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31. Add Two Numbers

mediumAsked at Reddit

Add two big integers represented as linked lists (least-significant-digit first). Reddit asks this to test linked-list arithmetic with carry — the same kind of pointer-walking they use to merge cross-shard counter chunks.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Reddit loops.

  • Glassdoor (2026-Q1)— Reddit on-site question for backend / infra roles.
  • Blind (2025-11)— Reported in Reddit comments-team rounds.

Problem

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 the sum as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Constraints

  • The number of nodes in each linked list is in the range [1, 100].
  • 0 <= Node.val <= 9
  • It is guaranteed that the list represents a number that does not have leading zeros.

Examples

Example 1

Input
l1 = [2,4,3], l2 = [5,6,4]
Output
[7,0,8]

Explanation: 342 + 465 = 807.

Example 2

Input
l1 = [0], l2 = [0]
Output
[0]

Example 3

Input
l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output
[8,9,9,9,0,0,0,1]

Approaches

1. Convert to numbers and back

Build the integers, add, then convert back to linked list.

Time
O(n)
Space
O(n)
function addTwoNumbers(l1, l2) {
  const toNum = (l) => { let n = 0n, p = 1n; while (l) { n += BigInt(l.val) * p; p *= 10n; l = l.next; } return n; };
  let sum = toNum(l1) + toNum(l2);
  const dummy = { next: null }; let cur = dummy;
  if (sum === 0n) return { val: 0, next: null };
  while (sum > 0n) { cur.next = { val: Number(sum % 10n), next: null }; cur = cur.next; sum /= 10n; }
  return dummy.next;
}

Tradeoff: Requires BigInt for big lists. Defeats the spirit of the problem.

2. Single-pass with carry (optimal)

Walk both lists simultaneously. At each step: digit sum + carry. New digit = sum % 10, new carry = sum / 10.

Time
O(max(n, m))
Space
O(max(n, m))
function addTwoNumbers(l1, l2) {
  const dummy = { val: 0, next: null };
  let tail = dummy;
  let 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);
    tail.next = { val: sum % 10, next: null };
    tail = tail.next;
    if (l1) l1 = l1.next;
    if (l2) l2 = l2.next;
  }
  return dummy.next;
}

Tradeoff: Linear time, single pass. The 'while (l1 || l2 || carry)' is the critical loop guard.

Reddit-specific tips

Reddit interviewers care about the carry-at-the-end case ([9,9,9] + [1] = [0,0,0,1]). Bonus signal: discuss what happens if the lists are stored most-significant-first (variant LC 445) — you'd reverse or use a stack.

Common mistakes

  • Forgetting carry in the loop condition (drops final carry).
  • Using parseInt — overflows for long lists.
  • Not advancing l1/l2 when one is null.

Follow-up questions

An interviewer at Reddit may pivot to one of these next:

  • Add Two Numbers II (LC 445) — most-significant-first.
  • What if the lists could be negative? Add a sign field.
  • Multiply two big integers represented as linked lists.

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Output

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FAQ

Why a dummy head?

Cleaner than tracking 'first iteration'. The tail always exists.

Why store in reverse order?

Reverse order makes carry propagation work without an extra pass — you start at the least significant digit.

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