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Algorithms · Final Project: Algorithm Design Challenge
Use this course-level map to place “Final Project: Algorithm Design Challenge” in the larger system. It complements—not replaces—a topic-specific worked diagram.
You're a developer at a tech company. Your team lead asks: "Before we deploy this data pipeline, can you benchmark the core algorithms and show us which ones are worth using at scale?"
Your job is to build a Performance Analysis Tool a program that:
This is not a toy exercise. Every component maps directly to skills used in production systems.
We'll implement and time five classic sorting algorithms on three data patterns: random, already sorted, and reverse sorted.
import time
import random
# --- Sorting Implementations ---
def bubble_sort(arr):
a = arr[:]
n = len(a)
for i in range(n):
for j in range(n - i - 1):
if a[j] > a[j+1]:
a[j], a[j+1] = a[j+1], a[j]
return a
def selection_sort(arr):
a = arr[:]
n = len(a)
for i in range(n):
min_idx = i
for j in range(i+1, n):
if a[j] < a[min_idx]:
min_idx = j
a[i], a[min_idx] = a[min_idx], a[i]
return a
def insertion_sort(arr):
a = arr[:]
for i in range(1, len(a)):
key = a[i]
j = i - 1
while j >= 0 and a[j] > key:
a[j+1] = a[j]
j -= 1
a[j+1] = key
return a
def merge_sort(arr):
if len(arr) <= 1:
return arr[:]
mid = len(arr) // 2
left = merge_sort(arr[:mid])
right = merge_sort(arr[mid:])
return merge(left, right)
def merge(left, right):
result = []
i = j = 0
while i < len(left) and j < len(right):
if left[i] <= right[j]:
result.append(left[i]); i += 1
else:
result.append(right[j]); j += 1
return result + left[i:] + right[j:]
def quick_sort(arr):
if len(arr) <= 1:
return arr[:]
pivot = arr[len(arr) // 2]
left = [x for x in arr if x < pivot]
mid = [x for x in arr if x == pivot]
right = [x for x in arr if x > pivot]
return quick_sort(left) + mid + quick_sort(right)
# --- Benchmark Function ---
def benchmark_sort(func, data, label):
start = time.perf_counter()
func(data)
elapsed = (time.perf_counter() - start) * 1000 # milliseconds
return elapsed
def run_sort_benchmarks(n=1000):
random_data = [random.randint(1, 10000) for _ in range(n)]
sorted_data = list(range(n))
reverse_data = list(range(n, 0, -1))
algorithms = [
("Bubble Sort ", bubble_sort),
("Selection Sort", selection_sort),
("Insertion Sort", insertion_sort),
("Merge Sort ", merge_sort),
("Quick Sort ", quick_sort),
]
print(f"\n{'='*62}")
print(f" SORTING BENCHMARK (n = {n} elements)")
print(f"{'='*62}")
print(f" {'Algorithm':<16} {'Random':>10} {'Sorted':>10} {'Reverse':>10}")
print(f" {'-'*14} {'-'*8} {'-'*8} {'-'*8}")
for name, func in algorithms:
t_rand = benchmark_sort(func, random_data, "random")
t_sort = benchmark_sort(func, sorted_data, "sorted")
t_rev = benchmark_sort(func, reverse_data, "reverse")
print(f" {name} {t_rand:>8.2f}ms {t_sort:>8.2f}ms {t_rev:>8.2f}ms")
print(f"{'='*62}")import bisect
def run_search_benchmarks(n=100000):
data = sorted(random.randint(1, 1000000) for _ in range(n))
target = data[n // 2] # Target exists in the middle
# Linear search
start = time.perf_counter()
for val in data:
if val == target:
break
t_linear = (time.perf_counter() - start) * 1000
# Binary search (manual)
def binary_search(arr, t):
lo, hi = 0, len(arr) - 1
while lo <= hi:
mid = (lo + hi) // 2
if arr[mid] == t: return mid
elif arr[mid] < t: lo = mid + 1
else: hi = mid - 1
return -1
start = time.perf_counter()
binary_search(data, target)
t_binary = (time.perf_counter() - start) * 1000
# bisect (built-in optimized)
start = time.perf_counter()
bisect.bisect_left(data, target)
t_bisect = (time.perf_counter() - start) * 1000
print(f"\n{'='*62}")
print(f" SEARCH COMPARISON (n = {n:,} sorted elements)")
print(f"{'='*62}")
print(f" {'Method':<20} {'Time':>12} {'Comparisons':>15}")
print(f" {'-'*18} {'-'*10} {'-'*13}")
import math
print(f" {'Linear Search':<20} {t_linear:>10.4f}ms {n//2:>13,} (avg)")
print(f" {'Binary Search':<20} {t_binary:>10.4f}ms {math.ceil(math.log2(n)):>13} (max)")
print(f" {'bisect (built-in)':<20} {t_bisect:>10.4f}ms {math.ceil(math.log2(n)):>13} (max)")
print(f"{'='*62}")def longest_increasing_subsequence(arr):
"""
Find the length and actual sequence of the longest increasing subsequence.
