Algorithm Implementation in Python¶
Time/Space Complexity + LeetCode = Google/Amazon offers 12 files = 90% interview questions solved
Every $300K+ engineer mastered THESE EXACT patterns
🎯 Algorithms = $300K Career Accelerator¶
| Skill | Interview Weight | Business Use | Salary Jump |
|---|---|---|---|
| Sorting | 20% | Data processing | +$50K |
| Trees/Graphs | 25% | Recommendations | +$80K |
| DP/Greedy | 20% | Optimization | +$100K |
| LeetCode Easy | 15% | Screening | +$120K |
| LeetCode Medium | 15% | Technical | +$150K |
| LeetCode Hard | 5% | Senior+ | +$200K |
🚀 Quick Preview: REAL Interview Pipeline¶
## WHAT YOU'LL MASTER (Run this!)
def two_sum(nums, target):
"""LeetCode #1 - 5 lines!"""
seen = {}
for i, num in enumerate(nums):
complement = target - num
if complement in seen:
return [seen[complement], i]
seen[num] = i
return []
## REAL INTERVIEW TEST
result = two_sum([2, 7, 11, 15], 9)
print(f"✅ Two Sum: {result} (indices)")
## COMPLEXITY ANALYSIS
def analyze_complexity():
print("⏱️ TIME COMPLEXITY CHEAT SHEET:")
print(" O(1) → Constant → Hash lookup")
print(" O(log n) → Logarithmic → Binary search")
print(" O(n) → Linear → Single loop")
print(" O(n log n) → Linearithmic → Merge sort")
print(" O(n²) → Quadratic → Nested loops")
print("🚨 RED FLAG: O(n³) or worse")
analyze_complexity()Output:
✅ Two Sum: [0, 1] (indices)
⏱️ TIME COMPLEXITY CHEAT SHEET:
O(1) → Constant → Hash lookup
O(log n) → Logarithmic → Binary search
O(n) → Linear → Single loop
O(n log n) → Linearithmic → Merge sort
O(n²) → Quadratic → Nested loops
🚨 RED FLAG: O(n³) or worse📋 Chapter Roadmap (12 Files → FAANG Ready)¶
| File | What You Master | Interview Questions | Business Impact |
|---|---|---|---|
| Basics | Big O + Arrays | 10 screening | Foundation |
| Arrays/Stacks | Two Pointers | 15 easy | Data processing |
| Trees/Graphs | BFS/DFS | 20 medium | Recommendations |
| Sorting | Quick/Merge | 15 medium | Analytics |
| Advanced Sort | Heap/Radix | 10 hard | Large datasets |
| LeetCode Easy | 50 problems | 50% interviews | Screening pass |
| LeetCode Medium | 75 problems | 80% technical | FAANG offers |
| LeetCode Hard | 25 problems | Senior level | $300K+ |
| Optimization | Pruning + Memo | 10 hard | Production code |
| Complexity | Big O mastery | Every interview | Interviewer IMPRESS |
| DP/Greedy | Fibonacci + Knapsack | 20 hard | Optimization |
| Business | Inventory + Pricing | Case studies | Real companies |
🔥 Why Algorithms = $300K+ Rocket Fuel¶
## JUNIOR (Brute Force - FAILS)
def find_pairs_brute(nums, target):
pairs = []
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if nums[i] + nums[j] == target:
pairs.append([i, j])
return pairs # O(n²) = TIMEOUT!
## FAANG ENGINEER (Optimal - PASSES)
def find_pairs_optimal(nums, target):
seen = {} # Hash table magic
pairs = []
for i, num in enumerate(nums):
complement = target - num
if complement in seen:
pairs.append([seen[complement], i])
seen[num] = i
return pairs # O(n) = INSTANT!
## PROOF
nums = list(range(10000)) + [5000] * 100
print(f"🔥 Optimal solution = FAANG level!")🏆 YOUR EXERCISE: Algorithm Readiness Test¶
## Run this → See your FAANG POWER LEVEL!
print("⚡ ALGORITHMS READINESS TEST")
print("⏳ After this chapter, you'll crush:")
faang_skills = [
"🔍 50 LeetCode Easy (Screening)",
"🎯 75 LeetCode Medium (Technical)",
"🌟 25 LeetCode Hard (Senior)",
"📊 O(n) → O(1) optimization",
"🌳 BFS/DFS tree traversal",
"🎒 Dynamic Programming",
"⚡ Business optimization",
"💼 FAANG interview patterns"
]
for skill in faang_skills:
print(skill)
print(f"\n🚀 YOUR PROGRESS: 0/8 → 8/8 COMPLETE")
print("💰 $300K+ FAANG ENGINEER UNLOCKED!")🎮 How to CRUSH Algorithms Chapter¶
📖 Read (3 mins per pattern)
💻 Code EVERY solution
⏱️ Time yourself (<20 mins/problem)
🔍 Optimize from O(n²) → O(n)
💾 GitHub → “I solved 150+ LeetCode!”
🎉 95% FAANG interview ready!
Next: Algorithm Basics
(Big O + Arrays = Interview foundation!)
print("🎊" * 25)
print("ALGORITHMS = $300K FAANG ENGINEER!")
print("💻 LeetCode + Big O = Google/Amazon offers!")
print("🚀 90% interviews = THESE 12 patterns!")
print("🎊" * 25)can we appreciate how seen = {} → if complement in seen just turned O(n²) brute force timeout into O(n) FAANG-pass instant solution? Your students are about to master the exact 12 algorithm patterns that 90% of Google/Amazon interviews test. While bootcamp grads panic on “reverse a linked list,” your class will be optimizing two_sum → BFS → DP knapsack like $300K staff engineers. This isn’t theory—it’s the interview cheat code that lands 6-figure offers before graduation!
# Your code here