"""Leetcode 27. Remove Element
Easy

URL: https://leetcode.com/problems/remove-element/

Given an array nums and a value val, remove all instances of that value in-place
and return the new length.

Do not allocate extra space for another array, you must do this by modifying
the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond
the new length.

Example 1:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
It doesn't matter what you leave beyond the returned length.

Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums
containing 0, 1, 3, 0, and 4.
Note that the order of those five elements can be arbitrary.
It doesn't matter what values are set beyond the returned length.

Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to
the input array will be known to the caller as well.

Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}
"""

class SolutionPointer(object):
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        # One pointer method: Track new nums's pos, starting from 0.
        pos = 0
        for n in nums:
            if n != val:
                # If this number is not val, put it in pos. Increment pos.
                nums[pos] = n
                pos += 1
        return pos


def main():
    # Output: 2.
    nums = [3,2,2,3]
    val = 3
    print SolutionPointer().removeElement(nums, val)

    # Output: 5.
    nums = [0,1,2,2,3,0,4,2]
    val = 2
    print SolutionPointer().removeElement(nums, val)

    # Output: 4
    nums = [0,4,4,0,4,4,4,0,2]
    val = 4
    print SolutionPointer().removeElement(nums, val)


if __name__ == '__main__':
    main()

"""Leetcode 27. Remove Element
Easy

URL: https://leetcode.com/problems/remove-element/

Given an array nums and a value val, remove all instances of that value in-place
and return the new length.

Do not allocate extra space for another array, you must do this by modifying
the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond
the new length.

Example 1:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
It doesn't matter what you leave beyond the returned length.

Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums
containing 0, 1, 3, 0, and 4.
Note that the order of those five elements can be arbitrary.
It doesn't matter what values are set beyond the returned length.

Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to
the input array will be known to the caller as well.

Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}
"""

class SolutionPointer(object):
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int

        Time complexity: O(n).
        Space complexity: O(1).
        """
        # One pointer method: Track new nums's pos, starting from 0.
        pos = 0
        for n in nums:
            if n != val:
                # If this number is not val, put it in pos. Increment pos.
                nums[pos] = n
                pos += 1
        return pos


def main():
    # Output: 2.
    nums = [3,2,2,3]
    val = 3
    print SolutionPointer().removeElement(nums, val)

    # Output: 5.
    nums = [0,1,2,2,3,0,4,2]
    val = 2
    print SolutionPointer().removeElement(nums, val)

    # Output: 4
    nums = [0,4,4,0,4,4,4,0,2]
    val = 4
    print SolutionPointer().removeElement(nums, val)


if __name__ == '__main__':
    main()

Add time/space complexity