Coin Change Total Number of Ways in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Coin Change problem about?

It is about finding how many different ways you can make a certain amount of money using given coin denominations.

intermediate

What does the dynamic programming table represent in the Coin Change problem?

Each cell represents the number of ways to make a certain amount using a subset of coins up to that point.

intermediate

In the Coin Change problem, why do we add dp[i][j - coin] to dp[i-1][j]?

Because dp[i-1][j] counts ways without the current coin, and dp[i][j - coin] counts ways including the current coin at least once.

beginner

What is the base case for the Coin Change dynamic programming solution?

There is exactly 1 way to make amount 0: by choosing no coins.

beginner

How does the order of coins affect the number of ways in the Coin Change problem?

The order does not affect the count because combinations are counted, not permutations.

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

A Maximum coins used to make amount j using first i coins

B Minimum coins needed to make amount j using first i coins

C Number of ways to make amount j using first i coins

D Number of coins available

dp[i][j] stores the count of ways to make amount j using the first i coins.

What is the initial value of dp[0][0] in the Coin Change DP table?

A 1

B Amount of first coin

C -1

D 0

There is exactly 1 way to make amount 0: by choosing no coins.

If you have coins [1, 2] and amount 3, how many ways are there to make 3?

A 4

B 1

C 3

D 2

Ways: (1+1+1), (1+2). So 2 ways.

Why do we use a 2D dp array in the Coin Change problem?

A To track ways for each amount and coin subset

B To store coin values

C To sort coins

D To count total coins

2D dp helps track ways for each amount using subsets of coins.

Which of these is NOT a valid step in the Coin Change DP solution?

A Include current coin and reduce amount

B Multiply ways of previous coins

C Exclude current coin and move to next

D Initialize dp[0][0] to 1

We add ways, not multiply, when combining counts.

Explain how dynamic programming solves the Coin Change Total Number of Ways problem.

Think about how to build solutions for smaller amounts and coins.

Describe the difference between counting combinations and permutations in the Coin Change problem.

Focus on whether order of coins matters.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?

easy

A. The total number of ways to make a given amount using unlimited coins of given denominations

B. The minimum number of coins needed to make a given amount

C. The maximum value of coins that can be used to make a given amount

D. The total number of coins available in the input array

Solution

1. Step 1: Understand the problem statement

   The problem asks for counting all possible ways to form a target amount using unlimited coins of given denominations.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, here we count ways, not minimize coins.

3. Final Answer:

   The total number of ways to make a given amount using unlimited coins of given denominations -> Option A

4. Quick Check:

   Count ways, not minimum coins [OK]

Hint: Count combinations, not minimum coins needed [OK]

Common Mistakes:

• Confusing with minimum coins problem
• Thinking about maximum coin value
• Counting coins instead of ways

2. Which of the following Java code snippets correctly initializes the DP array for the Coin Change Total Number of Ways problem for amount 5?

easy

A. int[] dp = new int[6]; dp[0] = 1;

B. int[] dp = new int[5]; dp[0] = 1;

C. int[] dp = new int[6]; dp[1] = 1;

D. int[] dp = new int[5]; dp[1] = 1;

Solution

1. Step 1: Determine DP array size

   For amount 5, dp array size should be 6 to include index 0 to 5.

2. Step 2: Initialize base case

   dp[0] = 1 means one way to make amount 0 (using no coins).

3. Final Answer:

   int[] dp = new int[6]; dp[0] = 1; -> Option A

4. Quick Check:

   dp size = amount+1, dp[0] = 1 [OK]

Hint: DP array size = amount + 1; dp[0] = 1 base case [OK]

Common Mistakes:

• Using dp array size equal to amount instead of amount+1
• Initializing dp[1] instead of dp[0]
• Forgetting to set base case dp[0] = 1

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

int[] coins = {1, 2};
int amount = 3;
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int coin : coins) {
  for (int i = coin; i <= amount; i++) {
    dp[i] += dp[i - coin];
  }
}
System.out.println(dp[amount]);

medium

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0] initially, dp[0] = 1.

2. Step 2: Process coin 1

   For i=1 to 3: dp[1]+=dp[0]=1, dp[2]+=dp[1]=1, dp[3]+=dp[2]=1; dp = [1,1,1,1]

3. Step 3: Process coin 2

   For i=2 to 3: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2; dp = [1,1,2,2]

4. Step 4: Final dp[3]

   dp[3] = 2 means 2 ways: {1+1+1, 1+2}

5. Final Answer:

   2 -> Option C

6. Quick Check:

   Ways to make 3 with coins 1,2 = 2 [OK]

Hint: Add ways for each coin from coin to amount [OK]

Common Mistakes:

• Not updating dp correctly inside nested loops
• Confusing dp[i] with dp[i-1]
• Miscounting combinations

4. Identify the error in the following Java code for Coin Change Total Number of Ways:

int[] coins = {1, 2, 3};
int amount = 4;
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int i = 1; i <= amount; i++) {
  for (int coin : coins) {
    if (i >= coin) {
      dp[i] += dp[i - coin];
    }
  }
}
System.out.println(dp[amount]);

medium

A. The dp array size is too small

B. dp[0] should be initialized to 0

C. The condition i >= coin is incorrect

D. The outer loop should iterate over coins, inner over amounts

Solution

1. Step 1: Analyze loop order

   Outer loop iterates over amounts, inner over coins, which counts permutations, not combinations.

2. Step 2: Correct loop order

   Outer loop should iterate over coins, inner over amounts to count combinations correctly.

3. Final Answer:

   The outer loop should iterate over coins, inner over amounts -> Option D

4. Quick Check:

   Loop order affects counting ways [OK]

Hint: Outer loop coins, inner loop amounts for combinations [OK]

Common Mistakes:

• Swapping loop order causing wrong counts
• Incorrect dp initialization
• Misusing conditions inside loops

5. Given coins = {1, 3, 5} and amount = 7, how many total ways are there to make amount 7 using unlimited coins? Use the Coin Change Total Number of Ways approach.

hard

A. 5

B. 4

C. 6

D. 7

Solution

1. Step 1: Initialize dp array

   dp size = 8, dp[0] = 1, others 0.

2. Step 2: Process coin 1

   Update dp for i=1 to 7: dp[i] += dp[i-1], dp = [1,1,1,1,1,1,1,1]

3. Step 3: Process coin 3

   For i=3 to 7: dp[3]+=dp[0]=2, dp[4]+=dp[1]=2, dp[5]+=dp[2]=2, dp[6]+=dp[3]=3, dp[7]+=dp[4]=3; dp = [1,1,1,2,2,2,3,3]

4. Step 4: Process coin 5

   For i=5 to 7: dp[5]+=dp[0]=3, dp[6]+=dp[1]=4, dp[7]+=dp[2]=4; dp = [1,1,1,2,2,3,4,4]

5. Final Answer:

   4 -> Option B

6. Quick Check:

   Count ways with coins 1,3,5 for amount 7 = 4 [OK]

Hint: Add ways coin by coin updating dp array [OK]

Common Mistakes:

• Miscounting dp updates for each coin
• Forgetting base case dp[0] = 1
• Mixing permutations with combinations