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1. Two Sum

easyAsked at Juniper Networks

Find two indices that add to a target. Juniper uses this as a first-round filter to check whether candidates think in O(n) hash-map terms rather than O(n²) nested loops — the same trade-off that matters when scanning packet headers in fast-path forwarding code.

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

Source citations

Public interview reports confirming this problem appears in Juniper Networks loops.

  • Glassdoor (2026-Q1)— Reported in Juniper SWE phone-screen summaries as a near-universal warm-up problem.
  • Blind (2025-11)— Multiple Juniper threads list Two Sum as the most common first-round coding question.

Problem

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.

Constraints

  • 2 <= nums.length <= 10^4
  • −10^9 <= nums[i] <= 10^9
  • −10^9 <= target <= 10^9
  • Only one valid answer exists.

Examples

Example 1

Input
nums = [2,7,11,15], target = 9
Output
[0,1]

Explanation: nums[0] + nums[1] = 2 + 7 = 9.

Example 2

Input
nums = [3,2,4], target = 6
Output
[1,2]

Explanation: nums[1] + nums[2] = 2 + 4 = 6.

Approaches

1. Brute force — O(n²)

Try every pair. Simple but too slow for large inputs.

Time
O(n²)
Space
O(1)
function twoSum(nums, target) {
  for (let i = 0; i < nums.length; i++) {
    for (let j = i + 1; j < nums.length; j++) {
      if (nums[i] + nums[j] === target) return [i, j];
    }
  }
  return [];
}

Tradeoff: Correct but O(n²). Juniper interviewers will ask you to improve it immediately.

2. Hash map — single pass O(n)

Store each number's index in a map. For every element check if its complement (target - nums[i]) already exists in the map.

Time
O(n)
Space
O(n)
function twoSum(nums, target) {
  const seen = new Map(); // value -> index
  for (let i = 0; i < nums.length; i++) {
    const complement = target - nums[i];
    if (seen.has(complement)) return [seen.get(complement), i];
    seen.set(nums[i], i);
  }
  return [];
}

Tradeoff: O(n) time and O(n) space. This is the expected answer. Explain that storing the complement check avoids the inner loop entirely — the same principle used in routing table lookups with a hash instead of a sequential scan.

Juniper Networks-specific tips

Juniper interviewers like asking what happens when the array is a stream of packets arriving one at a time — this maps perfectly to the single-pass hash map approach where you process each element exactly once. Mention the analogy: fast-path ASIC logic avoids double-scanning packet fields for the same reason. Also be ready to handle duplicate values — clarify whether two copies of the same value at different indices count.

Common mistakes

  • Using the brute-force loop without mentioning its time complexity first.
  • Returning values instead of indices — the problem asks for indices.
  • Checking seen.has(nums[i]) instead of seen.has(complement) — the complement is what you are looking for, not the current number.
  • Not clarifying whether the array contains duplicate values before coding.

Follow-up questions

An interviewer at Juniper Networks may pivot to one of these next:

  • Two Sum II (sorted array, LC 167) — use two pointers instead of a hash map.
  • Three Sum (LC 15) — sort then reduce to Two Sum with two pointers.
  • What if the input is a data stream and you cannot store all values?

Solve it now

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Output

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FAQ

Can I use the same element twice?

No — the problem states you may not. The hash map approach naturally handles this because you insert nums[i] after checking for the complement, so an element cannot match itself.

What if there are multiple valid pairs?

The problem guarantees exactly one solution, so you don't need to handle ties. Clarify this assumption with the interviewer anyway.

Why does Juniper ask Two Sum in phone screens?

It quickly distinguishes candidates who instinctively reach for O(n) hash-map lookups versus those who default to nested loops. In networking code, choosing the right lookup structure is critical for performance.

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