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* Retranslate the complete English edition from Simplified Chinese * Polish three English explanations after reader-focused review
61 lines
2.0 KiB
Python
61 lines
2.0 KiB
Python
"""
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File: coin_change.py
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Created Time: 2023-07-10
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Author: krahets (krahets@163.com)
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"""
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def coin_change_dp(coins: list[int], amt: int) -> int:
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"""Coin change: dynamic programming"""
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n = len(coins)
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MAX = amt + 1
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# Initialize the dp table
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dp = [[0] * (amt + 1) for _ in range(n + 1)]
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# State transitions: first row and first column
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for a in range(1, amt + 1):
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dp[0][a] = MAX
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# State transitions: remaining rows and columns
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for i in range(1, n + 1):
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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# If the coin value exceeds the current amount a, do not choose coin i
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dp[i][a] = dp[i - 1][a]
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else:
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# Take the smaller result of not choosing and choosing coin i
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dp[i][a] = min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1)
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return dp[n][amt] if dp[n][amt] != MAX else -1
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def coin_change_dp_comp(coins: list[int], amt: int) -> int:
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"""Coin change: space-optimized dynamic programming"""
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n = len(coins)
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MAX = amt + 1
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# Initialize the dp table
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dp = [MAX] * (amt + 1)
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dp[0] = 0
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# State transitions
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for i in range(1, n + 1):
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# Traverse in increasing order
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for a in range(1, amt + 1):
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if coins[i - 1] > a:
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# If the coin value exceeds the current amount a, do not choose coin i
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dp[a] = dp[a]
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else:
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# Take the smaller result of not choosing and choosing coin i
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dp[a] = min(dp[a], dp[a - coins[i - 1]] + 1)
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return dp[amt] if dp[amt] != MAX else -1
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"""Driver code"""
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if __name__ == "__main__":
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coins = [1, 2, 5]
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amt = 4
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# Dynamic programming
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res = coin_change_dp(coins, amt)
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print(f"Minimum number of coins needed to make the target amount: {res}")
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# Space-optimized dynamic programming
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res = coin_change_dp_comp(coins, amt)
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print(f"Minimum number of coins needed to make the target amount: {res}")
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