LeetCode with Corgis and Kittens — Visual Guide in Solving the Add Two Numbers Problem !!!
LeetCode with Corgis Comic GIF🎁 Issue #2 — Visual guide in solving the Add Two Numbers Problem
Hello 👋💻🌎 Tech World,
LeetCode with Corgis and Kittens Problem #2— Add Two Numbers
MAKING LEARNING HOW TO CODE
✧・゚:* CUTE(◕‿◕✿) and INFORMATIVEᕙ(⇀‸↼‶)ᕗ!!!
Hope you enjoy it!
Problem Statement
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contains a single digit. Add the two numbers and return them as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8Explanation: 342 + 465 = 807.
Add Two Numbers Animation
Each node contains a single digit and the digits are stored in reverse order.
Algorithm
Just like on a piece of paper📄, sum two numbers we start to begin by summing the least-significant digits, which is the head of l1 (linked list 1 ) and l2 (linked list 2). Since each digit is in the range of 0…9, summing two digits may “overflow”. For example 5 + 7 = 12. In this case, we set the current digit to 2 and bring over the carry = 1 to the next iteration. carry variable must be either 0 or 1 because the largest possible sum of two digits (including the carry) is 9+9+1=19.
The pseudocode (recipe guide) is as follows:
- Initialize the current node to dummy head of the returning list.
- Initialize carry to 0.
- Initialize p and q to head of l1 and l2 respectively.
- Loop through lists l1 and l2 until you reach both ends.
- Set x to node p’s value. If p has reached the end of l1, set to 0
- Set y to node q’s value. If q has reached the end of l2, set to 0.
- Set sum = x + y + carry.
- Update carry = sum/10.
- Create a new node with the digit value of (sum mod 10) and set it to the current node’s next, then advance the current node to next.
- Advance both p and q.
- Check if carry=1, if so append a new node with digit 11 to the returning list.
- Return dummy head’s next node.
Note: that we use a dummy head to simplify the code. Without a dummy head, you would have to write extra conditional statements to initialize the head’s value.
Take extra caution of the following cases:
The sum could have an extra carry of one at the end, which is easy to forget.
Follow up
What if the digits in the linked list are stored in a non-reversed order? For example:
(3→4→2) + (4→6→5) = 8→0→7
Approach 1: Elementary Math ➕
Java☕
JavaScript💛🍦
Python 🐍
Complexity Analysis
- Time complexity ⏰ : O(max(m,n)) Assume that m and n represent the length of l1 and l2 respectively, the algorithm above iterates at most max(m,n) times.
- Space complexity🚀 : O(max(m,n)) The length of the new list is at most max(m,n)+1.
Please clap if it helps … Thank you for reading ❤ ,
Kodie The Coding Corgi & Bits The A.I.
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