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7. Plus One

easyAsked at Vercel

Given an array of digits representing a non-negative integer, add one and return the result as an array. Vercel uses this as a 'do you handle carries cleanly' screen — the same skill you need to bump build numbers or version tags in their deployment graph.

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

Source citations

Public interview reports confirming this problem appears in Vercel loops.

  • Glassdoor (2025-11)— Vercel screen for new grads; carry propagation expected.
  • LeetCode Discuss (2026-Q1)— Mentioned in Vercel interview prep guide.

Problem

You are given a large integer represented as an integer array digits, where each digits[i] is the i-th digit of the integer. The digits are ordered from most significant to least significant. Increment the integer by one and return the resulting array of digits.

Constraints

  • 1 <= digits.length <= 100
  • 0 <= digits[i] <= 9
  • digits does not contain any leading 0s.

Examples

Example 1

Input
digits = [1,2,3]
Output
[1,2,4]

Example 2

Input
digits = [9]
Output
[1,0]

Explanation: 9 + 1 = 10.

Example 3

Input
digits = [9,9,9]
Output
[1,0,0,0]

Approaches

1. Convert to BigInt, add, convert back

Join digits into a string, parse as BigInt, add 1n, split into digits.

Time
O(n)
Space
O(n)
function plusOne(digits) {
  const n = BigInt(digits.join('')) + 1n;
  return String(n).split('').map(Number);
}

Tradeoff: Works but dodges the algorithm. Some interviewers reject this because they want to see the carry logic.

2. In-place carry propagation (optimal)

Walk right-to-left. If digit < 9, increment and return. If digit == 9, set to 0 and continue. If we fall off the front, prepend 1.

Time
O(n)
Space
O(1) when no extra digit; O(n) when [9,9,...,9]
function plusOne(digits) {
  for (let i = digits.length - 1; i >= 0; i--) {
    if (digits[i] < 9) {
      digits[i]++;
      return digits;
    }
    digits[i] = 0;
  }
  return [1, ...digits];
}

Tradeoff: Early return on the first non-9 digit makes the average case O(1) amortized. The only allocation happens when input is all 9s.

Vercel-specific tips

Vercel grades the in-place carry logic and the early-return optimization. Bonus signal: explaining that the only case requiring a new array is when the input is all 9s, and that this generalizes to base-N arithmetic, which they use in build-number versioning. Avoid the BigInt shortcut unless they explicitly say it's allowed.

Common mistakes

  • Forgetting the all-9s case — returns `[0,0,0]` for input `[9,9,9]` instead of `[1,0,0,0]`.
  • Iterating left-to-right — carries propagate right-to-left, this just doesn't work.
  • Mutating digits AND building a new array — wasteful; pick one and stick with it.

Follow-up questions

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

  • Add Two Numbers as linked lists (LC 2) — same carry pattern in a different shape.
  • Multiply Strings (LC 43) — generalized carry-and-shift.
  • Add Strings (LC 415) — same problem with two operands instead of one.

Solve it now

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Output

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FAQ

Why right-to-left?

Carries propagate from the least significant digit upward, exactly the way we add by hand. Going left-to-right would require a second pass to fix the carry chain.

Why not use Array.unshift(1)?

unshift is O(n) on most engines because it shifts every existing element. The spread `[1, ...digits]` is equivalent but reads more cleanly; either works.

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