Coin Change Minimum Coins in DSA Rust - Execution Trace

Concept Flow - Coin Change Minimum Coins

Start with amount = target

↓

Initialize dp array with large values

↓

Set dp[0

↓

For each coin

↓

For each amount from coin to target

↓

Update dp[amount

↓

After filling dp, check dp[target

↓

If dp[target

↓

End

We build a table dp where dp[i] stores the minimum coins needed for amount i. We update dp by trying each coin for all amounts up to target.

Execution Sample

DSA Rust

fn coin_change(coins: &[i32], amount: i32) -> i32 {
    let mut dp = vec![amount + 1; (amount + 1) as usize];
    dp[0] = 0;
    for &coin in coins {
        for x in coin..=amount {
            dp[x as usize] = dp[x as usize].min(dp[(x - coin) as usize] + 1);
        }
    }
    if dp[amount as usize] > amount { -1 } else { dp[amount as usize] }
}

This code finds the minimum number of coins needed to make the target amount using given coin denominations.

Execution Table

Step	Operation	Amount Index	Coin Used	dp Array State	Explanation
0	Initialize dp	-	-	[0, 8, 8, 8, 8, 8, 8, 8]	dp[0]=0, others set to amount+1=8 (no solution yet)
1	Use coin 1	1	1	[0, 1, 8, 8, 8, 8, 8, 8]	dp[1] = min(8, dp[0]+1)=1
2	Use coin 1	2	1	[0, 1, 2, 8, 8, 8, 8, 8]	dp[2] = min(8, dp[1]+1)=2
3	Use coin 1	3	1	[0, 1, 2, 3, 8, 8, 8, 8]	dp[3] = min(8, dp[2]+1)=3
4	Use coin 1	4	1	[0, 1, 2, 3, 4, 8, 8, 8]	dp[4] = min(8, dp[3]+1)=4
5	Use coin 1	5	1	[0, 1, 2, 3, 4, 5, 8, 8]	dp[5] = min(8, dp[4]+1)=5
6	Use coin 1	6	1	[0, 1, 2, 3, 4, 5, 6, 8]	dp[6] = min(8, dp[5]+1)=6
7	Use coin 1	7	1	[0, 1, 2, 3, 4, 5, 6, 7]	dp[7] = min(8, dp[6]+1)=7
8	Use coin 3	3	3	[0, 1, 2, 1, 4, 8, 8, 8]	dp[3] = min(3, dp[0]+1)=1 (better)
9	Use coin 3	4	3	[0, 1, 2, 1, 2, 8, 8, 8]	dp[4] = min(4, dp[1]+1)=2
10	Use coin 3	5	3	[0, 1, 2, 1, 2, 3, 8, 8]	dp[5] = min(8, dp[2]+1)=3
11	Use coin 3	6	3	[0, 1, 2, 1, 2, 3, 2, 8]	dp[6] = min(8, dp[3]+1)=2
12	Use coin 3	7	3	[0, 1, 2, 1, 2, 3, 2, 3]	dp[7] = min(8, dp[4]+1)=3
13	Use coin 4	4	4	[0, 1, 2, 1, 1, 3, 2, 3]	dp[4] = min(2, dp[0]+1)=1 (better)
14	Use coin 4	5	4	[0, 1, 2, 1, 1, 2, 2, 3]	dp[5] = min(3, dp[1]+1)=2
15	Use coin 4	6	4	[0, 1, 2, 1, 1, 2, 2, 3]	dp[6] = min(2, dp[2]+1)=2 (no change)
16	Use coin 4	7	4	[0, 1, 2, 1, 1, 2, 2, 2]	dp[7] = min(3, dp[3]+1)=2 (better)
17	End	-	-	[0, 1, 2, 1, 1, 2, 2, 2]	dp[7]=2 means minimum 2 coins for amount 7

💡 Finished updating dp array; dp[7] = 2 is minimum coins needed

Variable Tracker

Variable	Start	After Step 7	After Step 12	After Step 16	Final
dp	[0, 8, 8, 8, 8, 8, 8, 8]	[0, 1, 2, 3, 4, 5, 6, 7]	[0, 1, 2, 1, 2, 3, 2, 3]	[0, 1, 2, 1, 1, 2, 2, 2]	[0, 1, 2, 1, 1, 2, 2, 2]
coin	-	1	3	4	-
amount index	-	7	7	7	-

Key Moments - 3 Insights

Why do we initialize dp array with amount+1?

Because amount+1 is impossible to reach with given coins, it acts like infinity. See step 0 in execution_table where dp is set to 8 (amount+1).

Why do we set dp[0] = 0 at the start?

Because zero coins are needed to make amount 0. This is the base case that allows building solutions for larger amounts. See step 0 in execution_table.

How does using coin 4 improve dp[4] from 2 to 1?

At step 13, dp[4] is updated to min(2, dp[0]+1) = 1 because using one coin of 4 directly makes amount 4 with fewer coins. See execution_table row 13.

Visual Quiz - 3 Questions

Look at the execution_table at step 10. What is dp[5] after using coin 3?

A 2

B 3

C 5

D 1

💡 Hint

Check the dp Array State column at step 10 in execution_table.

At which step does dp[4] first become 1?

A Step 13

B Step 9

C Step 4

D Step 16

💡 Hint

Look for dp[4] updates in execution_table rows; step 13 shows dp[4] = 1.

If we remove coin 1 from coins, what happens to dp[1] after step 1?

A dp[1] becomes 0

B dp[1] becomes 1

C dp[1] stays 8 (no update)

D dp[1] becomes -1

💡 Hint

Without coin 1, dp[1] cannot be updated from initial large value; see initialization in variable_tracker.

