Coin Change II
Detailed guide and Python implementation for the 'Coin Change II' problem.
1. Concept Overview
The 'Coin Change II' problem is a key challenge in the 2D DP section.
This implementation focuses on medium-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 II.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Coin Change II 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
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.
You may assume that you have an infinite number of each kind of coin.
Write a function change(amount: int, coins: List[int]) -> int.
- •1 <= len(coins) <= 300
- •1 <= coins[i] <= 5000
- •0 <= amount <= 5000
Examples
amount = 5, coins = [1,2,5]
4
There are four ways to make up the amount: 5, 2+2+1, 2+1+1+1, 1+1+1+1+1.
amount = 3, coins = [2]
0
The amount of 3 cannot be made up with just 2s.
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 change_opt(amount, coins):
dp = [0] * (amount + 1); dp[0] = 1
for coin in coins:
for x in range(coin, amount + 1):
dp[x] += dp[x - coin]
return dp[amount]Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def change_brute(amount, coins):
def solve(i, cur):
if cur == amount: return 1
if i == len(coins) or cur > amount: return 0
return solve(i, cur + coins[i]) + solve(i + 1, cur)
return solve(0, 0)Algorithm Pattern Checklist
When dealing with 2D DP data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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