Coin Change

Problem

Given different coins of infinite amount, find the fewest coins to make up an amount.

Solution

  • DP.
  • Can be done recursively or iteratively.
  • Iterative code is much shorter.
  • Top-down recursion: Start with amount and use one coin and find the min for the remaining recursively.
  • Use memoization.
  • Iterative: For 1 to amount find the min coin using min of (amount - coin_i) for each coin.
  • Don't forget to initialize inf and 0 properly.
  • For loops can be swapped!

# DP: Iterative

from typing import List

def coinChange(self, coins: List[int], amount: int) -> int:                
   
    import math
    dp = [math.inf] * (amount + 1)
    dp[0] = 0

    for x in range(1, amount + 1):
        for coin in coins:
            if x >= coin:
                dp[x] = min(dp[x], dp[x - coin] + 1)

    return dp[amount] if dp[amount] != math.inf else -1







# DP: Recursive:

def coinChange(self, coins: List[int], amount: int) -> int:

        def minCoins(coins, rem, memoized):
            if rem < 0:
                return -1
            if rem == 0:
                return 0

            if (rem-1) in memoized:
                return memoized[rem-1]

            tree_min = None
            for coin in coins:
                sub_tree_min = minCoins(coins, rem - coin, memoized)
                if sub_tree_min >=0:
                    if not tree_min or sub_tree_min < tree_min:
                        tree_min = sub_tree_min + 1
           
            memoized[rem-1] = tree_min if tree_min else -1
            return memoized[rem-1]

        if amount < 1:
            return 0

        memoized = {}
        return minCoins(coins, amount, memoized)

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