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

Recall & Review

beginner

What does the 'Coin Change Total Number of Ways' problem ask you to find?

It asks for the total number of different ways to make a target amount using given coin denominations, where coins can be used unlimited times.

beginner

In the coin change problem, why do we use dynamic programming?

Because it helps us build the solution step-by-step by storing results of smaller amounts to avoid repeated calculations, making the solution efficient.

beginner

What does the dp array represent in the coin change total number of ways solution?

dp[i] represents the total number of ways to make the amount i using the given coins.

intermediate

How do you update the dp array when processing a coin of value c?

For each amount i from c to target, add dp[i - c] to dp[i], because ways to make i include ways to make i - c plus this coin.

beginner

What is the base case for the dp array in the coin change total number of ways problem?

dp[0] = 1, meaning there is exactly one way to make amount 0 — by choosing no coins.

What does dp[0] equal in the coin change total number of ways problem?

A 1

B 0

C Number of coins

D Undefined

dp[0] = 1 because there is exactly one way to make amount 0: by choosing no coins.

When updating dp for a coin c, which dp value do you add to dp[i]?

A dp[i + c]

B dp[c - i]

C dp[i - c]

D dp[i]

We add dp[i - c] to dp[i] because ways to make i include ways to make i - c plus the coin c.

If coins = [1, 2] and target = 3, how many ways are there to make 3?

A 3

B 2

C 1

D 4

There are 2 ways: (1+1+1) and (1+2). Order does not matter, so (2+1) is the same as (1+2).

Which approach is best for solving the coin change total number of ways problem?

A Dynamic programming

B Sorting coins only

C Backtracking without memoization

D Greedy algorithm

Dynamic programming efficiently counts all ways by building up solutions for smaller amounts.

In the dp array, what does dp[i] represent?

A Sum of coin values used

B Maximum coins to make amount i

C Minimum coins to make amount i

D Total number of ways to make amount i

dp[i] stores the total number of ways to make amount i using the given coins.

Explain how dynamic programming solves the coin change total number of ways problem step-by-step.

Think about building solutions from smaller amounts to bigger amounts.

Describe the role of the base case in the coin change total number of ways problem and why dp[0] is set to 1.

Why is amount zero special in counting ways?

Practice

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

easy

A. The minimum number of coins needed to make a target amount.

B. The total number of ways to make a target amount using given coin denominations.

C. The maximum value of coins that can be used without exceeding the target.

D. The total sum of all coin denominations.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible combinations of coins that sum up to the target amount.

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 target amount using given coin denominations. -> Option B

4. Quick Check:

   Count ways, not minimum coins [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins needed
• Thinking about maximum coin value instead of count
• Ignoring that order of coins does not matter

2. Which of the following Rust code snippets correctly initializes a vector to store ways to make amounts up to `target` with 0 ways initially except for 0 amount which has 1 way?

easy

A. let mut ways = vec![0; target + 1]; ways[0] = 1;

B. let mut ways = vec![1; target]; ways[0] = 0;

C. let mut ways = vec![0; target]; ways[0] = 1;

D. let mut ways = vec![1; target + 1]; ways[0] = 0;

Solution

1. Step 1: Initialize vector size correctly

   Vector size must be target + 1 to include amount 0 to target.

2. Step 2: Set base case for amount 0

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

3. Final Answer:

   let mut ways = vec![0; target + 1]; ways[0] = 1; -> Option A

4. Quick Check:

   Vector size = target+1, ways[0] = 1 [OK]

Hint: ways vector size = target+1, ways[0] = 1 [OK]

Common Mistakes:

• Using vector size target instead of target+1
• Setting ways[0] to 0 instead of 1
• Initializing all elements to 1 incorrectly

3. What is the output of the following Rust code snippet for coins = [1, 2] and target = 3?

let coins = vec![1usize, 2usize];
let target: usize = 3;
let mut ways = vec![0; target + 1];
ways[0] = 1;
for &coin in &coins {
    for amount in coin..=target {
        ways[amount] += ways[amount - coin];
    }
}
println!("{}", ways[target]);

medium

A. 4

B. 1

C. 2

D. 3

Solution

1. Step 1: Calculate ways using coin 1

   ways after coin 1: ways[0]=1, ways[1]=1, ways[2]=1, ways[3]=1

2. Step 2: Update ways using coin 2

   ways[2] += ways[0] => 1+1=2; ways[3] += ways[1] => 1+1=2; final ways = [1,1,2,2]

3. Final Answer:

   2 -> Option C

4. Quick Check:

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

Hint: Add ways for each coin incrementally [OK]

Common Mistakes:

• Not updating ways correctly for each coin
• Confusing ways count with minimum coins
• Off-by-one errors in loops

4. Identify the bug in this Rust code snippet for coin change ways calculation:

let coins = vec![1usize, 2usize, 3usize];
let target: usize = 4;
let mut ways = vec![0; target + 1];
ways[0] = 1;
for &coin in &coins {
    for amount in 1usize..=target {
        if amount >= coin {
            ways[amount] += ways[amount - coin];
        }
    }
}
println!("{}", ways[target]);

medium

A. The inner loop should start from coin, not 1.

B. ways[0] should be initialized to 0.

C. coins vector should be mutable.

D. The outer loop should iterate over amounts, not coins.

Solution

1. Step 1: Check inner loop range

   Starting inner loop from 1 causes unnecessary checks; it should start from coin value to avoid redundant iterations.

2. Step 2: Confirm initialization and loops

   ways[0] = 1 is correct; coins vector does not need to be mutable; outer loop over coins is correct.

3. Final Answer:

   The inner loop should start from coin, not 1. -> Option A

4. Quick Check:

   Inner loop start = coin [OK]

Hint: Inner loop starts at coin value to optimize [OK]

Common Mistakes:

• Starting inner loop from 1 instead of coin
• Incorrect initialization of ways[0]
• Misunderstanding loop roles

5. Given coins = [1, 3, 5] and target = 7, how many total ways are there to make the target amount? Use the coin change total number of ways approach.

hard

A. 6

B. 5

C. 7

D. 4

Solution

1. Step 1: Initialize ways vector

   ways = [1,0,0,0,0,0,0,0] for amounts 0 to 7.

2. Step 2: Update ways with coin 1

   ways after coin 1: ways[i] = 1 for all i from 0 to 7 (only using 1s).

3. Step 3: Update ways with coin 3

   ways[3]+=ways[0]=1, ways[4]+=ways[1]=1, ways[5]+=ways[2]=1, ways[6]+=ways[3]=2, ways[7]+=ways[4]=2; ways now: [1,1,1,2,2,2,3,3]

4. Step 4: Update ways with coin 5

   ways[5] += ways[0] => 2 + 1 = 3; ways[6] += ways[1] => 3 + 1 = 4; ways[7] += ways[2] => 3 + 1 = 4; final ways: [1,1,1,2,2,3,4,4]

5. Final Answer:

   4 -> Option D

6. Quick Check:

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

Hint: Add ways incrementally for each coin [OK]

Common Mistakes:

• Not updating ways correctly for each coin
• Forgetting to include ways[0] = 1
• Mixing up order of loops