Coin Change Minimum Coins in DSA C# - Execution Trace

Concept Flow - Coin Change Minimum Coins

Start with amount

↓

Initialize DP array with max values

↓

Set dp[0

↓

For each coin

↓

For each amount from coin to target

↓

Update dp[amount

↓

After all coins processed

↓

Check dp[target

↓

If dp[target

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 C#

int[] coins = {1, 3, 4};
int amount = 6;
int[] dp = new int[amount + 1];
for (int i = 1; i <= amount; i++) dp[i] = amount + 1;
foreach (int coin in coins)
  for (int i = coin; i <= amount; i++)
    dp[i] = Math.Min(dp[i], dp[i - coin] + 1);

This code finds the minimum number of coins needed to make amount 6 using coins 1, 3, and 4.

Execution Table

Step	Operation	Amount Index (i)	Coin Used	dp Array State	Explanation
0	Initialize dp	-	-	[0, 7, 7, 7, 7, 7, 7]	dp[0]=0, others set to amount+1=7 (infinity)
1	Use coin=1	1	1	[0, 1, 7, 7, 7, 7, 7]	dp[1] = min(7, dp[0]+1)=1
2	Use coin=1	2	1	[0, 1, 2, 7, 7, 7, 7]	dp[2] = min(7, dp[1]+1)=2
3	Use coin=1	3	1	[0, 1, 2, 3, 7, 7, 7]	dp[3] = min(7, dp[2]+1)=3
4	Use coin=1	4	1	[0, 1, 2, 3, 4, 7, 7]	dp[4] = min(7, dp[3]+1)=4
5	Use coin=1	5	1	[0, 1, 2, 3, 4, 5, 7]	dp[5] = min(7, dp[4]+1)=5
6	Use coin=1	6	1	[0, 1, 2, 3, 4, 5, 6]	dp[6] = min(7, dp[5]+1)=6
7	Use coin=3	3	3	[0, 1, 2, 1, 4, 7, 6]	dp[3] = min(3, dp[0]+1)=1
8	Use coin=3	4	3	[0, 1, 2, 1, 2, 7, 6]	dp[4] = min(4, dp[1]+1)=2
9	Use coin=3	5	3	[0, 1, 2, 1, 2, 3, 6]	dp[5] = min(5, dp[2]+1)=3
10	Use coin=3	6	3	[0, 1, 2, 1, 2, 3, 2]	dp[6] = min(6, dp[3]+1)=2
11	Use coin=4	4	4	[0, 1, 2, 1, 1, 3, 2]	dp[4] = min(2, dp[0]+1)=1
12	Use coin=4	5	4	[0, 1, 2, 1, 1, 2, 2]	dp[5] = min(3, dp[1]+1)=2
13	Use coin=4	6	4	[0, 1, 2, 1, 1, 2, 2]	dp[6] = min(2, dp[2]+1)=2 (no change)
14	End	-	-	[0, 1, 2, 1, 1, 2, 2]	Minimum coins for amount 6 is dp[6]=2

💡 All coins processed; dp[6] = 2 means minimum 2 coins needed

Variable Tracker

Variable	Start	After Step 6	After Step 10	After Step 13	Final
dp	[0,7,7,7,7,7,7]	[0,1,2,3,4,5,6]	[0,1,2,1,2,3,2]	[0,1,2,1,1,2,2]	[0,1,2,1,1,2,2]
coin	N/A	1	3	4	4
i (amount index)	N/A	6	6	6	6

Key Moments - 3 Insights

Why do we initialize dp array with amount+1 (7) instead of zero?

Because dp[i] stores minimum coins needed, initializing with amount+1 (which is impossible) acts like infinity, so any real solution will be smaller. See step 0 in execution_table.

Why is dp[0] set to 0 at the start?

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

Why does dp[6] update from 6 to 2 after using coins 3 and 4?

Because using bigger coins reduces the number of coins needed. For example, dp[6] = min(dp[6], dp[6-3]+1) = min(6, dp[3]+1) = min(6,1+1)=2. See steps 10 and 13.