Concept Snapshot

Coin Change Minimum Coins:
- Use dp array of size amount+1, dp[0]=0
- Initialize others with amount+1 (infinity)
- For each coin, update dp[x] = min(dp[x], dp[x-coin]+1)
- dp[amount] gives min coins or -1 if no solution
- Bottom-up dynamic programming approach

Full Transcript

This visualization shows how to find the minimum number of coins needed to make a target amount using given coin denominations. We create a dp array where dp[i] means minimum coins for amount i. Initially, dp[0] is 0 because no coins are needed for zero amount. All other dp values start at amount+1, representing impossible states. We then try each coin for all amounts from coin value up to target. For each amount, we update dp[amount] to the minimum between current dp[amount] and dp[amount - coin] + 1. This means we try to use the current coin and add one coin to the best solution for the smaller amount. After processing all coins, dp[target] holds the minimum coins needed or remains large if no solution exists. The execution table traces each update step, showing how dp array improves over time. Key moments clarify why dp is initialized with amount+1, why dp[0] is zero, and how using larger coins can improve solutions. The quiz tests understanding of dp updates and effects of removing coins.

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 list

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 interpretation

   Among the options, only the smallest number of coins needed matches the problem's goal.

3. Final Answer:

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

4. Quick Check:

   Minimum coins = smallest count [OK]

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

Common Mistakes:

• Confusing total coin count with minimum coins
• Thinking about coin values instead of counts
• Ignoring the target amount

2. Which Rust code snippet correctly initializes a vector of coin denominations for the coin change problem?

easy

A. let coins = {1, 2, 5};

B. let coins = [1, 2, 5];

C. let coins = vec![1, 2, 5];

D. let coins = (1, 2, 5);

Solution

1. Step 1: Identify Rust vector initialization syntax

   In Rust, vectors are created using vec![elements], which allows dynamic sizing.

2. Step 2: Compare options

   Only let coins = vec![1, 2, 5]; uses the vec! macro correctly. let coins = [1, 2, 5]; creates an array with fixed size. let coins = (1, 2, 5); creates a tuple. let coins = {1, 2, 5}; has invalid syntax.

3. Final Answer:

   let coins = vec![1, 2, 5]; -> Option C

4. Quick Check:

   vec! macro creates vector [OK]

Hint: Use vec![...] to create dynamic coin lists [OK]

Common Mistakes:

• Using tuple or array instead of vector
• Using curly braces which are invalid here
• Forgetting the vec! macro

3. What is the output of the following Rust code snippet for minimum coins to make amount 11 with coins [1, 2, 5]?

fn min_coins(coins: &[i32], amount: i32) -> i32 {
    let mut dp = vec![amount + 1; (amount + 1) as usize];
    dp[0] = 0;
    for i in 1..=amount {
        for &coin in coins {
            if coin <= i {
                dp[i as usize] = dp[i as usize].min(dp[(i - coin) as usize] + 1);
            }
        }
    }
    if dp[amount as usize] > amount { -1 } else { dp[amount as usize] }
}

fn main() {
    let coins = vec![1, 2, 5];
    println!("{}", min_coins(&coins, 11));
}

medium

A. 5

B. 3

C. 4

D. -1

Solution

1. Step 1: Understand the dynamic programming array dp

   dp[i] stores minimum coins needed for amount i. Initialized with amount+1 (impossible high value), dp[0]=0.

2. Step 2: Trace dp updates for amount 11

   Using coins 1,2,5, minimum coins for 11 is 3 (5+5+1). The code finds dp[11] = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   11 = 5+5+1 uses 3 coins [OK]

Hint: Check dp array for minimum coins at target amount [OK]

Common Mistakes:

• Confusing dp index with coin values
• Not initializing dp[0] to zero
• Returning amount instead of -1 when no solution

4. Identify the bug in this Rust function for minimum coins:

fn min_coins(coins: &[i32], amount: i32) -> i32 {
    let mut dp = vec![0; (amount + 1) as usize];
    for i in 1..=amount {
        dp[i as usize] = i;
        for &coin in coins {
            if coin <= i {
                dp[i as usize] = dp[i as usize].min(dp[(i - coin) as usize] + 1);
            }
        }
    }
    dp[amount as usize]
}

medium

A. Loop range should start from 0, not 1

B. Coins vector should be mutable

C. Missing return statement

D. dp vector initialized with zeros instead of large values

Solution

1. Step 1: Check dp initialization

   dp is initialized with zeros, which causes min comparisons to fail because 0 is less than any valid coin count.

2. Step 2: Correct initialization

   dp should be initialized with a large value (like amount+1) to represent impossible states, allowing min to work correctly.

3. Final Answer:

   dp vector initialized with zeros instead of large values -> Option D

4. Quick Check:

   dp init affects min logic [OK]

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

Common Mistakes:

• Starting loop at 0 instead of 1
• Forgetting to return dp[amount]
• Making coins mutable unnecessarily

5. Given coin denominations [1, 3, 4] and amount 6, what is the minimum number of coins needed? Use the coin change minimum coins approach.

hard

A. 2

B. 3

C. 4

D. -1

Solution

1. Step 1: Analyze possible combinations for amount 6

   Using coins 1,3,4, possible sums: 3+3=6 (2 coins), 4+1+1=6 (3 coins), 1+1+1+1+1+1=6 (6 coins).

2. Step 2: Find minimum coins

   The minimum coins needed is 2 (two 3-value coins).

3. Final Answer:

   2 -> Option A

4. Quick Check:

   6 = 3+3 uses 2 coins [OK]

Hint: Try largest coins first to minimize count [OK]

Common Mistakes:

• Choosing 4+1+1 instead of 3+3
• Assuming no solution exists
• Counting coins incorrectly