🏷️ Category: HashMap / Algorithms | 🎯 Difficulty:
Easy| 🔥 Frequency: ★★★★★
📋 Problem Statement
Given a list of integers and a target, return the indices of two numbers that add up to target using the concept of Two Sum.
📥 Input → nums = [2, 7, 11, 15], target = 9
📤 Output → [0, 1] # 2 + 7 = 9
📥 Input → nums = [3, 2, 4], target = 6
📤 Output → [1, 2] # 2 + 4 = 6
📥 Input → nums = [3, 3], target = 6
📤 Output → [0, 1]
🧠 What the Interviewer is Testing
| Concept | Why It Matters |
|---|---|
| HashMap usage | Core data structure skill |
| O(n) thinking | Optimization over brute force |
| Complement logic | Mathematical reasoning |
⚠️ Warning: Never start with O(n²) brute force. Go straight to the hashmap solution.
💡 Approach
a + b = target → b = target - a (complement)
For every number, check if its complement already exists in the hashmap.
If yes → found the pair. If no → store current number and index.
✅ Optimal Solution
def two_sum(nums: list, target: int) -> list:
seen = {} # { number: index }
for i, num in enumerate(nums):
complement = target - num
if complement in seen:
return [seen[complement], i]
seen[num] = i
return []
# ── Test Cases ─────────────────────────────────
print(two_sum([2, 7, 11, 15], 9)) # ✅ [0, 1]
print(two_sum([3, 2, 4], 6)) # ✅ [1, 2]
print(two_sum([3, 3], 6)) # ✅ [0, 1]
🔍 Step-by-Step Walkthrough
nums = [2, 7, 11, 15], target = 9:
| Step | i | num | complement | seen | Action |
|---|---|---|---|---|---|
| 1 | 0 | 2 | 7 | {} | Store 2→0 |
| 2 | 1 | 7 | 2 | {2:0} | ✅ Found! Return [0,1] |
📊 Complexity Analysis
┌──────────────────┬────────────┬──────────────────────────┐
│ Approach │ Time │ Space │
├──────────────────┼────────────┼──────────────────────────┤
│ Brute Force │ O(n²) │ O(1) │
│ HashMap ✅ │ O(n) │ O(n) │
└──────────────────┴────────────┴──────────────────────────┘
🎤 How to Explain in the Interview
“I’ll use a hashmap. For each number I calculate its complement — target minus current number. If that complement already exists in the map, I return both indices. Otherwise I store the current number. This gives O(n) time at the cost of O(n) space.”
“I’ll use a hashmap. For each number I calculate its complement — target minus current number. If that complement already exists in the map, I return both indices. Otherwise I store the current number. This gives O(n) time at the cost of O(n) space.”
🔁 Common Follow-up Questions
Q1 → Multiple valid pairs?
results = []
for i, num in enumerate(nums):
complement = target - num
if complement in seen:
results.append([seen[complement], i])
seen[num] = i
return results
Q2 → Numbers can be negative?
Same solution works — hashmap handles negatives fine.
📁 Set 1 of 4 — Python Interview Prep Series — devprepguide.com