Coin Change Total Number of Ways in DSA C++ - Execution Trace

Concept Flow - Coin Change Total Number of Ways

Start with empty ways array

↓

For each coin

↓

For each amount from coin to target

↓

Update ways[amount

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Repeat for all coins

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Result: ways[target

We start with an array representing ways to make each amount. For each coin, we update this array by adding ways to make amounts that include this coin. Finally, the target amount's value is the total ways.

Execution Sample

DSA C++

int ways[6] = {1,0,0,0,0,0};
int coins[] = {1,2,3};
for (int c=0; c<3; c++) {
  for (int amt=coins[c]; amt<=5; amt++) {
    ways[amt] += ways[amt - coins[c]];
  }
}

This code calculates the number of ways to make amounts 0 to 5 using coins 1, 2, and 3.

Execution Table

Step	Operation	Current Coin	Amount Updated	ways Array State	Explanation
0	Initialize ways array	N/A	N/A	[1,0,0,0,0,0]	ways[0] = 1 means one way to make 0 amount (no coins)
1	Process coin 1	1	1	[1,1,0,0,0,0]	ways[1] += ways[0] → 0 + 1 = 1
2	Process coin 1	1	2	[1,1,1,0,0,0]	ways[2] += ways[1] → 0 + 1 = 1
3	Process coin 1	1	3	[1,1,1,1,0,0]	ways[3] += ways[2] → 0 + 1 = 1
4	Process coin 1	1	4	[1,1,1,1,1,0]	ways[4] += ways[3] → 0 + 1 = 1
5	Process coin 1	1	5	[1,1,1,1,1,1]	ways[5] += ways[4] → 0 + 1 = 1
6	Process coin 2	2	2	[1,1,2,1,1,1]	ways[2] += ways[0] → 1 + 1 = 2
7	Process coin 2	2	3	[1,1,2,2,1,1]	ways[3] += ways[1] → 1 + 1 = 2
8	Process coin 2	2	4	[1,1,2,2,3,1]	ways[4] += ways[2] → 1 + 2 = 3
9	Process coin 2	2	5	[1,1,2,2,3,3]	ways[5] += ways[3] → 1 + 2 = 3
10	Process coin 3	3	3	[1,1,2,3,3,3]	ways[3] += ways[0] → 2 + 1 = 3
11	Process coin 3	3	4	[1,1,2,3,4,3]	ways[4] += ways[1] → 3 + 1 = 4
12	Process coin 3	3	5	[1,1,2,3,4,5]	ways[5] += ways[2] → 3 + 2 = 5
13	End	N/A	N/A	[1,1,2,3,4,5]	All coins processed, ways[5] = 5 total ways

💡 All coins processed and amounts updated up to target 5

Variable Tracker

Variable	Start	After Step 5	After Step 9	After Step 12	Final
ways	[1,0,0,0,0,0]	[1,1,1,1,1,1]	[1,1,2,2,3,3]	[1,1,2,3,4,5]	[1,1,2,3,4,5]
Current Coin	N/A	1	2	3	N/A
Amount Updated	N/A	5	5	5	N/A

Key Moments - 3 Insights

Why does ways[0] start at 1 instead of 0?

ways[0] = 1 means there is exactly one way to make amount 0: by choosing no coins. This is the base case and allows building up ways for other amounts (see Step 0 in execution_table).

Why do we add ways[amount - coin] to ways[amount]?

Adding ways[amount - coin] means we count all ways to make the smaller amount (amount - coin) and then add the current coin to reach amount. This accumulates all combinations including the current coin (see Steps 1-12).

Why do we process coins one by one instead of all at once?

Processing coins one by one ensures we count unique combinations without duplicates. Each coin updates ways array cumulatively, avoiding counting same sets multiple times (see coin changes at Steps 1-5, 6-9, 10-12).

Visual Quiz - 3 Questions

Look at the execution_table at Step 6. What is the value of ways[2] after processing coin 2?

A 1

B 2

C 0

D 3

💡 Hint

Check the ways array state column at Step 6 in execution_table

At which step does ways[5] first become greater than 1?

A Step 9

B Step 5

C Step 12

D Step 1

💡 Hint

Look at ways[5] values in execution_table rows for Steps 5, 9, and 12

If we remove coin 1 from the coins array, how would ways[5] change after all processing?

