Coin Change Total Number of Ways in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Coin Change Total Number of Ways problem?

To find how many different ways we can make a target amount using given coin denominations, where order does not matter and coins can be used unlimited times.

intermediate

In the Coin Change problem, why do we use dynamic programming?

Because the problem has overlapping subproblems and optimal substructure, dynamic programming helps store intermediate results to avoid repeated calculations and efficiently find the total number of ways.

beginner

What does the DP array represent in the Coin Change Total Number of Ways solution?

Each index in the DP array represents the number of ways to make that amount using the given coins processed so far.

beginner

Why do we initialize dp[0] = 1 in the Coin Change problem?

Because there is exactly one way to make amount 0 — by choosing no coins at all.

intermediate

How does the order of processing coins affect the Coin Change Total Number of Ways solution?

Processing coins in order ensures combinations are counted without duplicates, as it avoids counting the same set of coins in different orders multiple times.

What does dp[i] represent in the Coin Change Total Number of Ways problem?

A Number of ways to make amount i using given coins

B Minimum coins needed to make amount i

C Maximum coins used to make amount i

D Number of coins available

dp[i] stores the total number of ways to make amount i using the coins considered so far.

Why is dp[0] initialized to 1 in the Coin Change problem?

A Because it represents the first coin

B Because amount 0 requires one coin

C Because dp array must start with 1

D Because amount 0 can be made in one way: no coins

There is exactly one way to make amount 0: by choosing no coins.

Which approach is used to avoid counting duplicate combinations in Coin Change Total Number of Ways?

A Count permutations instead of combinations

B Process coins one by one and update dp array

C Sort coins in descending order

D Use recursion without memoization

Processing coins one by one ensures combinations are counted uniquely without duplicates.

If coins = [1, 2] and amount = 3, how many ways are there to make 3?

A 2

B 3

C 1

D 4

Ways: (1+1+1), (1+2), (2+1) but since order does not matter, (1+2) and (2+1) count as one combination, so total 2 ways.

What is the time complexity of the dynamic programming solution for Coin Change Total Number of Ways?

A O(n^2)

B O(amount^2)

C O(n * amount), where n is number of coins

D O(n + amount)

The solution iterates over each coin and for each amount up to target, resulting in O(n * amount) complexity.

Explain how dynamic programming is used to solve the Coin Change Total Number of Ways problem.

Think about how you build up solutions from smaller amounts.

Describe the role of the order in which coins are processed in the Coin Change problem.

Why does order matter when counting sets?

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?
Given coins and a total amount, how many ways can you make that amount?

easy

A. The total number of ways to make the amount using given coins

B. The minimum number of coins needed to make the amount

C. The maximum coin value used to make the amount

D. The total sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for the count of different ways to combine coins to reach a target amount.

2. Step 2: Differentiate from similar problems

   This is not about minimum coins or maximum coin value, but total ways to form the amount.

3. Final Answer:

   The total number of ways to make the amount using given coins -> Option A

4. Quick Check:

   Count ways, not minimum or max coin [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins needed
• Thinking it's about maximum coin value
• Mixing sum of coins with number of ways

2. Which of the following is the correct way to initialize the DP array for the Coin Change Total Number of Ways problem in C++?

int amount = 5;
vector<int> dp(amount + 1, 0);
dp[0] = ?;

easy

A. 1

B. 0

C. -1

D. amount

Solution

1. Step 1: Understand base case initialization

   dp[0] represents ways to make amount 0, which is 1 (using no coins).

2. Step 2: Set dp[0] to 1

   Setting dp[0] = 1 allows building solutions for other amounts correctly.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   dp[0] = 1 means one way to make zero amount [OK]

Hint: dp[0] always 1 for coin change ways [OK]

Common Mistakes:

• Setting dp[0] to 0 causes no valid ways
• Using negative values for dp initialization
• Confusing dp array meaning with minimum coins

3. What is the output of the following C++ code?

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> coins = {1, 2};
    int amount = 3;
    vector<int> dp(amount + 1, 0);
    dp[0] = 1;
    for (int coin : coins) {
        for (int x = coin; x <= amount; x++) {
            dp[x] += dp[x - coin];
        }
    }
    cout << dp[amount] << endl;
    return 0;
}

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Trace dp array updates for coins {1, 2} and amount 3

   Initially dp = [1,0,0,0]. After coin=1: dp = [1,1,1,1]. After coin=2: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2, so dp = [1,1,2,2].

2. Step 2: Final dp[3] value

   dp[3] = 2 means there are 2 ways to make amount 3 with coins 1 and 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Ways to make 3 with {1,2} = 2 [OK]

Hint: Add dp[x - coin] for each coin and amount [OK]

Common Mistakes:

• Mixing order of loops causing wrong counts
• Not initializing dp[0] to 1
• Confusing dp[x] update with subtraction

4. Find the error in this C++ code snippet for Coin Change Total Number of Ways:

vector<int> dp(amount + 1, 0);
dp[0] = 1;
for (int x = 1; x <= amount; x++) {
    for (int coin : coins) {
        if (coin <= x) {
            dp[x] += dp[x - coin];
        }
    }
}

medium

A. The dp array size is too small

B. dp[0] should be initialized to 0

C. The condition if (coin <= x) is incorrect

D. The outer loop should iterate over coins, inner over amounts

Solution

1. Step 1: Analyze loop order for correct counting

   Outer loop must be over coins to avoid counting duplicates; current code counts duplicates.

2. Step 2: Correct loop order

   Change outer loop to for (int coin : coins) and inner loop to for (int x = coin; x <= amount; x++) to count unique combinations.

3. Final Answer:

   The outer loop should iterate over coins, inner over amounts -> Option D

4. Quick Check:

   Loop order affects counting unique ways [OK]

Hint: Outer loop coins, inner loop amounts for unique ways [OK]

Common Mistakes:

• Swapping loops causes counting duplicates
• Incorrect dp[0] initialization
• Wrong dp array size causing out-of-bounds

5. You have coins {1, 3, 5} and want to find the total number of ways to make amount 7. Which approach correctly computes this using dynamic programming?

hard

A. Sort coins descending and greedily subtract largest coin until amount is zero

B. Use a 2D dp array with rows as coins and columns as amounts, fill with minimum coins needed

C. Use a 1D dp array, initialize dp[0]=1, then for each coin, update dp[x] += dp[x - coin] for x from coin to amount

D. Use recursion without memoization to count ways

Solution

1. Step 1: Identify correct DP approach for total ways

   Using 1D dp array with dp[0]=1 and updating dp[x] for each coin counts total ways efficiently.

2. Step 2: Reject other approaches

   2D dp for minimum coins is different problem; greedy fails for total ways; recursion without memo is inefficient.

3. Final Answer:

   Use a 1D dp array, initialize dp[0]=1, then for each coin, update dp[x] += dp[x - coin] for x from coin to amount -> Option C

4. Quick Check:

   1D dp with coin loops counts ways correctly [OK]

Hint: 1D dp with coin loops counts ways, not min coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins problem
• Using greedy which fails for counting ways
• Ignoring memoization causing slow recursion