Coin Change Minimum Coins in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the goal of the Coin Change Minimum Coins problem?

To find the minimum number of coins needed to make a given amount using available coin denominations.

beginner

Which algorithmic approach is commonly used to solve the Coin Change Minimum Coins problem efficiently?

Dynamic Programming is commonly used to solve it efficiently by building up solutions for smaller amounts to reach the target amount.

beginner

In the dynamic programming solution for Coin Change Minimum Coins, what does the dp array represent?

dp[i] represents the minimum number of coins needed to make the amount i.

intermediate

Why do we initialize the dp array with a large number (like amount+1) in Coin Change Minimum Coins problem?

Because we want to represent that initially, the amount cannot be made with any coins, so we use a large number as 'infinity' to find minimum later.

intermediate

What is the time complexity of the dynamic programming solution for the Coin Change Minimum Coins problem?

The time complexity is O(n * amount), where n is the number of coin denominations and amount is the target amount.

What does dp[i] represent in the Coin Change Minimum Coins dynamic programming solution?

A Total value of coins used

B Maximum coins needed to make amount i

C Number of coin types used

D Minimum coins needed to make amount i

dp[i] stores the minimum number of coins needed to make the amount i.

Which of the following is a valid base case initialization for dp array in Coin Change Minimum Coins?

A dp[0] = 0

B dp[0] = 1

C dp[0] = amount

D dp[0] = -1

dp[0] = 0 means zero coins are needed to make amount zero.

If no combination of coins can make the target amount, what should the algorithm return?

A 0

B -1

C amount

D 1

Returning -1 indicates it is impossible to make the amount with given coins.

What is the main idea behind using dynamic programming for Coin Change Minimum Coins?

A Sort coins and pick the largest first

B Try all permutations of coins

C Build solutions for smaller amounts to solve larger amounts

D Use recursion without memoization

Dynamic programming solves smaller subproblems first to build up the answer for the target amount.

What is the time complexity of the dynamic programming solution for Coin Change Minimum Coins?

A O(n * amount)

B O(n + amount)

C O(n^2)

D O(amount^2)

The solution iterates over all coins and amounts, resulting in O(n * amount) complexity.

Explain how dynamic programming solves the Coin Change Minimum Coins problem step-by-step.

Think about building solutions from amount 0 up to the target amount.

Describe how you would handle the case when it is impossible to make the target amount with given coins.

Consider what dp[target] means after filling the dp array.

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