Visual Quiz - 3 Questions

Look at the execution_table at step 7, what is the value of dp[3] after using coin 3?

A 7

B 3

C 1

D 0

💡 Hint

Check the dp Array State column at step 7 in execution_table

At which step does dp[4] first get updated to 1?

A Step 4

B Step 11

C Step 8

D Step 2

💡 Hint

Look for dp[4] value changes in dp Array State column in execution_table

If we remove coin 4 from the coins array, what would be the minimum coins for amount 6?

A 2

B 3

C 6

D 1

💡 Hint

check dp Array State column at step 10 in execution_table

Concept Snapshot

Coin Change Minimum Coins:
- Use dp array where dp[i] = min coins for amount i
- Initialize dp with amount+1 (infinity), dp[0]=0
- For each coin, update dp[i] = min(dp[i], dp[i-coin]+1)
- Result is dp[amount], if <= amount; else no solution
- Bottom-up dynamic programming approach

Full Transcript

The Coin Change Minimum Coins problem finds the least number of coins needed to make a target amount from given coin denominations. We use a dp array where dp[i] holds the minimum coins needed for amount i. Initially, dp[0] is zero because no coins are needed for zero amount, and all other dp values are set to amount+1 to represent infinity. We then iterate over each coin and for each amount from coin value up to target amount, we update dp[i] to the minimum between its current value and dp[i - coin] + 1. This process builds up solutions for all amounts until we reach the target. The final answer is dp[amount], which gives the minimum coins needed or indicates no solution if it remains infinity.

Practice

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

easy

A. The smallest number of coins needed to make the amount

B. The total value of all coins

C. The maximum number of coins that can be used

D. The number of different coin types

Solution

1. Step 1: Understand the problem goal

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

2. Step 2: Clarify what is asked

   We want the smallest count of coins that sum exactly to the amount, not total value or max coins.

3. Final Answer:

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

4. Quick Check:

   Minimum coins = smallest count [OK]

Hint: Focus on minimum count, not total value or types [OK]

Common Mistakes:

• Confusing minimum coins with total coin value
• Thinking about maximum coins instead of minimum
• Counting coin types instead of coin quantity

2. Which of the following C# code snippets correctly initializes an array to store minimum coins for amounts 0 to 5, setting 0 coins for amount 0 and a large number for others?

easy

A. int[] minCoins = new int[5] {0, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue};

B. int[] minCoins = new int[6] {0, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue};

C. int[] minCoins = new int[6]; for(int i=1; i<=5; i++) minCoins[i] = 0;

D. int[] minCoins = new int[6]; minCoins[0] = int.MaxValue;

Solution

1. Step 1: Check array size and initialization

   We need an array of size 6 for amounts 0 to 5. Amount 0 requires 0 coins, others start with a large number.

2. Step 2: Verify each option

   int[] minCoins = new int[6] {0, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue}; correctly initializes size 6 with 0 at index 0 and int.MaxValue for others. Others either have wrong size or wrong initialization.

3. Final Answer:

   int[] minCoins = new int[6] {0, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue}; -> Option B

4. Quick Check:

   Array size 6, 0 at start, max values after [OK]

Hint: Array size = amount + 1, 0 coins for amount 0 [OK]

Common Mistakes:

• Using wrong array size
• Setting 0 coins for all amounts
• Setting int.MaxValue for amount 0

3. What is the output of the following C# code?

int[] coins = {1, 3, 4};
int amount = 6;
int[] minCoins = new int[amount + 1];
for (int i = 1; i <= amount; i++) minCoins[i] = int.MaxValue;

for (int i = 1; i <= amount; i++) {
    foreach (int coin in coins) {
        if (coin <= i && minCoins[i - coin] != int.MaxValue) {
            minCoins[i] = Math.Min(minCoins[i], minCoins[i - coin] + 1);
        }
    }
}
Console.WriteLine(minCoins[amount]);

medium

A. 3

B. int.MaxValue

C. 4

D. 2

Solution

1. Step 1: Initialize minCoins array

   minCoins[0] = 0 (default), others set to int.MaxValue initially.

