing indexes operate as a half-open range — the first item is included, the last is not:

>>> l = [1,2,3,4]
>>> l[0:1]
[1]

• to insert at a given point, start pointing to the item you want the new structure to start with. then, use the same index value after
• e.g. [0:0] to insert a sublist of a random length in the very beginning of a list
• the identical values in ([0:0]) are not specifying the span, they only point to the point of entry
• in a stupid example below I want to add 666 at the beginning of the list [1,2,3,4]

>>> l = [1,2,3,4]
>>> l[0:0] = [6,6,6,]
>>> l
[6, 6, 6, 1, 2, 3, 4]
>>>

EXAMPLE: LEET CODE’S DUPLICATE ZEROS PUZZLE

• I’ve used the technique to easily solve https://leetcode.com/problems/duplicate-zeros/ which seems rather challenging in an official solution
• It goes like this: Given a fixed-length integer array arr, duplicate each occurrence of zero, shifting the remaining elements to the right.
• Note that elements beyond the length of the original array are not written.
• Do the above modifications to the input array in place and do not return anything.

# EXAMPLE
Input: arr = [1,0,2,3,0,4,5,0]
Output: [1,0,0,2,3,0,0,4]
Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]

• I only insert at the point where the zero is found with inserting with slices
• Then, I remove the last item by with pop()
• Finally, I move the i pointer by two to reflect the insertion as we're mutating a collection that's also being iterated upon

# SOLUTION
def duplicateZeros(self, arr: List[int]) -> None:
  i = 0
  while i < len(arr):
    if arr[i] == 0:
      arr[i:i] = [0]
      arr.pop()
      i += 2
    else:
      i += 1