Move Zeroes – Solution & Complexity

Q: What is the time and space complexity of Move Zeroes?

The optimal Move Zeroes solution runs in `O(n)` time and `O(1)` space. Use a write pointer to pack non-zero values forward, then fill the rest with zeros.

Q: What is the most efficient way to solve Move Zeroes?

Use a write pointer to pack non-zero values forward, then fill the rest with zeros. This gives `O(n)` time complexity, and is a classic arrays and two pointers problem.

Q: How do you solve Move Zeroes in Python & JavaScript?

We provide a worked Move Zeroes solution in Python & JavaScript with a step-by-step explanation above. You can also open the interactive editor to write and run your own Python solution against the test cases in the browser.

Q: Is Move Zeroes an easy or hard interview question?

Move Zeroes is generally rated **Easy**. It tests arrays and two pointers, which are common coding-interview patterns.

1. Understand the Goal

Move every 0 to the end of the array.
Preserve the relative order of all non-zero values.
The function returns the resulting array.

2. Simple Extra-Array Approach

Collect all non-zero values first.
Count how many zeroes are needed at the end.
This is easy to understand but uses extra space.

def move_zeroes(nums):
    non_zeroes = [num for num in nums if num != 0]
    zero_count = len(nums) - len(non_zeroes)
    return non_zeroes + [0] * zero_count

3. Use a Write Pointer

Keep a pointer for where the next non-zero value should be placed.
Scan the array from left to right.
Every non-zero value is written to the next available position, preserving order.

4. Fill the Remaining Positions with Zeroes

After all non-zero values are written, everything from the write pointer to the end should be 0.
This produces the same array shape without needing a second array.
Finally, return nums.

def move_zeroes(nums):
    write = 0
    for num in nums:
        if num != 0:
            nums[write] = num
            write += 1
    for i in range(write, len(nums)):
        nums[i] = 0
    return nums

5. Final Solution and Complexity

The solution keeps non-zero elements in their original relative order.
Time complexity is O(n).
Space complexity is O(1) besides the input array.

def move_zeroes(nums: list[int]) -> list[int]:
    write = 0
    for num in nums:
        if num != 0:
            nums[write] = num
            write += 1
    for i in range(write, len(nums)):
        nums[i] = 0
    return nums

Move Zeroes – Solution & Complexity

1. Understand the Goal

2. Simple Extra-Array Approach

3. Use a Write Pointer

4. Fill the Remaining Positions with Zeroes

5. Final Solution and Complexity

FAQ

What is the time and space complexity of Move Zeroes?

What is the most efficient way to solve Move Zeroes?

How do you solve Move Zeroes in Python & JavaScript?

Is Move Zeroes an easy or hard interview question?