2. Step 2: Calculate minCoins for amounts 1 to 6

   For amount 6, check coins 1,3,4:
   - Using coin 1: minCoins[5] + 1
   - Using coin 3: minCoins[3] + 1
   - Using coin 4: minCoins[2] + 1
   Calculations:
   minCoins[1] = 1 (coin 1)
   minCoins[2] = 2 (1+1)
   minCoins[3] = 1 (coin 3)
   minCoins[4] = 1 (coin 4)
   minCoins[5] = 2 (3+1 or 4+1)
   minCoins[6] = min(minCoins[5]+1, minCoins[3]+1, minCoins[2]+1) = min(3, 2, 3) = 2

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Minimum coins for 6 = 2 (3+3 or 4+1+1) [OK]

Hint: Trace minCoins array updates stepwise [OK]

Common Mistakes:

• Not updating minCoins correctly
• Ignoring int.MaxValue checks
• Miscounting coin combinations

4. Identify the error in this C# code snippet for minimum coin change and select the fix:

int[] coins = {1, 2, 5};
int amount = 11;
int[] minCoins = new int[amount + 1];
for (int i = 1; i <= amount; i++) minCoins[i] = int.MaxValue;

for (int i = 1; i <= amount; i++) {
    foreach (int coin in coins) {
        if (coin < i && minCoins[i - coin] != int.MaxValue) {
            minCoins[i] = Math.Min(minCoins[i], minCoins[i - coin] + 1);
        }
    }
}
Console.WriteLine(minCoins[amount]);

medium

A. Change loop to start from i = 0 instead of 1

B. Initialize minCoins array with zeros instead of int.MaxValue

C. Change 'coin < i' to 'coin <= i' in the if condition

D. Remove the if condition inside the foreach loop

Solution

1. Step 1: Analyze the if condition

   The condition 'coin < i' excludes cases where coin equals i, missing direct matches.

2. Step 2: Correct the condition

   Changing to 'coin <= i' allows using coins that exactly match the current amount, fixing the logic.

3. Final Answer:

   Change 'coin < i' to 'coin <= i' in the if condition -> Option C

4. Quick Check:

   Include coin equal to amount for correct min coins [OK]

Hint: Use <= to include exact coin matches [OK]

Common Mistakes:

• Using < instead of <= excludes valid coins
• Initializing minCoins incorrectly
• Removing necessary condition checks

5. You have coin denominations {2, 5, 10} and want to find the minimum coins for amount 7. Which approach correctly solves this using dynamic programming in C#?

hard

A. Initialize minCoins with 0 for amount 0 and int.MaxValue for others; update minCoins[i] by checking coins <= i and minCoins[i - coin] + 1

B. Initialize minCoins with zeros for all amounts; update minCoins[i] by adding 1 for each coin

C. Use recursion without memoization to find minimum coins

D. Initialize minCoins with int.MaxValue for all amounts including 0; update minCoins[i] by subtracting coin values

Solution

1. Step 1: Proper initialization

   Set minCoins[0] = 0 because zero coins are needed for amount 0; others set to int.MaxValue to represent unreachable amounts initially.

2. Step 2: Update minCoins using dynamic programming

   For each amount i, check each coin less than or equal to i. If minCoins[i - coin] is reachable, update minCoins[i] to the minimum of current value and minCoins[i - coin] + 1.

3. Final Answer:

   Initialize minCoins with 0 for amount 0 and int.MaxValue for others; update minCoins[i] by checking coins <= i and minCoins[i - coin] + 1 -> Option A

4. Quick Check:

   DP with correct initialization and update finds minimum coins [OK]

Hint: Initialize 0 for amount 0, update with min of previous +1 [OK]

Common Mistakes:

• Initializing all minCoins to zero
• Using recursion without memoization causing inefficiency
• Incorrectly updating minCoins by subtraction