Coin Change Problem
Detailed guide and Python implementation for the 'Coin Change Problem' problem.
1. Concept Overview
The 'Coin Change Problem' problem is a key challenge in the Dynamic Programming section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Coin Change Problem.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Coin Change Problem carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function count_coin_change(coins, n) that returns the number of distinct ways to make change for a target amount n using an unlimited supply of the given coin denominations coins.
- •1 <= len(coins) <= 100
- •1 <= coins[i] <= 1000
- •1 <= n <= 1000
Examples
count_coin_change([1, 2, 3], 4)
4
There are 4 ways: {1,1,1,1}, {1,1,2}, {2,2}, {1,3}.
count_coin_change([2, 5, 3, 6], 10)
5
There are 5 ways: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5}, {5,5}.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def coin_change_opt(coins, n):
dp = [0] * (n + 1); dp[0] = 1
for c in coins:
for i in range(c, n + 1): dp[i] += dp[i-c]
return dp[n]Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def coin_change_brute(coins, n):
def solve(i, cur):
if cur == 0: return 1
if cur < 0 or i == len(coins): return 0
return solve(i, cur - coins[i]) + solve(i + 1, cur)
return solve(0, n)Algorithm Pattern Checklist
When dealing with Dynamic Programming data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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