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

Recall & Review

beginner

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

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

beginner

Which data structure is commonly used to store intermediate results in the Coin Change Minimum Coins problem?

An array (or list) where each index represents an amount and stores the minimum coins needed to make that amount.

intermediate

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

Because it represents an impossible high number of coins, so any real solution will be smaller and can replace it.

beginner

What does dp[0] = 0 represent in the Coin Change Minimum Coins problem?

It means zero coins are needed to make the amount zero.

intermediate

How do you update the dp array when considering a coin of value c for amount i?

If i >= c, update dp[i] = min(dp[i], dp[i - c] + 1), meaning either keep current or use one more coin c.

What does dp[i] represent in the Coin Change Minimum Coins problem?

A Number of coin types available

B Maximum coins needed to make amount i

C Total value of coins used

D Minimum coins needed to make amount i

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

What is the base case value for dp[0] in the Coin Change Minimum Coins problem?

A 0

B 1

C amount

D -1

Zero coins are needed to make amount zero, so dp[0] = 0.

If no combination of coins can make the amount, what should the result be?

A -1

B 0

C amount

D Infinity

Return -1 to indicate no solution exists.

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

A Greedy algorithm

B Dynamic programming

C Brute force recursion without memoization

D Sorting coins only

Dynamic programming efficiently stores intermediate results to avoid repeated work.

When updating dp[i], why do we add 1 to dp[i - coin]?

A Because we reset the count

B Because we remove one coin

C Because we use one more coin of the current denomination

D Because we double the coins

Adding 1 accounts for using one coin of the current denomination.

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

Think about building the solution from smaller amounts to larger amounts.

Describe what the dp array looks like after processing coins for amount 5 with coins [1, 3, 4].

Build dp array incrementally considering each coin.

Practice

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

easy

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

B. The total number of different coins available

C. The largest coin value in the set

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

3. Final Answer:

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

4. Quick Check:

   Minimum coins = smallest count needed [OK]

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

Common Mistakes:

• Confusing total coin count with minimum coins
• Thinking about coin values instead of counts
• Mixing up largest coin with minimum coins

2. Which of the following is the correct way to initialize the DP array for the Coin Change Minimum Coins problem in C?

easy

A. int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = 0;

B. int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = -1;

C. int dp[amount+1] = {0};

D. int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = INT_MAX;

Solution

1. Step 1: Understand DP initialization

   We initialize dp array with a large number (like INT_MAX) to represent impossible states initially.

2. Step 2: Identify correct initialization

   int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = INT_MAX; correctly sets all dp values to INT_MAX, except dp[0] which is usually set to 0 later.

3. Final Answer:

   int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = INT_MAX; -> Option D

4. Quick Check:

   Initialize dp with max values for minimum search [OK]

Hint: Use INT_MAX to mark unreachable amounts initially [OK]

Common Mistakes:

• Initializing dp with zeros causes wrong minimum calculations
• Using -1 without handling leads to errors
• Not setting dp[0] = 0 explicitly

3. What is the output of the following code snippet for coins = {1, 3, 4} and amount = 6?

int coins[] = {1, 3, 4};
int amount = 6;
int dp[7];
for(int i=0; i<=amount; i++) dp[i] = amount+1;
dp[0] = 0;
for(int i=1; i<=amount; i++) {
  for(int j=0; j<3; j++) {
    if(coins[j] <= i) {
      dp[i] = dp[i] < dp[i - coins[j]] + 1 ? dp[i] : dp[i - coins[j]] + 1;
    }
  }
}
printf("%d\n", dp[amount]);

medium

A. 3

B. 4

C. 2

D. 5

Solution

1. Step 1: Trace dp array updates for amount 6

   Calculate minimum coins for each amount: - dp[0]=0 - dp[1]=1 (1 coin) - dp[2]=2 (1+1) - dp[3]=1 (3 coin) - dp[4]=1 (4 coin) - dp[5]=2 (4+1) - dp[6]=2 (3+3)

2. Step 2: Final dp[6] value

   dp[6] = 2, minimum coins needed to make 6.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Minimum coins for 6 = 2 [OK]

Hint: Trace dp array from 0 to amount stepwise [OK]

Common Mistakes:

• Forgetting to update dp[i] with minimum value
• Incorrectly handling coin indices
• Assuming dp[i] equals i always

4. Identify the bug in this code snippet for Coin Change Minimum Coins:

int coins[] = {1, 2, 5};
int amount = 11;
int dp[12];
for(int i=0; i<=amount; i++) dp[i] = 0;
dp[0] = 0;
for(int i=1; i<=amount; i++) {
  for(int j=0; j<3; j++) {
    if(coins[j] <= i) {
      dp[i] = dp[i - coins[j]] + 1;
    }
  }
}
printf("%d\n", dp[amount]);

medium

A. dp array is not initialized with large values

B. dp[0] should be set to 1

C. The inner loop should run from j=1 to 3

D. The condition coins[j] <= i is incorrect

Solution

1. Step 1: Check dp initialization

   dp array is initialized with zeros, which incorrectly suggests zero coins needed for all amounts.

2. Step 2: Correct initialization needed

   dp should be initialized with a large number (like INT_MAX) to represent unreachable states, then updated with minimum coins.

3. Final Answer:

   dp array is not initialized with large values -> Option A

4. Quick Check:

   Initialize dp with max values to find minimum [OK]

Hint: Initialize dp with large values, not zeros [OK]

Common Mistakes:

• Setting dp to zero causes wrong minimum calculations
• Not updating dp[i] with minimum of previous values
• Incorrect loop ranges

5. Given coins = {2, 5, 10} and amount = 7, what is the minimum number of coins needed? If it is not possible, return -1. Use the Coin Change Minimum Coins logic.

hard

A. -1

B. 2

C. 3

D. 1

Solution

1. Step 1: Check possible combinations for amount 7

   Coins are 2, 5, 10. Possible sums: - 2+5=7 (2 coins) - 5+2=7 (2 coins) - 10 is too large So minimum coins = 2.

2. Step 2: Confirm no smaller combination

   Since 1-coin denominations are not available, 2 coins is minimum.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   7 = 2 + 5, minimum coins = 2 [OK]

Hint: Try all coin combos to find minimum or detect impossibility [OK]

Common Mistakes:

• Assuming 7 cannot be formed without 1-coin denomination
• Returning -1 incorrectly
• Ignoring order of coins