Move all zeroes to end of array using List Comprehension in Python

Given an array of random numbers, Push all the zeros of a given array to the end of the array. For example, if the given arrays is {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0}, it should be changed to {1, 9, 8, 4, 2, 7, 6, 0, 0, 0, 0}. The order of all other elements should be same. Expected time complexity is O(n) and extra space is O(1). Examples:

Input :  arr = [1, 2, 0, 4, 3, 0, 5, 0]
Output : arr = [1, 2, 4, 3, 5, 0, 0, 0]

Input : arr  = [1, 2, 0, 0, 0, 3, 6]
Output : arr = [1, 2, 3, 6, 0, 0, 0]

We have existing solution for this problem please refer Move all zeroes to end of array link. We will solve this problem in python using List Comprehension in a single line of code.

Implementation:

Python

# Function to append all zeros at the end
# of array
def moveZeros(arr):
     
    # first expression returns a list of
    # all non zero elements in arr in the
    # same order they were inserted into arr
    # second expression returns a list of
    # zeros present in arr
    return [nonZero for nonZero in arr if nonZero!=0] + \
        [Zero for Zero in arr if Zero==0]
 
# Driver function
if __name__ == "__main__":
    arr = [1, 2, 0, 4, 3, 0, 5, 0]
    print (moveZeros(arr))

Output

[1, 2, 4, 3, 5, 0, 0, 0]

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Approach#2: Using while loop

This approach moves all zeroes in an input array to the end of the array while keeping the non-zero elements in their original order. It uses two pointers to keep track of the positions of the zeroes and non-zeroes in the array, and swaps them as necessary.

Algorithm

1. Initialize two pointers i and j to 0.
2. Traverse the array with j and for each non-zero element, swap arr[i] and arr[j] and increment i.
3. Return the resulting array.

Python3

def move_zeroes(arr):
    i, j = 0, 0
    while j < len(arr):
        if arr[j] != 0:
            arr[i], arr[j] = arr[j], arr[i]
            i += 1
        j += 1
    return arr
arr=[1, 2, 0, 4, 3, 0, 5, 0]
print(move_zeroes(arr))