Coin Change II (Count Ways)

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Coin Change II (Count Ways)

Edge cases:

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IDE

def change(amount: int, coins: list[int]) -> int:public int change(int amount, int[] coins)int change(int amount, vector<int>& coins)function change(amount, coins)

def change(amount, coins):
    # Write your solution here
    pass

class Solution {
    public int change(int amount, int[] coins) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int change(int amount, vector<int>& coins) {
    // Write your solution here
    return 0;
}

function change(amount, coins) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: Counting permutations instead of combinations (too large count)Looping amounts outer and coins inner, causing order-sensitive counting.✅ Change loop order: iterate coins outer loop, amounts inner loop to count combinations.

Wrong: Zero ways returned for amount=0Not initializing dp[0] = 1 to represent empty combination.✅ Initialize dp[0] = 1 before DP iteration.

Wrong: Non-zero ways returned when coins array is emptyNot handling empty coins input properly.✅ Return 0 for any positive amount if coins array is empty.

Wrong: Treating problem as 0/1 knapsack limiting coin usage to onceIncorrect DP update disallowing multiple usage of same coin.✅ Allow unlimited usage by updating dp in coin outer loop and amounts inner loop.

Wrong: TLE on large inputsUsing brute force recursion or exponential approach.✅ Implement bottom-up DP with O(n*amount) time complexity.

✓

Test Cases

Focus on handling base cases like zero amount and empty coin sets correctly.

Beware of common algorithmic traps like counting permutations or limiting coin usage.

Optimize your solution to use bottom-up DP with O(n*amount) complexity to pass performance tests.

t1_01basic

Input{"amount":5,"coins":[1,2,5]}

Expected4

⏱ Performance - must finish in 2000ms

The four combinations are: 5=5; 5=2+2+1; 5=2+1+1+1; 5=1+1+1+1+1 (order does not matter).

💡 Think about how to count combinations using dynamic programming.

💡 Use a 1D dp array where dp[x] represents ways to make amount x.

💡 Update dp by iterating coins and for each coin, add dp[x - coin] to dp[x].

Why it failed: Incorrect count likely due to mixing permutations and combinations. Fix by iterating coins outer loop and amounts inner loop to avoid counting order.

✓ Correctly counts combinations using unbounded knapsack DP.

t1_02basic

Input{"amount":3,"coins":[2]}

Expected0

⏱ Performance - must finish in 2000ms

No combination of coin 2 can make amount 3.

💡 Check if your code handles amounts that cannot be formed by given coins.

💡 Ensure dp[0] is initialized to 1 to represent empty combination.

💡 Verify that dp values for unreachable amounts remain zero.

Why it failed: Returned non-zero ways for unreachable amount. Fix by initializing dp[0]=1 and only updating dp[x] when x >= coin value.

✓ Correctly returns zero when no combinations possible.

t2_01edge

Input{"amount":0,"coins":[1,2]}

Expected1

⏱ Performance - must finish in 2000ms

Only one way to make amount 0: use no coins (empty combination).

💡 Remember that amount zero always has one combination: empty set.

💡 Initialize dp[0] = 1 regardless of coins.

💡 Check that your code returns 1 for amount zero even if coins are present.

Why it failed: Returned 0 for amount=0. Fix by initializing dp[0]=1 to count empty combination.

✓ Correctly counts empty combination for zero amount.

t2_02edge

Input{"amount":3,"coins":[]}

Expected0

⏱ Performance - must finish in 2000ms

No coins means no way to form any positive amount.

💡 Check behavior when coins array is empty.

💡 Ensure dp array is initialized but no updates occur without coins.

💡 Verify that dp[amount] remains zero for positive amount with no coins.

Why it failed: Returned non-zero ways with empty coins. Fix by handling empty coins input and returning 0 for positive amounts.

✓ Correctly returns zero ways when no coins are available.

t2_03edge

Input{"amount":10,"coins":[10]}

Expected1

⏱ Performance - must finish in 2000ms

Only one combination: using the single coin once to make amount 10.

💡 Check if single coin equal to amount returns 1.

