Sure, here is a step-by-step solution using Dynamic Programming to find the total number of ways you can change the given amount of money (W=5) using given coins(2,3,5).

Step 1: Initialize a list (dp) of size W+1 with all elements as 0. This list will be used to store the number of ways for each amount from 0 to W. Set dp[0] = 1 because there is exactly 1 way to make change for 0 which is using no coin.

Step 2: Loop over each coin. For each coin, we update the dp list from the coin value to W. For each amount in this range, we add the number of ways to make change without this coin (which is dp[amount]) and the number of ways to make change with this coin (which is dp[amount - coin value]).

Step 3: After we finish looping over all coins, dp[W] will be the total number of ways to make change for W.

Here is the Python code for the above steps:

```python
def countWays(coins, W):
dp = [0] * (W + 1)
dp[0] = 1

for coin in coins:
for amount in range(coin, W + 1):
dp[amount] += dp[amount - coin]

return dp[W]

coins = [2, 3, 5]
W = 5
print(countWays(coins, W)) # Output: 3
```

The output is 3, which means there are 3 ways to make change for 5 using coins 2, 3, and 5. They are (2, 2, 1), (2, 3), and (5).

Question

Sure, here is a step-by-step solution using Dynamic Programming to find the total number of ways you can change the given amount of money (W=5) using given coins(2,3,5).

Step 1: Initialize a list (dp) of size W+1 with all elements as 0. This list will be used to store the number of ways for each amount from 0 to W. Set dp[0] = 1 because there is exactly 1 way to make change for 0 which is using no coin.

Step 2: Loop over each coin. For each coin, we update the dp list from the coin value to W. For each amount in this range, we add the number of ways to make change without this coin (which is dp[amount]) and the number of ways to make change with this coin (which is dp[amount - coin value]).

Step 3: After we finish looping over all coins, dp[W] will be the total number of ways to make change for W.

Here is the Python code for the above steps:

```python
def countWays(coins, W):
    dp = [0] * (W + 1)
    dp[0] = 1

for coin in coins:
        for amount in range(coin, W + 1):
            dp[amount] += dp[amount - coin]

return dp[W]

coins = [2, 3, 5]
W = 5
print(countWays(coins, W))  # Output: 3
```

The output is 3, which means there are 3 ways to make change for 5 using coins 2, 3, and 5. They are (2, 2, 1), (2, 3), and (5).

Knowee AI · Accepted Answer

Sure, here is a step-by-step solution using Dynamic Programming to find the total number of ways you can change the given amount of money (W=5) using given coins(2,3,5).

Step 1: Initialize a list (dp) of size W+1 with all elements as 0. This list will be used to store the number of ways for each amount from 0 to W. Set dp[0] = 1 because there is exactly 1 way to make change for 0 which is using no coin.

Step 2: Loop over each coin. For each coin, we update the dp list from the coin value to W. For each amount in this range, we add the number of ways to make change without this coin (which is dp[amount]) and the number of ways to make change with this coin (which is dp[amount - coin value]).

Step 3: After we finish looping over all coins, dp[W] will be the total number of ways to make change for W.

Here is the Python code for the above steps:

```python
def countWays(coins, W):
    dp = [0] * (W + 1)
    dp[0] = 1

for coin in coins:
        for amount in range(coin, W + 1):
            dp[amount] += dp[amount - coin]

return dp[W]

coins = [2, 3, 5]
W = 5
print(countWays(coins, W))  # Output: 3
```

The output is 3, which means there are 3 ways to make change for 5 using coins 2, 3, and 5. They are (2, 2, 1), (2, 3), and (5).

Use Dynamic Programming to find the total number of ways you can change the given amount of money (W=5) using given coins(2,3,5).

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