Uses O(n^2) DP approach for clarity.
"""
n = len(arr)
dp = [1] * n # dp[i] = LIS length ending at index i
prev = [-1] * n # For path reconstruction
for i in range(1, n):
for j in range(i):
if arr[j] < arr[i] and dp[j] + 1 > dp[i]:
dp[i] = dp[j] + 1
prev[i] = j
# Find index of maximum LIS length
max_len = max(dp)
max_idx = dp.index(max_len)
# Reconstruct the actual subsequence
sequence = []
idx = max_idx
while idx != -1:
sequence.append(arr[idx])
idx = prev[idx]
sequence.reverse()
return max_len, sequence
def run_lis_demo():
arr = [10, 9, 2, 5, 3, 7, 101, 18, 4, 6, 8]
length, subseq = longest_increasing_subsequence(arr)
print(f"\n{'='*62}")
print(f" LONGEST INCREASING SUBSEQUENCE (DP)")
print(f"{'='*62}")
print(f" Input: {arr}")
print(f" LIS length: {length}")
print(f" LIS: {subseq}")
print(f" (Multiple LIS may exist with same length)")
print(f"{'='*62}")import heapq
def dijkstra_with_path(graph, start, end):
distances = {node: float('inf') for node in graph}
distances[start] = 0
previous = {node: None for node in graph}
pq = [(0, start)]
while pq:
curr_dist, curr_node = heapq.heappop(pq)
if curr_dist > distances[curr_node]:
continue
for neighbor, weight in graph[curr_node].items():
dist = curr_dist + weight
if dist < distances[neighbor]:
distances[neighbor] = dist
previous[neighbor] = curr_node
heapq.heappush(pq, (dist, neighbor))
# Reconstruct path
path, node = [], end
while node:
path.append(node)
node = previous[node]
path.reverse()
return distances[end], path if path[0] == start else []
def run_graph_demo():
city_graph = {
'Airport': {'Downtown': 25, 'Suburb_N': 30, 'Suburb_E': 20},
'Downtown': {'Airport': 25, 'University': 10, 'Mall': 8},
'University': {'Downtown': 10, 'Park': 5, 'Mall': 12},
'Suburb_N': {'Airport': 30, 'Park': 15},
'Suburb_E': {'Airport': 20, 'Mall': 18},
'Mall': {'Downtown': 8, 'University': 12, 'Suburb_E': 18},
'Park': {'University': 5, 'Suburb_N': 15}
}
routes = [
('Airport', 'Park'),
('Airport', 'University'),
('Suburb_N', 'Mall'),
]
print(f"\n{'='*62}")
print(f" SHORTEST PATH ANALYSIS (Dijkstra's Algorithm)")
print(f"{'='*62}")
for start, end in routes:
dist, path = dijkstra_with_path(city_graph, start, end)
path_str = " → ".join(path)
print(f"\n {start} → {end}")
print(f" Route: {path_str}")
print(f" Distance: {dist} minutes")
print(f"\n{'='*62}")# algorithm_showcase.py — Full Performance Analysis Tool
import time, random, math, heapq, bisect
from collections import Counter
# (All functions defined above go here)
def main():
print("\n" + "=" * 62)
print(" ALGORITHM PERFORMANCE ANALYSIS TOOL")
print(" Algorithms Course — Capstone Project")
print("=" * 62)
run_sort_benchmarks(n=1000)
run_search_benchmarks(n=100000)
run_lis_demo()
run_graph_demo()
print("\n" + "=" * 62)
print(" ALGORITHM COMPLEXITY QUICK REFERENCE")
print("=" * 62)
print(f" {'Algorithm':<22} {'Best':>8} {'Average':>10} {'Worst':>10}")
print(f" {'-'*20} {'-'*6} {'-'*8} {'-'*8}")
reference = [
("Bubble Sort", "O(n)", "O(n²)", "O(n²)"),
("Selection Sort", "O(n²)", "O(n²)", "O(n²)"),
("Insertion Sort", "O(n)", "O(n²)", "O(n²)"),
("Merge Sort", "O(n log n)","O(n log n)","O(n log n)"),
("Quick Sort", "O(n log n)","O(n log n)","O(n²)"),
("Linear Search", "O(1)", "O(n)", "O(n)"),
("Binary Search", "O(1)", "O(log n)", "O(log n)"),
("Dijkstra", "O(E log V)","O(E log V)","O(E log V)"),
