Duplicate Zeros

'''
 T.C O(n) two pass, space complexity O(1)
 Logic:
 We can find that what are the elments will be inside array after modification.
 so found the last index of element which will be present in array after first
 loop.

[8,4,5,0,0,0,0,7]   example of k>arr.size()
one edge case, is there gonna be only one zero at end , if only one place empty at last?
'''

class Solution:
    def duplicateZeros(self, arr):
        n = len(arr)
        k = 0
        i = 0

        # First loop: find last valid index to process
        while i < n:
            if arr[i] == 0:
                k += 2
            else:
                k += 1

            if k >= n:
                break
            i += 1

        j = n - 1

        # Special case when last zero fits partially
        if k > n:
            arr[j] = arr[i]
            j -= 1
            i -= 1

        # Second loop: fill from back
        while i >= 0:
            if arr[i] == 0:
                arr[j] = 0
                j -= 1
                arr[j] = 0
                j -= 1
            else:
                arr[j] = arr[i]
                j -= 1
            i -= 1