💡 Ensure dp updates correctly when coin equals amount.

💡 Verify no extra combinations are counted.

Why it failed: Returned incorrect count for single coin equal to amount. Fix by correctly updating dp for coin == amount.

✓ Correctly counts single combination for coin equal to amount.

t2_04edge

Input{"amount":7,"coins":[2,4]}

Expected0

⏱ Performance - must finish in 2000ms

No combination of 2 and 4 can make 7.

💡 Verify your code returns zero when no combination possible.

💡 Check that dp array does not accumulate invalid counts.

💡 Fix by initializing dp properly and updating only valid states.

Why it failed: Returned non-zero ways for impossible amount. Fix by ensuring dp updates only valid sums.

✓ Correctly returns zero for impossible sums.

t3_01corner

Input{"amount":5,"coins":[1,3,4]}

Expected3

⏱ Performance - must finish in 2000ms

Combinations: 1+1+1+1+1, 1+1+3, 1+4, 3+3 is invalid (exceeds amount). Valid are 4 combinations total.

💡 Beware of counting permutations instead of combinations.

💡 Use coin-first outer loop to avoid overcounting order.

💡 Fix by iterating coins outer loop and amounts inner loop.

Why it failed: Counting permutations instead of combinations. Fix by changing loop order: coins outer, amounts inner.

✓ Correctly counts combinations, not permutations.

t3_02corner

Input{"amount":4,"coins":[1,2]}

Expected3

⏱ Performance - must finish in 2000ms

Combinations: 1+1+1+1, 1+1+2, 2+2

💡 Check if your solution mistakenly treats this as 0/1 knapsack.

💡 Ensure unlimited usage of each coin is allowed.

💡 Fix by allowing multiple counts of same coin in dp updates.

Why it failed: Mistakenly treated as 0/1 knapsack, limiting coin usage to once. Fix by allowing repeated usage in dp.

✓ Correctly allows unlimited coin usage.

t3_03corner

Input{"amount":7,"coins":[2,4]}

Expected0

⏱ Performance - must finish in 2000ms

No combination of 2 and 4 can make 7.

💡 Verify your code returns zero when no combination possible.

💡 Check that dp array does not accumulate invalid counts.

💡 Fix by initializing dp properly and updating only valid states.

Why it failed: Returned non-zero ways for impossible amount. Fix by ensuring dp updates only valid sums.

✓ Correctly returns zero for impossible sums.

t4_01performance

Input{"amount":10000,"coins":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100]}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=100 coins and amount=10000; O(n*amount) DP must complete within 2s.

💡 Use bottom-up DP with space optimization to handle large inputs.

💡 Avoid exponential brute force recursion to prevent TLE.

💡 Implement dp array of size amount+1 and update iteratively.

Why it failed: TLE due to exponential brute force or inefficient recursion. Fix by using O(n*amount) DP approach.

✓ Efficient DP solution runs within time limit.

Practice

1. Consider the following code snippet implementing the minimum cost for tickets problem. What is the value of dp[0] after the loop completes for the input days = [1,4,6] and costs = [2,7,15]?

easy

A. 6

B. 7

C. 9

D. 4

Solution

1. Step 1: Trace dp from the end

   dp[3] = 0 (base case). For i=2 (day=6): - 1-day ticket: cost=2 + dp[3]=2 - 7-day ticket: cost=7 + dp[3]=7 - 30-day ticket: cost=15 + dp[3]=15 Minimum = 2 -> dp[2]=2

2. Step 2: Calculate dp[1] and dp[0]

   i=1 (day=4): - 1-day: cost=2 + dp[2]=4 - 7-day: cost=7 + dp[3]=7 - 30-day: cost=15 + dp[3]=15 Minimum=4 -> dp[1]=4 i=0 (day=1): - 1-day: cost=2 + dp[1]=6 - 7-day: cost=7 + dp[3]=7 - 30-day: cost=15 + dp[3]=15 Minimum=6 -> dp[0]=6

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   dp[0] correctly computed as 6 [OK]

Hint: Trace dp from end to start carefully [OK]