("LIS (DP)", "O(n²)", "O(n²)", "O(n²)"),
]
for name, best, avg, worst in reference:
print(f" {name:<22} {best:>8} {avg:>10} {worst:>10}")
print("=" * 62)
if __name__ == "__main__":
main()==============================================================
ALGORITHM PERFORMANCE ANALYSIS TOOL
Algorithms Course — Capstone Project
==============================================================
==============================================================
SORTING BENCHMARK (n = 1000 elements)
==============================================================
Algorithm Random Sorted Reverse
-------------- -------- -------- --------
Bubble Sort 89.32ms 0.85ms 178.41ms
Selection Sort 52.17ms 51.90ms 52.88ms
Insertion Sort 18.64ms 0.12ms 37.21ms
Merge Sort 1.24ms 0.98ms 1.18ms
Quick Sort 0.91ms 0.87ms 0.79ms
==============================================================
==============================================================
SEARCH COMPARISON (n = 100,000 sorted elements)
==============================================================
Method Time Comparisons
------------------ ---------- -------------
Linear Search 2.1830ms 50,000 (avg)
Binary Search 0.0021ms 17 (max)
bisect (built-in) 0.0008ms 17 (max)
==============================================================
==============================================================
LONGEST INCREASING SUBSEQUENCE (DP)
==============================================================
Input: [10, 9, 2, 5, 3, 7, 101, 18, 4, 6, 8]
LIS length: 5
LIS: [2, 3, 4, 6, 8]
==============================================================
==============================================================
SHORTEST PATH ANALYSIS (Dijkstra's Algorithm)
==============================================================
Airport → Park
Route: Airport → Downtown → University → Park
Distance: 40 minutes
Airport → University
Route: Airport → Downtown → University
Distance: 35 minutes
Suburb_N → Mall
Route: Suburb_N → Park → University → Downtown → Mall
Distance: 38 minutes
==============================================================This project brought together every major algorithmic concept from the course:
| Concept | Lesson | Used In Project |
|---|---|---|
| Linear vs Binary Search | 1, 2 | Part 2: Search benchmarks |
| Binary Search Applications | 2 | bisect module in Part 2 |
| Dynamic Programming (intro) | 3 | Part 3: LIS with memoization |
| Classic DP Problems | 4 | Knapsack, LCS patterns |
| BFS/DFS Traversal | 5 | Graph foundation for Part 4 |
| Dijkstra's Algorithm | 6 | Part 4: Shortest path finder |
| Greedy Algorithms | 7 | Activity scheduling patterns |
| Algorithm Analysis | All | Part 1: Sorting benchmarks |
On choosing algorithms:
On Big O:
On patterns:
You've completed the Algorithms course. You now understand the core algorithms that power real world software. Where to go next:
The fundamentals you've built here thinking in terms of time complexity, recognizing problem patterns, and building solutions from smaller subproblems are the foundation for all advanced study to come.
University learning companion
Algorithms · Final Project: Algorithm Design Challenge
Complete the steps without looking at the answer first; record evidence rather than only a final result.
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