Approach Summary
Optimal solution for Add Two Numbers II with max(n, m) time complexity and max(n, m) space complexity.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n^2) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | max(n, m) | max(n, m) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n^2)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
Stack<Integer> s1 = new Stack<>();
Stack<Integer> s2 = new Stack<>();
while(l1 != null){
s1.add(l1.val);
l1 = l1.next;
}
while(l2 != null){
s2.add(l2.val);
l2 = l2.next;
}
int carry = 0;
ListNode head = null;
while(!s1.isEmpty() || !s2.isEmpty() || carry > 0){
int sum = carry;
if(!s1.isEmpty()){
sum = sum + s1.pop();
}
if(!s2.isEmpty()){
sum = sum + s2.pop();
}
carry = sum / 10;
ListNode newNode = new ListNode(sum % 10);
newNode.next = head;
head = newNode;
}
return head;
}
}class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
stack<int> s1;
stack<int> s2;
while(l1 != NULL){
s1.push(l1->val);
l1 = l1->next;
}
while(l2 != NULL){
s2.push(l2->val);
l2 = l2->next;
}
int carry = 0;
ListNode* head = NULL;
while(!s1.empty() || !s2.empty() || carry > 0){
int sum = carry;
if(!s1.empty()){
sum = sum + s1.top();
s1.pop();
}
if(!s2.empty()){
sum = sum + s2.top();
s2.pop();
}
carry = sum / 10;
ListNode* newNode = new ListNode(sum % 10);
newNode->next = head;
head = newNode;
}
return head;
}
};class Solution:
def add_two_numbers_ii(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Add Two Numbers II
*/
function addTwoNumbersIi(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: max(n, m)
Space: max(n, m)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
Stack<Integer> s1 = new Stack<>();
Stack<Integer> s2 = new Stack<>();
while(l1 != null){
s1.add(l1.val);
l1 = l1.next;
}
while(l2 != null){
s2.add(l2.val);
l2 = l2.next;
}
int carry = 0;
ListNode head = null;
while(!s1.isEmpty() || !s2.isEmpty() || carry > 0){
int sum = carry;
if(!s1.isEmpty()){
sum = sum + s1.pop();
}
if(!s2.isEmpty()){
sum = sum + s2.pop();
}
carry = sum / 10;
ListNode newNode = new ListNode(sum % 10);
newNode.next = head;
head = newNode;
}
return head;
}
}class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
stack<int> s1;
stack<int> s2;
while(l1 != NULL){
s1.push(l1->val);
l1 = l1->next;
}
while(l2 != NULL){
s2.push(l2->val);
l2 = l2->next;
}
int carry = 0;
ListNode* head = NULL;
while(!s1.empty() || !s2.empty() || carry > 0){
int sum = carry;
if(!s1.empty()){
sum = sum + s1.top();
s1.pop();
}
if(!s2.empty()){
sum = sum + s2.top();
s2.pop();
}
carry = sum / 10;
ListNode* newNode = new ListNode(sum % 10);
newNode->next = head;
head = newNode;
}
return head;
}
};class Solution:
def add_two_numbers_ii(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Add Two Numbers II
*/
function addTwoNumbersIi(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Linked List Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Linked List pattern.
Framework: Linked List • Interview Prep: Tested at FAANG •
Explore all Linked List problems & invariants →
How to Recognize This Pattern
- Common LinkedList interview pattern
- Optimal time: max(n, m)
Complexity Analysis
Time Complexity
max(n, m)
Space Complexity
max(n, m)
Tags
Linked List
Frequently Asked Questions
What is the optimal time and space complexity for Add Two Numbers II?
The optimal solution for Add Two Numbers II achieves max(n, m) runtime complexity and requires max(n, m) auxiliary space. Time Complexity is max(n, m) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is max(n, m) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Add Two Numbers II requires the Linked List pattern?
In technical interviews, look for key problem signals: Common LinkedList interview pattern; Optimal time: max(n, m). Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Linked List strategy.
What critical edge cases must be handled in Add Two Numbers II?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Add Two Numbers II?
Add Two Numbers II is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Google, Microsoft.