Common Mistakes:

• Off-by-one in dp indexing
• Miscomputing next_i with bisect
• Confusing costs and dp sums

2. You are given a set of positive integers and need to partition it into two subsets such that the absolute difference of their sums is minimized. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Greedy algorithm that sorts and assigns elements alternately to subsets

B. Dynamic programming using a subset-sum style approach to find achievable sums

C. Divide and conquer by recursively splitting the array into halves

D. Sorting and then pairing elements from opposite ends to balance sums

Solution

1. Step 1: Understand problem constraints

   The problem requires minimizing the difference between sums of two subsets, which is a classic partition problem variant.

2. Step 2: Identify algorithmic pattern

   Greedy or divide-and-conquer approaches do not guarantee minimal difference. Dynamic programming, specifically subset-sum style DP, can find all achievable sums up to total_sum, enabling minimal difference calculation.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   DP subset-sum approach guarantees optimal partition [OK]

Hint: Minimal difference requires subset-sum DP, not greedy [OK]

Common Mistakes:

• Assuming greedy or sorting suffices for minimal difference

3. You need to find the minimum number of perfect square numbers that sum to a given integer n. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Dynamic programming using unbounded knapsack pattern with bottom-up tabulation

B. Greedy approach picking the largest square first until the sum is reached

C. Simple recursion without memoization exploring all combinations

D. Sorting all squares and using binary search to find the closest sum

Solution

1. Step 1: Understand problem constraints

   The problem requires the minimum count of perfect squares summing to n, which is a classic unbounded knapsack variant where items (squares) can be reused.

2. Step 2: Identify algorithm that guarantees optimality

   Greedy fails because picking largest squares first can be suboptimal (e.g., n=12). Simple recursion is correct but inefficient. Binary search does not solve the combinatorial optimization. Bottom-up DP with unbounded knapsack pattern systematically explores all combinations and ensures optimality.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP approach always finds minimum count [OK]

Hint: Greedy fails; DP unbounded knapsack guarantees optimal [OK]

Common Mistakes:

• Assuming greedy works for minimum perfect squares
• Confusing top-down recursion with guaranteed efficiency

4. You are given an array of integers and a target integer. You want to assign either a plus or minus sign to each integer such that the resulting sum equals the target. Which algorithmic approach guarantees finding the total number of such assignments efficiently?

easy

A. Sorting the array and using two pointers to find pairs that sum to the target

B. Greedy algorithm that picks signs based on local sum minimization

C. Dynamic programming using a 0/1 knapsack-like approach with state representing achievable sums

D. Divide and conquer by splitting the array and combining results without memoization

Solution

1. Step 1: Understand problem constraints

   The problem requires counting all sign assignments to reach a target sum, which involves exploring exponential combinations.

2. Step 2: Identify suitable algorithmic pattern

   Greedy and two-pointer approaches fail because they do not consider all combinations. Divide and conquer without memoization is exponential. DP with states representing achievable sums efficiently counts all valid assignments.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP handles overlapping subproblems and sums efficiently [OK]

Hint: Counting sign assignments requires DP over sums [OK]

Common Mistakes:

• Thinking greedy or sorting solves counting sign assignments

5. Consider the following buggy code for lastStoneWeightII. Which line contains the subtle bug that causes incorrect results on some inputs?

medium

A. Line 4: dp[0] = true is missing

B. Line 7: Outer loop over stones

C. Line 8: Inner loop iterates backwards

D. Line 10: Return statement calculation

Solution

1. Step 1: Identify dp initialization

   dp[0] must be true to represent sum zero achievable with no stones; missing this means no sums are marked achievable.

2. Step 2: Consequence of missing dp[0] = true

   Without dp[0] = true, dp array remains false, so no sums can be formed, causing the function to fail to find any valid partition.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Initializing dp[0] is critical base case for subset sum DP [OK]

Hint: dp[0] = true is base case for subset sums [OK]

Common Mistakes:

• Forgetting dp[0] initialization
• Iterating forward in inner loop causing double counting
• Miscomputing return value