LeetCode #283. Move Zeroes

Extra Space solution

Time Complexity: O(n)

First loop: Iterates through all n elements of the input array once
Second loop: Iterates through all n elements to copy back to the original array

Total: O(n) + O(n) = O(n)

Space Complexity: O(n)

Creates a new array newArray of size n (same as input array)
Uses constant extra space for variables i and j
Total: O(n) auxiliary space

-

class Solution {
    public void moveZeroes(int[] nums) {
        int[] newArray = new int[nums.length];
        int j = 0;

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0)
                continue;
            newArray[j] = nums[i];
            j++;
        }  

        for (int i = 0; i < newArray.length; i++) {
            nums[i] = newArray[i];
        }
    }
}

-

In-place solution

Time Complexity: O(n)

First loop: Iterates through all n elements once - O(n)
Second loop: In worst case, fills remaining positions with zeros

Maximum iterations: n - (number of non-zeros)
Still bounded by O(n)

Total: O(n) + O(n) = O(n)

Space Complexity: O(1)

No auxiliary data structures created
Only uses a constant amount of extra variables:

overwrite (integer)
i (loop counter integer)

class Solution {
    public void moveZeroes(int[] nums) {
        int overwrite = 0;

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != 0) {
                nums[overwrite] = nums[i];
                overwrite++;
            }
        }

        while (overwrite < nums.length) {
            nums[overwrite] = 0;
            overwrite++;
        }
    }
}