Coin Change Minimum Coins in DSA Java - 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

What does a dynamic programming solution for Coin Change Minimum Coins use to store intermediate results?

An array where each index represents the minimum coins needed to make that amount.

intermediate

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

To represent that initially, the amount cannot be made with the given coins until proven otherwise.

beginner

In the Coin Change Minimum Coins problem, what does dp[0] = 0 represent?

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

intermediate

What is the time complexity of the dynamic programming solution for Coin Change Minimum Coins with n coins and amount m?

O(n * m), where n is the number of coin types and m is the amount.

What does the dp array store in the Coin Change Minimum Coins problem?

A Minimum coins needed for each amount

B Maximum coins needed for each amount

C Number of coin types

D Sum of all coin values

The dp array stores the minimum number of coins needed to make each amount.

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

A -1

B 1

C amount

D 0

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

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

A 0

B amount

C -1

D 1

Return -1 to indicate the amount cannot be made with given coins.

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

A Dynamic Programming

B Brute Force

C Greedy Algorithm

D Divide and Conquer

Dynamic Programming efficiently solves this by building solutions for smaller amounts.

What does it mean if dp[amount] is still a large number after processing?

A Amount is zero

B Amount cannot be formed with given coins

C Amount can be formed with one coin

D Coins are unlimited

A large number means no valid combination was found to form the amount.

Explain how dynamic programming is used to find the minimum number of coins for a given amount.

Think about building the solution from smaller amounts to the target amount.

Describe the base case and initialization steps in the Coin Change Minimum Coins dynamic programming solution.

Focus on how the dp array is prepared before the main loop.

Practice

1. What does the Coin Change Minimum Coins problem ask you to find?
Given coins of different denominations and a total amount, find the minimum number of coins needed to make that amount.

easy

A. Minimum number of coins needed to make the amount

B. Maximum number of coins needed to make the amount

C. Number of ways to make the amount

D. Sum of all coin denominations

Solution

1. Step 1: Understand the problem statement

   The problem asks for the least number of coins needed to reach a given amount using available coin denominations.

2. Step 2: Differentiate from similar problems

   It is not about counting ways or maximum coins, but the minimum coins required.

3. Final Answer:

   Minimum number of coins needed to make the amount -> Option A

4. Quick Check:

   Minimum coins = minimum coins needed [OK]

Hint: Focus on 'minimum coins' needed, not ways or sums [OK]

Common Mistakes:

• Confusing minimum coins with number of ways
• Thinking about maximum coins instead of minimum
• Ignoring the total amount requirement

2. Which of the following Java method signatures correctly represents a function to find minimum coins for a given amount?

public static int minCoins(int[] coins, int amount)

easy

A. public int minCoins(int coins, int[] amount)

B. public static int minCoins(int[] coins, int amount)

C. static void minCoins(int[] coins, int amount)

D. public static int minCoins(int amount, int[] coins)

Solution

1. Step 1: Check parameter types and order

   The method should accept an array of coin denominations and an integer amount, in that order.

2. Step 2: Check return type and modifiers

   It should return an int representing minimum coins and be static for utility use.

3. Final Answer:

   public static int minCoins(int[] coins, int amount) -> Option B

4. Quick Check:

   Correct signature = public static int minCoins(int[] coins, int amount) [OK]

Hint: Array first, amount second; return int minimum coins [OK]

Common Mistakes:

• Swapping parameter order
• Wrong return type (void instead of int)
• Missing static keyword if utility method

3. What is the output of the following Java code?

public class CoinChange {
    public static int minCoins(int[] coins, int amount) {
        int max = amount + 1;
        int[] dp = new int[amount + 1];
        java.util.Arrays.fill(dp, max);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (coin <= i) {
                    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }

    public static void main(String[] args) {
        int[] coins = {1, 3, 4};
        int amount = 6;
        System.out.println(minCoins(coins, amount));
    }
}

medium

A. -1

B. 3

C. 1

D. 2

Solution

1. Step 1: Understand the DP array filling

   dp[i] stores minimum coins for amount i. Initialized with max (7 here) except dp[0]=0.

2. Step 2: Calculate dp values for amount=6

   Using coins {1,3,4}: - dp[1] = 1 (1 coin of 1) - dp[2] = 2 (1+1) - dp[3] = 1 (coin 3) - dp[4] = 1 (coin 4) - dp[5] = 2 (4+1) - dp[6] = min(dp[6-1]+1=dp[5]+1=3, dp[6-3]+1=dp[3]+1=2, dp[6-4]+1=dp[2]+1=3) = 2

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Minimum coins for 6 = 2 [OK]

Hint: Trace dp array updates for each amount [OK]

Common Mistakes:

• Not initializing dp array correctly
• Returning dp[amount] without checking if solution exists
• Confusing dp indices or coin subtraction

4. Find the error in the following Java code snippet for minimum coin change:

public static int minCoins(int[] coins, int amount) {
    int[] dp = new int[amount + 1];
    Arrays.fill(dp, Integer.MAX_VALUE);
    dp[0] = 0;
    for (int i = 1; i <= amount; i++) {
        for (int coin : coins) {
            if (coin <= i) {
                dp[i] = Math.min(dp[i], dp[i - coin] + 1);
            }
        }
    }
    return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
}

medium

A. Using Integer.MAX_VALUE causes overflow when adding 1

B. dp array size should be amount, not amount + 1

C. dp[0] should be Integer.MAX_VALUE, not 0

D. The for loop should start from 0, not 1

Solution

1. Step 1: Identify the problem with Integer.MAX_VALUE

   Adding 1 to Integer.MAX_VALUE causes integer overflow, leading to incorrect dp values.

2. Step 2: Explain correct initialization

   Use a large number less than Integer.MAX_VALUE (like amount+1) to avoid overflow when adding 1.

3. Final Answer:

   Using Integer.MAX_VALUE causes overflow when adding 1 -> Option A

4. Quick Check:

   Integer.MAX_VALUE + 1 overflows [OK]

Hint: Avoid Integer.MAX_VALUE for dp initialization to prevent overflow [OK]

Common Mistakes:

• Ignoring overflow when adding 1 to max value
• Wrong dp array size
• Incorrect dp[0] initialization

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

medium

A. 1

B. 3

C. 2

D. -1

Solution

1. Step 1: Check possible combinations for amount 7

   Coins: 2, 5, 10 - 2+5=7 (2 coins) - 5+2=7 (2 coins) - 10 is too big But 2+5=7 is valid, so minimum coins = 2.

2. Step 2: Verify if amount 7 can be formed

   Actually, 2+5=7 is valid, so minimum coins = 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   7 = 5 + 2 = 2 coins [OK]

Hint: Try largest coin first, then smaller coins [OK]

Common Mistakes:

• Assuming no solution without checking combinations
• Ignoring order of coins
• Returning -1 prematurely