DPChange (M, c, d): bestNumCoins [0] - 0 for m - 1 to M bestNumcoins (m] = inf for i - 1 to d if m >- c[i] if bestNumcoins (m-c[i]] + 1 < bestNumcoins (m] bestNumCoins (m] = bestNumCoins (m-c[i]] + 1 return bestNumCoins (M) a) Convert this pseudocode to Python. b) What is the runtime of this algorithm?
DPChange (M, c, d): bestNumCoins [0] - 0 for m - 1 to M bestNumcoins (m] = inf for i - 1 to d if m >- c[i] if bestNumcoins (m-c[i]] + 1 < bestNumcoins (m] bestNumCoins (m] = bestNumCoins (m-c[i]] + 1 return bestNumCoins (M) a) Convert this pseudocode to Python. b) What is the runtime of this algorithm?
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![### DPChange(M, c, d) Pseudocode
This pseudocode is used to find the minimum number of coins needed to make change for a given amount \( M \) using a list of coin denominations.
#### Pseudocode Explanation:
1. **Initialization:**
- `bestNumCoins[0] = 0`: Set the base case for making change for 0 amount.
2. **Outer Loop (1 to M):**
- For each amount \( m \) from 1 to \( M \), do the following:
- Initialize `bestNumCoins[m] = inf` (infinity) to represent initially an impossible large number of coins.
3. **Inner Loop (1 to d):**
- Iterate over each coin denomination.
- **Check Condition:**
- If the current amount \( m \) is greater than or equal to the coin denomination \( c[i] \):
- Compare and update the best number of coins:
```
if bestNumCoins[m-c[i]] + 1 < bestNumCoins[m]:
bestNumCoins[m] = bestNumCoins[m-c[i]] + 1
```
- This checks if using the current coin \( c[i] \) results in fewer coins than previously computed for amount \( m \).
4. **Return Result:**
- Return `bestNumCoins[M]` which contains the minimum number of coins needed for amount \( M \).
### Questions:
a) **Convert this pseudocode to Python.**
```python
def DPChange(M, c):
bestNumCoins = [float('inf')] * (M + 1)
bestNumCoins[0] = 0
for m in range(1, M + 1):
for coin in c:
if m >= coin:
if bestNumCoins[m - coin] + 1 < bestNumCoins[m]:
bestNumCoins[m] = bestNumCoins[m - coin] + 1
return bestNumCoins[M]
```
b) **What is the runtime of this algorithm?**
The runtime of this algorithm is \( O(M \cdot d) \) where \( M \) is the amount for which we are making change and \( d \) is the number of different coin denominations. This is because for each amount up to \( M \), we iterate through each denomination.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12c18db6-f6e8-4129-b933-2f1b3972f7de%2F0619b341-f7c0-415c-91fd-cf3775ef7b99%2F16x3xla_processed.png&w=3840&q=75)
Transcribed Image Text:### DPChange(M, c, d) Pseudocode
This pseudocode is used to find the minimum number of coins needed to make change for a given amount \( M \) using a list of coin denominations.
#### Pseudocode Explanation:
1. **Initialization:**
- `bestNumCoins[0] = 0`: Set the base case for making change for 0 amount.
2. **Outer Loop (1 to M):**
- For each amount \( m \) from 1 to \( M \), do the following:
- Initialize `bestNumCoins[m] = inf` (infinity) to represent initially an impossible large number of coins.
3. **Inner Loop (1 to d):**
- Iterate over each coin denomination.
- **Check Condition:**
- If the current amount \( m \) is greater than or equal to the coin denomination \( c[i] \):
- Compare and update the best number of coins:
```
if bestNumCoins[m-c[i]] + 1 < bestNumCoins[m]:
bestNumCoins[m] = bestNumCoins[m-c[i]] + 1
```
- This checks if using the current coin \( c[i] \) results in fewer coins than previously computed for amount \( m \).
4. **Return Result:**
- Return `bestNumCoins[M]` which contains the minimum number of coins needed for amount \( M \).
### Questions:
a) **Convert this pseudocode to Python.**
```python
def DPChange(M, c):
bestNumCoins = [float('inf')] * (M + 1)
bestNumCoins[0] = 0
for m in range(1, M + 1):
for coin in c:
if m >= coin:
if bestNumCoins[m - coin] + 1 < bestNumCoins[m]:
bestNumCoins[m] = bestNumCoins[m - coin] + 1
return bestNumCoins[M]
```
b) **What is the runtime of this algorithm?**
The runtime of this algorithm is \( O(M \cdot d) \) where \( M \) is the amount for which we are making change and \( d \) is the number of different coin denominations. This is because for each amount up to \( M \), we iterate through each denomination.
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