Use when designing algorithms: greedy, DP, divide-conquer.
Scanned 9/10/2026
Install to Claude Code
npx -y skills add LoopyLuci/Skills --skill algorithm-design-techniques --agent claude-codeInstalls into .claude/skills of the current project.
Are you the author of Algorithm Design Techniques?
Add the live security badge to your README — it updates automatically with every re-scan.
[](https://www.skillsdirectory.com/skills/loopyluci-algorithm-design-techniques)More formats (shields.io, HTML) on the badges page.
---
name: algorithm-design-techniques
description: "Use when designing algorithms: greedy, DP, divide-conquer."
category: software-development
tags: [algorithms, design, greedy, dynamic-programming, backtracking]
---
# Algorithm Design Techniques
Core algorithm design paradigms.
## Greedy Algorithms
```python
# Make locally optimal choice at each step
# Works when local optimum = global optimum
# Activity Selection
def max_activities(intervals: list[tuple[int,int]]):
intervals.sort(key=lambda x: x[1]) # sort by end time
selected = [intervals[0]]
for start, end in intervals[1:]:
if start >= selected[-1][1]:
selected.append((start, end))
return selected
```
## Divide and Conquer
```python
# 1. Divide into subproblems
# 2. Conquer subproblems recursively
# 3. Combine results
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
result.extend(left[i:] or right[j:])
return result
```
## Dynamic Programming
```python
# Optimal substructure + overlapping subproblems
# Top-down (memoization) vs Bottom-up (tabulation)
# Longest Common Subsequence
def lcs(text1: str, text2: str) -> int:
m, n = len(text1), len(text2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
if text1[i-1] == text2[j-1]:
dp[i][j] = dp[i-1][j-1] + 1
else:
dp[i][j] = max(dp[i-1][j], dp[i][j-1])
return dp[m][n]
```
## Backtracking
```python
# Explore all candidates, abandon (prune) when impossible
# N-Queens
def solve_n_queens(n: int) -> list[list[str]]:
result = []
cols, diag1, diag2 = set(), set(), set()
def backtrack(row, board):
if row == n:
result.append(["".join(r) for r in board])
return
for col in range(n):
if col in cols or (row - col) in diag1 or (row + col) in diag2:
continue
cols.add(col); diag1.add(row - col); diag2.add(row + col)
board[row][col] = 'Q'
backtrack(row + 1, board)
board[row][col] = '.'
cols.remove(col); diag1.remove(row - col); diag2.remove(row + col)
backtrack(0, [['.']*n for _ in range(n)])
return result
```
## Binary Search (Advanced)
```python
# Search in sorted rotated array
def search_rotated(nums: list[int], target: int) -> int:
l, r = 0, len(nums) - 1
while l <= r:
mid = (l + r) // 2
if nums[mid] == target: return mid
if nums[l] <= nums[mid]: # left half sorted
if nums[l] <= target < nums[mid]:
r = mid - 1
else:
l = mid + 1
else: # right half sorted
if nums[mid] < target <= nums[r]:
l = mid + 1
else:
r = mid - 1
return -1
```
## Pitfalls
- Greedy fails when choices have future consequences
- DP needs well-defined state and transition — wrong state = wrong answer
- Backtracking without pruning is brute force — always prune
- Binary search edge cases: off-by-one on l/r bounds
- Divide and conquer has log n depth; recursion may overflow for large n
Is this your skill, or is something wrong with this listing? Request removal or report an issue. Author removals are honored within 72 hours.
No comments yet. Be the first to comment!