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 an array dp of size W+1 with all elements as 0. This array will be used to store the number of ways to make change for all amounts from 0 to W. Set dp[0] = 1, because there is exactly 1 way to make change for 0 (that is, using no coins).

Step 2: Loop over each coin. For each coin, loop over all amounts from the value of the coin to W. For each amount, add the number of ways to make change without the current coin (which is dp[amount]) and the number of ways to make change with the current coin (which is dp[amount - coin]). Store this sum in dp[amount].

Step 3: After looping over all coins and all amounts, 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
```

This code will output 3, which means there are 3 ways to make change for 5 using coins of 2, 3, and 5. These ways 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 an array dp of size W+1 with all elements as 0. This array will be used to store the number of ways to make change for all amounts from 0 to W. Set dp[0] = 1, because there is exactly 1 way to make change for 0 (that is, using no coins).

Step 2: Loop over each coin. For each coin, loop over all amounts from the value of the coin to W. For each amount, add the number of ways to make change without the current coin (which is dp[amount]) and the number of ways to make change with the current coin (which is dp[amount - coin]). Store this sum in dp[amount].

Step 3: After looping over all coins and all amounts, 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
```

This code will output 3, which means there are 3 ways to make change for 5 using coins of 2, 3, and 5. These ways 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 an array dp of size W+1 with all elements as 0. This array will be used to store the number of ways to make change for all amounts from 0 to W. Set dp[0] = 1, because there is exactly 1 way to make change for 0 (that is, using no coins).

Step 2: Loop over each coin. For each coin, loop over all amounts from the value of the coin to W. For each amount, add the number of ways to make change without the current coin (which is dp[amount]) and the number of ways to make change with the current coin (which is dp[amount - coin]). Store this sum in dp[amount].

Step 3: After looping over all coins and all amounts, 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
```

This code will output 3, which means there are 3 ways to make change for 5 using coins of 2, 3, and 5. These ways 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).1 point1234Clear selection

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