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Coin Change in Python
45 Lectures 4.5 hours
66 Lectures 5.5 hours
Suppose we have coins of different denominations and a total amount of money amount. We have to define one function to compute the fewest number of coins that we need to make up that amount. When that amount of money cannot be accommodated by any combination of the coins, return -1. So if the input is [1,2,5], and the amount is 11, the output is 3. This is formed using 5 + 5 + 1 = 11.
To solve this, we will follow these steps −
- if amount = 0, then return 0
- if minimum of coins array > amount, then return -1
- define one array called dp, of size amount + 1, and fill this with -1
- for i in range coins array
- if i > length of dp – 1, then skip the next part, go for the next iteration
- dp[i] := 1
- for j in range i + 1 to amount
- if dp[j – 1] = -1, then skip the next part, go for the next iteration
- otherwise if dp[j] = -1, then dp[j] := dp[j - i] + 1
- otherwise dp[j] := minimum of dp[j] and dp[j – i] + 1
- return dp[amount]
Let us see the following implementation to get better understanding −
class Solution(object): def coinChange(self, coins, amount): if amount == 0 : return 0 if min(coins) > amount: return -1 dp = [-1 for i in range(0, amount + 1)] for i in coins: if i > len(dp) - 1: continue dp[i] = 1 for j in range(i + 1, amount + 1): if dp[j - i] == -1: continue elif dp[j] == -1: dp[j] = dp[j - i] + 1 else: dp[j] = min(dp[j], dp[j - i] + 1) #print(dp) return dp[amount] ob1 = Solution() print(ob1.coinChange([1,2,5], 11))
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