A 3

B 6

C 1

D 0

💡 Hint

Without coin 1 (start [1,0,0,0,0,0]): After coin 2 [1,0,1,0,1,0] (ways[5]=0); after coin 3 ways[5] += ways[2] → 0 + 1 = 1 (one 2 + one 3)

Concept Snapshot

Coin Change Total Number of Ways:
- Use a ways array of size (target+1), ways[0]=1
- For each coin, update ways[amount] += ways[amount - coin]
- After all coins, ways[target] holds total combinations
- Counts combinations, order-independent
- Dynamic programming approach, O(coins*target) time

Full Transcript

This visualization shows how to find the total number of ways to make a target amount using given coins. We start with an array 'ways' where ways[0] = 1, meaning one way to make zero amount. For each coin, we update the ways array for amounts from coin value up to target by adding ways[amount - coin]. This accumulates all combinations including the current coin. The execution table traces each update step by step, showing how the ways array changes. Key moments clarify why ways[0] starts at 1, why we add ways[amount - coin], and why coins are processed one by one. The visual quiz tests understanding of array states at specific steps and effects of removing coins. The concept snapshot summarizes the approach as a dynamic programming solution counting combinations efficiently.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?
Given coins and a total amount, how many ways can you make that amount?

easy

A. The total number of ways to make the amount using given coins

B. The minimum number of coins needed to make the amount

C. The maximum coin value used to make the amount

D. The total sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for the count of different ways to combine coins to reach a target amount.

2. Step 2: Differentiate from similar problems

   This is not about minimum coins or maximum coin value, but total ways to form the amount.

3. Final Answer:

   The total number of ways to make the amount using given coins -> Option A

4. Quick Check:

   Count ways, not minimum or max coin [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins needed
• Thinking it's about maximum coin value
• Mixing sum of coins with number of ways

2. Which of the following is the correct way to initialize the DP array for the Coin Change Total Number of Ways problem in C++?

int amount = 5;
vector<int> dp(amount + 1, 0);
dp[0] = ?;

easy

A. 1

B. 0

C. -1

D. amount

Solution

1. Step 1: Understand base case initialization

   dp[0] represents ways to make amount 0, which is 1 (using no coins).

2. Step 2: Set dp[0] to 1

   Setting dp[0] = 1 allows building solutions for other amounts correctly.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   dp[0] = 1 means one way to make zero amount [OK]

Hint: dp[0] always 1 for coin change ways [OK]

Common Mistakes:

• Setting dp[0] to 0 causes no valid ways
• Using negative values for dp initialization
• Confusing dp array meaning with minimum coins

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

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> coins = {1, 2};
    int amount = 3;
    vector<int> dp(amount + 1, 0);
    dp[0] = 1;
    for (int coin : coins) {
        for (int x = coin; x <= amount; x++) {
            dp[x] += dp[x - coin];
        }
    }
    cout << dp[amount] << endl;
    return 0;
}

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Trace dp array updates for coins {1, 2} and amount 3

   Initially dp = [1,0,0,0]. After coin=1: dp = [1,1,1,1]. After coin=2: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2, so dp = [1,1,2,2].

2. Step 2: Final dp[3] value

   dp[3] = 2 means there are 2 ways to make amount 3 with coins 1 and 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

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

Hint: Add dp[x - coin] for each coin and amount [OK]

Common Mistakes:

• Mixing order of loops causing wrong counts
• Not initializing dp[0] to 1
• Confusing dp[x] update with subtraction

4. Find the error in this C++ code snippet for Coin Change Total Number of Ways:

vector<int> dp(amount + 1, 0);
dp[0] = 1;
for (int x = 1; x <= amount; x++) {
    for (int coin : coins) {
        if (coin <= x) {
            dp[x] += dp[x - coin];
        }
    }
}

medium

A. The dp array size is too small

B. dp[0] should be initialized to 0

C. The condition if (coin <= x) is incorrect

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

Solution

1. Step 1: Analyze loop order for correct counting

   Outer loop must be over coins to avoid counting duplicates; current code counts duplicates.

2. Step 2: Correct loop order

   Change outer loop to for (int coin : coins) and inner loop to for (int x = coin; x <= amount; x++) to count unique combinations.

3. Final Answer:

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

4. Quick Check:

   Loop order affects counting unique ways [OK]

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

Common Mistakes:

• Swapping loops causes counting duplicates
• Incorrect dp[0] initialization
• Wrong dp array size causing out-of-bounds

5. You have coins {1, 3, 5} and want to find the total number of ways to make amount 7. Which approach correctly computes this using dynamic programming?

hard

A. Sort coins descending and greedily subtract largest coin until amount is zero

B. Use a 2D dp array with rows as coins and columns as amounts, fill with minimum coins needed

C. Use a 1D dp array, initialize dp[0]=1, then for each coin, update dp[x] += dp[x - coin] for x from coin to amount

D. Use recursion without memoization to count ways

Solution

1. Step 1: Identify correct DP approach for total ways

   Using 1D dp array with dp[0]=1 and updating dp[x] for each coin counts total ways efficiently.

2. Step 2: Reject other approaches

   2D dp for minimum coins is different problem; greedy fails for total ways; recursion without memo is inefficient.

3. Final Answer:

   Use a 1D dp array, initialize dp[0]=1, then for each coin, update dp[x] += dp[x - coin] for x from coin to amount -> Option C

4. Quick Check:

   1D dp with coin loops counts ways correctly [OK]

Hint: 1D dp with coin loops counts ways, not min coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins problem
• Using greedy which fails for counting ways
• Ignoring memoization causing slow recursion