LeetCode: Arrays 101: Introduction – Solutions


LeetCode is a good place to practice programming by solving problems of all levels!

After you solve a problem, you’ll get access to all submitted solutions, sorted by time and memory usage. That’s a nice way to compare your code to others and see what you did well and where you can improve.

Here’s my solutions for the Arrays 101: Introduction card:

Max Consecutive Ones


class Solution:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
        max_counter = 0
        counter = 0
        for i in nums:
            if i == 1:
                counter += 1
                if counter > max_counter:
                    max_counter = counter
                counter = 0
        if counter > max_counter:
            max_counter = counter
        return max_counter

Find Numbers with Even Number of Digits


class Solution:

    def is_even(value):
        return len(str(value)) % 2 == 0

    def findNumbers(self, nums: List[int]) -> int:
        return_value = 0
        for number in nums:
            if self.is_even(number):
                return_value += 1
        return return_value

Squares of a Sorted Array


When writing code like this be careful because you’re altering the A list in place.

class Solution:
    def sortedSquares(self, A: List[int]) -> List[int]:
        for index, num in enumerate(A):
            A[index] = num ** 2
        return sorted(A)

Thanks for reading!

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LeetCode: Add Two Numbers Python 3 iterative solution

[Problem Link]


Here’s my solution for the add two numbers problem on LeetCode.


Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

When doing problems on websites such as LeetCode, I like to code in my own IDE, that’s why you’ll see some additional helper functions besides the solution code.

class Solution(object):
    def _get_append_func(self, root):
        _root = root
        def __add_to_list(value):
            nonlocal _root
            _root.next = ListNode(value)
            _root = _root.next
        return __add_to_list

    def _compute(self, result, base=10):
        carry = result // base
        rest = result % base
        return carry, rest

    def addTwoNumbers(self, l1, l2):
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        root_node = None
        root_append_func = None

        term_1 = l1
        term_2 = l2
        saved_carry = 0
        while True:
            # end of expression ; add cary if any
            if term_1 is None and term_2 is None:
                if saved_carry > 0:

            # We haven't reached the end, we assume that whatever term is left is 0
            if term_1 is None:
                term_1 = ListNode(0)
            if term_2 is None:
                term_2 = ListNode(0)

            # add the two terms
            term_sum_carry, term_sum_no_carry = self._compute(term_1.val + term_2.val)

            # add sum to current carry
            next_carry, total_with_carry = self._compute(term_sum_no_carry + saved_carry)

            # save carry
            saved_carry = term_sum_carry + next_carry

            # save total
            if not root_node:
                root_node = ListNode(total_with_carry)
                root_append_func = self._get_append_func(root_node)

            # move on
            term_1 = term_1.next
            term_2 = term_2.next

        return root_node

My solution isn’t that great because of the _add_to_list function which iterates the root_node to append the next value. I have no idea how to optimize that right now.


I’ve learnt about nonlocal and found a way to slightly optimize my code:

    def _get_append_func(self, root):
        _root = root
        def __add_to_list(value):
            nonlocal _root
            _root.next = ListNode(value)
            _root = _root.next
        return __add_to_list

Thanks for reading!