13. Add Two Numbers
mediumAsked at RampAdd two numbers whose digits are stored in reverse-order linked lists.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each node contains a single digit. Add the two numbers and return the sum as a linked list.
Constraints
Number of nodes in each list is in the range [1, 100]0 <= Node.val <= 9Neither list has leading zeros except the number 0 itself
Examples
Example 1
l1 = [2,4,3], l2 = [5,6,4][7,0,8]Example 2
l1 = [9,9,9,9], l2 = [9,9,9][8,9,9,0,1]Approaches
1. Brute force via integer reconstruction
Walk both lists into BigInts, add, split back into a linked list.
- Time
- O(n)
- Space
- O(n)
function addTwoNumbers(l1, l2) {
const toBig = (n) => { let s = '', p = 10n ** 0n; while (n) { s = n.val + s; n = n.next; } return BigInt(s); };
let sum = (toBig(l1) + toBig(l2)).toString().split('').reverse();
let head = null;
for (const d of sum) head = { val: +d, next: head };
return reverse(head);
}Tradeoff:
2. Single-pass carry
Walk both lists in lockstep, summing nodes with a carry; create a new node per digit.
- Time
- O(max(m, n))
- Space
- O(max(m, n))
function addTwoNumbers(l1, l2) {
const dummy = { val: 0, next: null };
let cur = dummy, carry = 0;
while (l1 || l2 || carry) {
const a = l1 ? l1.val : 0;
const b = l2 ? l2.val : 0;
const sum = a + b + carry;
carry = Math.floor(sum / 10);
cur.next = { val: sum % 10, next: null };
cur = cur.next;
if (l1) l1 = l1.next;
if (l2) l2 = l2.next;
}
return dummy.next;
}Tradeoff:
Ramp-specific tips
Ramp pushes back on the BigInt-conversion shortcut because their corporate card rules engine processes money in minor units; the carry-propagation pattern signals you respect integer-precision boundaries.
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