Coin Change Minimum Coins in DSA C++ - Build from Scratch

Coin Change Minimum Coins

📖 Scenario: You are helping a cashier who needs to give change using the fewest coins possible. The cashier has coins of certain fixed denominations.

🎯 Goal: Build a program that finds the minimum number of coins needed to make a given amount of money using the available coin denominations.

📋 What You'll Learn

Create a vector called coins with the exact values: 1, 5, 10, 25

Create an integer variable called amount and set it to 37

Create a vector called minCoins to store the minimum coins needed for each amount from 0 to amount

Use a loop to fill minCoins with the minimum coins needed for each value

Print the minimum number of coins needed for amount

💡 Why This Matters

🌍 Real World

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

💼 Career

Understanding dynamic programming and optimization problems is useful for software engineers working on finance, retail, or embedded systems.

1

Create the coin denominations vector and amount variable

Create a std::vector<int> called coins with these exact values: 1, 5, 10, 25. Then create an int variable called amount and set it to 37.

DSA C++

#include 
#include 

int main() {
    // Create coins vector and amount variable
    // Your code here
}

Hint

Use std::vector<int> coins = {1, 5, 10, 25}; to create the coins vector.

Set int amount = 37; for the amount.

2

Create the minCoins vector and initialize it

Create a std::vector<int> called minCoins with size amount + 1. Initialize all elements to amount + 1 except minCoins[0] which should be 0.

DSA C++

#include 
#include 

int main() {
    std::vector coins = {1, 5, 10, 25};
    int amount = 37;

    // Create minCoins vector and initialize
    // Your code here
}

Hint

Use std::vector<int> minCoins(amount + 1, amount + 1); to create and initialize the vector.

Set minCoins[0] = 0; because zero coins are needed to make amount zero.

3

Fill minCoins vector with minimum coins needed for each amount

Use a for loop with variable i from 1 to amount. Inside it, use another for loop with variable coin iterating over coins. If coin <= i, update minCoins[i] to the minimum of its current value and minCoins[i - coin] + 1.

DSA C++

#include 
#include 

int main() {
    std::vector coins = {1, 5, 10, 25};
    int amount = 37;

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

    // Fill minCoins with minimum coins needed
    // Your code here
}

Hint

Use nested loops: outer loop from 1 to amount, inner loop over coins.

Update minCoins[i] with the minimum value between current and minCoins[i - coin] + 1.

4

Print the minimum number of coins needed for the amount

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

DSA C++

#include 
#include 
#include 

int main() {
    std::vector coins = {1, 5, 10, 25};
    int amount = 37;

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

    for (int i = 1; i <= amount; ++i) {
        for (int coin : coins) {
            if (coin <= i) {
                minCoins[i] = std::min(minCoins[i], minCoins[i - coin] + 1);
            }
        }
    }

    // Print the minimum coins needed for amount
    // Your code here
}

Hint

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

Practice

1. What does the Coin Change Minimum Coins problem ask you to find?

easy

A. The fewest number of coins needed to make a given amount

B. The total number of different coins available

C. The maximum amount of money you can make with given coins

D. The sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of coins needed to make a specific amount using given coin denominations.

2. Step 2: Identify the correct description

   The fewest number of coins needed to make a given amount correctly describes this goal, while others describe unrelated concepts.

3. Final Answer:

   The fewest number of coins needed to make a given amount -> Option A

4. Quick Check:

   Minimum coins = fewest coins needed [OK]

Hint: Focus on 'minimum coins' needed for the amount [OK]

Common Mistakes:

• Confusing minimum coins with total coin types
• Thinking about maximum amount instead of minimum coins
• Mixing sum of coins with count of coins

2. Which of the following is the correct C++ syntax to declare a vector named dp of size amount+1 initialized with INT_MAX?

easy

A. vector<int> dp(amount, 0);

B. std::vector dp(amount+1, INT_MAX);

C. std::vector<int> dp(amount+1, INT_MAX);

D. vector<int> dp = new vector(amount+1, INT_MAX);

Solution

1. Step 1: Check vector declaration syntax

   Correct syntax uses std::vector<int> dp(size, initial_value); std::vector<int> dp(amount+1, INT_MAX); matches this exactly.

