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31. Add Two Numbers

mediumAsked at Datadog

Add two non-negative integers represented as reversed-order linked lists. Datadog uses this to test carry-propagation logic — the same logic as their atomic-counter rollup across pipeline stages.

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

Source citations

Public interview reports confirming this problem appears in Datadog loops.

  • Glassdoor (2026-Q1)— Datadog phone screen linked-list arithmetic.
  • Blind (2025-11)— Recurring at Datadog NYC.

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 = [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 BigInt, add, rebuild

Decode both lists to BigInts, add, encode back.

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

Tradeoff: Works but signals you avoid the digit-by-digit logic. Interviewers see this as a dodge.

2. Digit-by-digit with carry (optimal)

Walk both lists in lockstep; sum digits + carry; emit (sum % 10), propagate carry. Handle uneven lengths and trailing carry.

Time
O(max(n, m))
Space
O(max(n, m)) output
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;
    tail.next = { val: sum % 10, next: null };
    tail = tail.next;
    carry = Math.floor(sum / 10);
    if (l1) l1 = l1.next;
    if (l2) l2 = l2.next;
  }
  return dummy.next;
}

Tradeoff: Single pass; handles uneven lengths and final carry. Direct carry-propagation pattern Datadog uses in their counter rollup.

Datadog-specific tips

Datadog interviewers grade on the three loop conditions: l1 OR l2 OR carry. Forgetting any one bombs an edge case. They'll follow up with: 'Now do it with digits stored MSB-first' — that's LC 445, requiring stacks or reversal.

Common mistakes

  • Loop condition while (l1 && l2) — misses uneven-length lists.
  • Loop condition while (l1 || l2) — misses trailing carry like [9,9] + [1] = [0,0,1].
  • Forgetting to advance l1/l2 only when they're non-null — crash on null deref.

Follow-up questions

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

  • Add Two Numbers II (LC 445) — digits stored MSB-first.
  • Plus One Linked List (LC 369) — increment by 1.
  • Add Strings (LC 415) — same pattern on strings.

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Output

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FAQ

What about the trailing carry?

If both lists end and carry is 1, you need one more node. That's why the loop condition is l1 || l2 || carry, not just l1 || l2.

Why reverse-order digits?

Carry propagates from least-significant to most-significant. Reverse order means the head IS the least-significant digit, so you can walk forward and propagate carry naturally.

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