Use Dynamic Programming to find the minimum number coins you can change for the given amount of money (W=7) using given coins(1,3,4,5).1 point0123

Sure, here is a step-by-step solution using Dynamic Programming to find the minimum number of coins you can change for the given amount of money (W=7) using given coins(1,3,4,5).

Step 1: Initialize an array dp[] of size W+1 with a maximum value. Here, W is the total amount of money. dp[i] will be storing the minimum number of coins required for i amount. So dp[W] will have the result.

W = 7
coins = [1,3,4,5]
dp = [float('inf')] * (W+1)

Step 2: Base Case: dp[0] = 0. To make the total 0, we don't need any coin.

dp[0] = 0

Step 3: Compute the values of dp[i] for all i from 1 to W.

for i in range(1, W+1):
    for coin in coins:
        if coin <= i:
            dp[i] = min(dp[i], dp[i-coin] + 1)

In the inner loop, we iterate through each coin. If the value of the coin is less than or equal to i, then we update dp[i].

Step 4: Finally, dp[W] will have the result, the minimum number of coins required for W amount.

print(dp[W])

If dp[W] is still infinity, it means we don't have any combination of coins that can sum up to W, so we return -1 or any indication for no solution.

This is a typical dynamic programming problem where we solve the problem by breaking it down into smaller subproblems. We use the solutions to the subproblems to build up the solution to the original problem.

Sure, here is a step-by-step solution using Dynamic Programming to find the minimum number of coins you can change for the given amount of money (W=7) using given coins(1,3,4,5).

Step 1: Initialize an array dp[] of size W+1 with a large value (Infinity). This array will store the minimum number of coins required to get the value of the index.

Step 2: Set dp[0] = 0, because to make the total 0, we don't need any coin.

Step 3: Now, start from the first index of the dp array, and for each index, go through each coin. If the value of the coin is less than or equal to the index value, then check if dp[index - coin value] + 1 is less than the current value at dp[index]. If it is, then update dp[index] with the new value.

Step 4: Continue this process until you have filled up the dp array completely. The value at dp[W] will be the minimum number of coins required to get the total W.

Here is the Python code for the above steps:

def minCoins(coins, W):
    dp = [float('inf')] * (W + 1)
    dp[0] = 0

    for i in range(1, W + 1):
        for coin in coins:
            if coin <= i:
                dp[i] = min(dp[i], dp[i - coin] + 1)

    return dp[W]

coins = [1, 3, 4, 5]
W = 7
print(minCoins(coins, W))

This code will output 2, which means we need a minimum of 2 coins to make the total 7 using the given coins.

Sure, here is a step-by-step solution using Dynamic Programming to find the minimum number of coins you can change for the given amount of money (W=7) using given coins(1,3,4,5).

Step 1: Initialize an array dp[] of size W+1 with a large value (Infinity). This array will store the minimum number of coins required to get the value of the index.

Step 2: Set dp[0] = 0, because to make the total 0, we don't need any coin.

Step 3: Now, start from the first coin and go up to the total W. For each coin, we try to pick it and update the remaining value in our dp[].

Step 4: For each coin, we go through all the higher amounts than itself. If the amount minus the current coin value is less than or equal to the current coin value, we update the dp value.

Step 5: For each coin, we calculate dp[j], for all j>=coins[i] where coins[i] is the coin of the ith denomination. And we update the dp[j] as dp[j] = min(dp[j], 1+dp[j-coins[i]]).

Step 6: Finally, dp[W] will hold our answer, if dp[W] is infinity then the answer is not possible.

Here is the Python code for the above steps:

def minCoins(coins, W):
    dp = [float('inf')] * (W + 1)
    dp[0] = 0
    for i in range(len(coins)):
        for j in range(coins[i], W + 1):
            dp[j] = min(dp[j], 1 + dp[j - coins[i]])
    return dp[W] if dp[W] != float('inf') else -1

coins = [1, 3, 4, 5]
W = 7
print(minCoins(coins, W))

This code will return 2, which means we need minimum 2 coins to make the change for 7.

Sure, here is a step-by-step solution using Dynamic Programming to find the minimum number of coins you can change for the given amount of money (W=7) using given coins(1,3,4,5).

Step 1: Initialize an array dp[] of size W+1 with a large value (Infinity). This array will store the minimum number of coins required to get the value of the index.

Step 2: Set dp[0] = 0, because to make the total 0, we don't need any coin.

Step 3: Now, iterate through the array dp[] from 1 to W (i.e., 1 to 7 in this case) and for each dp[i], calculate the minimum number of coins required.

Step 4: For each coin, if the coin value is less than or equal to i, then we will keep the minimum value between the current value of dp[i] and dp[i-coin]+1.

Step 5: Finally, dp[W] will have the result.

Here is the Python code for the above steps:

def minCoins(coins, W):
    dp = [float('inf')] * (W + 1)
    dp[0] = 0
    for i in range(1, W + 1):
        for coin in coins:
            if coin <= i:
                dp[i] = min(dp[i], dp[i - coin] + 1)
    return dp[W]

coins = [1, 3, 4, 5]
W = 7
print(minCoins(coins, W))

This code will return 2 as the minimum number of coins required to make 7 with given coins (1,3,4,5) is 2 (3+4).