Coin Change Minimum Coins in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with a large number.

DSA C++

int dp[n + 1];
for (int i = 0; i <= n; i++) {
    dp[i] = [1];
}
dp[0] = 0;

Drag options to blanks, or click blank then click option'

A0

B1

C-1

DINT_MAX

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing dp with 0 causes incorrect minimum calculations.

Using negative numbers leads to wrong results.

We initialize dp array with INT_MAX to represent that initially, no amount can be formed except 0.

2fill in blank

medium

Complete the code to iterate over all coin denominations.

DSA C++

for (int i = 1; i <= n; i++) {
    for (int j = 0; j < [1]; j++) {
        if (coins[j] <= i) {
            dp[i] = std::min(dp[i], dp[i - coins[j]] + 1);
        }
    }
}

Drag options to blanks, or click blank then click option'

Adp[i]

Bm

Cn

Dcoins.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using n instead of m causes out-of-bound errors.

Using dp[i] as loop limit is incorrect.

We loop over all m coin denominations to check which coins can contribute to amount i.

3fill in blank

hard

Fix the error in the condition to update dp[i].

DSA C++

if (coins[j] <= i && dp[i - coins[j]] != [1]) {
    dp[i] = std::min(dp[i], dp[i - coins[j]] + 1);
}

Drag options to blanks, or click blank then click option'

AINT_MAX

B0

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Not checking dp[i - coins[j]] leads to overflow or wrong results.

Checking against 0 or -1 is incorrect here.

We check dp[i - coins[j]] != INT_MAX to ensure the previous amount is reachable before adding a coin.

4fill in blank

hard

Fill both blanks to return the minimum coins or -1 if not possible.

DSA C++

return (dp[n] == [1]) ? [2] : dp[n];

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AINT_MAX

B-1

C0

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of -1 for no solution.

Not checking dp[n] leads to wrong output.

If dp[n] is INT_MAX, amount n cannot be formed, so return -1; otherwise, return dp[n].

5fill in blank

hard

Fill all three blanks to complete the function signature and initialization.

DSA C++

int coinChange(std::vector<int>& coins, int [1]) {
    int [2] = [1];
    std::vector<int> dp(n + 1, [3]);
    dp[0] = 0;

Drag options to blanks, or click blank then click option'

An

Bamount

CINT_MAX

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong parameter name causes confusion.

Initializing dp with 0 causes incorrect results.

Function takes amount as parameter, uses n as amount variable, and initializes dp with INT_MAX.

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