2. Step 2: Identify errors in other options

   vector<int> dp(amount, 0); initializes with 0 and wrong size; std::vector dp(amount+1, INT_MAX); misses template brackets; vector<int> dp = new vector(amount+1, INT_MAX); uses incorrect new syntax.

3. Final Answer:

   std::vector<int> dp(amount+1, INT_MAX); -> Option C

4. Quick Check:

   Correct vector declaration = std::vector<int> dp(amount+1, INT_MAX); [OK]

Hint: Use std::vector<Type> name(size, value) for initialization [OK]

Common Mistakes:

• Omitting template brackets <>
• Using wrong initialization syntax
• Confusing size and initial value parameters

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

int coins[] = {1, 2, 5};
int amount = 11;
std::vector<int> dp(amount+1, INT_MAX);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
  for (int c : coins) {
    if (i - c >= 0 && dp[i - c] != INT_MAX) {
      dp[i] = std::min(dp[i], dp[i - c] + 1);
    }
  }
}
std::cout << dp[amount] << std::endl;

medium

A. 11

B. 3

C. 5

D. INT_MAX

Solution

1. Step 1: Understand the DP approach

   The code finds minimum coins to make amount 11 using coins {1,2,5} by building dp array.

2. Step 2: Calculate minimum coins for 11

   11 = 5 + 5 + 1 uses 3 coins, which is minimum. dp[11] = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Minimum coins for 11 = 3 [OK]

Hint: Try largest coins first to minimize count [OK]

Common Mistakes:

• Assuming dp[amount] remains INT_MAX
• Counting total coin value instead of count
• Ignoring dp initialization and updates

4. Identify the error in this C++ code snippet for Coin Change Minimum Coins:

int coins[] = {1, 3, 4};
int amount = 6;
std::vector<int> dp(amount+1, INT_MAX);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
  for (int c : coins) {
    if (i - c >= 0) {
      dp[i] = std::min(dp[i], dp[i - c] + 1);
    }
  }
}
std::cout << dp[amount] << std::endl;

medium

A. dp[0] should be initialized to INT_MAX

B. Wrong initialization of dp vector size

C. Incorrect coin denominations array syntax

D. Missing check if dp[i - c] is INT_MAX before adding 1

Solution

1. Step 1: Analyze dp update condition

   The code updates dp[i] without checking if dp[i - c] is INT_MAX, which can cause overflow.

2. Step 2: Identify missing condition

   Must check dp[i - c] != INT_MAX before adding 1 to avoid invalid calculations.

3. Final Answer:

   Missing check if dp[i - c] is INT_MAX before adding 1 -> Option D

4. Quick Check:

   Check dp[i - c] != INT_MAX before update [OK]

Hint: Always check dp[i-c] != INT_MAX before adding 1 [OK]

Common Mistakes:

• Skipping dp[i-c] validity check
• Incorrect dp vector size
• Wrong initialization of dp[0]

5. Given coin denominations {2, 5, 10} and amount 7, what is the minimum number of coins needed? If it is not possible, return -1.

hard

A. 2

B. 3

C. -1

D. 1

Solution

1. Step 1: Check possible combinations for amount 7

   Coins are 2, 5, 10. 10 > 7 so ignore. Try 5 + 2 = 7 (2 coins). But 5 + 2 is valid, so minimum coins = 2.

2. Step 2: Verify if 5 + 2 is allowed

   Since 5 and 2 are in denominations, 7 can be made with 2 coins.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   7 = 5 + 2 uses 2 coins [OK]

Hint: Try combinations starting from largest coin [OK]

Common Mistakes:

• Assuming no solution because 7 is odd
• Ignoring valid combinations with smaller coins
• Returning -1 incorrectly