Coin Change Total Number of Ways in DSA C++ - Build from Scratch

Coin Change Total Number of Ways

📖 Scenario: You are helping a cashier who wants to find out how many different ways they can give change for a certain amount using specific coin denominations.

🎯 Goal: Build a program that calculates the total number of ways to make change for a given amount using a list of coin denominations.

📋 What You'll Learn

Create a vector called coins with the exact values 1, 2, and 5

Create an integer variable called amount and set it to 5

Use a vector called ways to store the number of ways to make each amount from 0 to amount

Use a nested loop to update ways based on each coin denomination

Print the total number of ways to make change for amount using the coins

💡 Why This Matters

🌍 Real World

Cashiers and vending machines use this logic to calculate how to give change efficiently.

💼 Career

Understanding dynamic programming and coin change problems is common in software engineering interviews and financial software development.

1

Create the coin denominations vector

Create a vector called coins with the exact values 1, 2, and 5.

DSA C++

#include 
#include 

int main() {
    // Create vector coins with values 1, 2, 5
    // Your code here

    return 0;
}

Hint

Use std::vector<int> coins = {1, 2, 5}; to create the vector.

2

Set the amount to make change for

Create an integer variable called amount and set it to 5.

DSA C++

#include 
#include 

int main() {
    std::vector coins = {1, 2, 5};
    // Create int variable amount and set it to 5
    // Your code here

    return 0;
}

Hint

Use int amount = 5; to create the variable.

3

Calculate the total number of ways to make change

Create a vector called ways of size amount + 1 initialized with 0. Set ways[0] to 1. Use a for loop with variable coin to iterate over coins. Inside it, use a for loop with variable j from coin to amount. Update ways[j] by adding ways[j - coin].

DSA C++

#include 
#include 

int main() {
    std::vector coins = {1, 2, 5};
    int amount = 5;

    // Create vector ways of size amount+1 with 0
    // Set ways[0] = 1
    // Use for loop with coin to iterate coins
    // Use for loop with j from coin to amount
    // Update ways[j] += ways[j - coin]
    // Your code here

    return 0;
}

Hint

Use dynamic programming with nested loops to update the ways vector.

4

Print the total number of ways to make change

Print the value of ways[amount] using std::cout.

DSA C++

#include 
#include 

int main() {
    std::vector coins = {1, 2, 5};
    int amount = 5;

    std::vector ways(amount + 1, 0);
    ways[0] = 1;

    for (int coin : coins) {
        for (int j = coin; j <= amount; ++j) {
            ways[j] += ways[j - coin];
        }
    }

    // Print the total number of ways to make change for amount
    // Your code here

    return 0;
}

Hint

Use std::cout << ways[amount] << std::endl; to print the result.

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