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322. Coin Change

LeetCode link: 322. Coin Change

LeetCode problem description

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

You may assume that you have an infinite number of each kind of coin.

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3

Explanation: 11 = 5 + 5 + 1
------------------------------------------------------------------------

Example 2:

Input: coins = [2], amount = 3
Output: -1
------------------------------------------------------------------------

Constraints:

1 <= coins.length <= 12
1 <= coins[i] <= 2**31 - 1
0 <= amount <= 10000

Thoughts

It is a Unbounded Knapsack Problem.

Detailed solutions will be given later, and now only the best practices in 7 languages are given.

Complexity

  • Time: O(n * m).
  • Space: O(n).

C#

public class Solution
{
    public int CoinChange(int[] coins, int amount)
    {
        int defaultValue = amount + 2; // As long as the value is greater than 'amount', it doesn't matter how much it is.
        int[] dp = Enumerable.Repeat(defaultValue, amount + 1).ToArray();
        dp[0] = 0;

        for (var i = 1; i < dp.Length; i++)
        {
            foreach (int coin in coins)
            {
                if (i >= coin)
                {
                    dp[i] = Math.Min(dp[i], dp[i - coin] + 1);
                }
            }
        }

        if (dp.Last() == defaultValue)
            return -1;

        return dp.Last();
    }
}

Python

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        default_value = amount + 2 # As long as the value is greater than 'amount', it doesn't matter how much it is.
        dp = [default_value] * (amount + 1)
        dp[0] = 0

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

        if dp[-1] == default_value:
            return -1

        return dp[-1]

C++

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        auto default_value = amount + 2; // As long as the value is greater than 'amount', it doesn't matter how much it is.
        auto dp = vector<int>(amount + 1, default_value);
        dp[0] = 0;

        for (auto i = 1; i < dp.size(); i++) {
            for (auto coin : coins) {
                if (i >= coin) {
                    dp[i] = min(dp[i], dp[i - coin] + 1);
                }
            }
        }

        if (dp.back() == default_value) {
            return -1;
        }
        return dp.back();
    }
};

Java

class Solution {
    public int coinChange(int[] coins, int amount) {
        var defaultValue = amount + 2; // As long as the value is greater than 'amount', it doesn't matter how much it is.

        var dp = new int[amount + 1];
        Arrays.fill(dp, defaultValue);
        dp[0] = 0;

        for (var i = 1; i < dp.length; i++) {
            for (var coin : coins) {
                if (i >= coin) {
                    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
                }
            }
        }

        var result = dp[dp.length - 1];
        if (result == defaultValue) {
            return -1;
        }
        return result;
    }
}

JavaScript

var coinChange = function (coins, amount) {
  const DEFAULT_VALUE = amount + 2 // As long as the value is greater than 'amount', it doesn't matter how much it is.
  const dp = Array(amount + 1).fill(DEFAULT_VALUE)
  dp[0] = 0

  for (let i = 1; i < dp.length; i++) {
    for (const coin of coins) {
      if (i >= coin) {
        dp[i] = Math.min(dp[i], dp[i - coin] + 1)
      }
    }
  }

  if (dp.at(-1) == DEFAULT_VALUE) {
    return -1
  }
  return dp.at(-1)
};

Go

func coinChange(coins []int, amount int) int {
    defaultValue := amount + 2 // As long as the value is greater than 'amount', it doesn't matter how much it is.
    dp := slices.Repeat([]int{defaultValue}, amount + 1)
    dp[0] = 0

    for i := 1; i < len(dp); i++ {
        for _, coin := range coins {
            if i >= coin {
                dp[i] = min(dp[i], dp[i - coin] + 1)
            }
        }
    }

    result := dp[len(dp) - 1]
    if result == defaultValue {
        return -1
    }
    return result
}

Ruby

def coin_change(coins, amount)
  default_value = amount + 2 # As long as the value is greater than 'amount', it doesn't matter how much it is.
  dp = Array.new(amount + 1, default_value)
  dp[0] = 0

  (1...dp.size).each do |i|
    coins.each do |coin|
      dp[i] = [ dp[i], dp[i - coin] + 1 ].min if i >= coin
    end
  end

  return -1 if dp[-1] == default_value

  dp[-1]
end

Rust

// Welcome to create a PR to complete the code of this language, thanks!

Other languages

// Welcome to create a PR to complete the code of this language, thanks!