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518. Coin Change II

LeetCode link: 518. Coin Change II

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 number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.

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

The answer is guaranteed to fit into a signed 32-bit integer.

Example 1:

Input: amount = 5, coins = [1,2,5]
Output: 4

Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
------------------------------------------------------------------------

Example 2:

Input: amount = 3, coins = [2]
Output: 0

Explanation: the amount of 3 cannot be made up just with coins of 2.
------------------------------------------------------------------------

Example 3:

Input: amount = 10, coins = [10]
Output: 1
------------------------------------------------------------------------

Constraints:

1 <= coins.length <= 300
1 <= coins[i] <= 5000
All the values of 'coins' are unique.
0 <= amount <= 5000

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 Change(int amount, int[] coins)
    {
        var dp = new int[amount + 1];
        dp[0] = 1;

        foreach (int coin in coins)
        {
            for (var j = coin; j < dp.Length; j++)
            {
                dp[j] += dp[j - coin];
            }
        }

        return dp.Last();
    }
}

Python

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        dp = [0] * (amount + 1)
        dp[0] = 1
        
        for coin in coins:
            for j in range(coin, len(dp)):
                dp[j] += dp[j - coin]

        return dp[-1]

C++

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        auto dp = vector<unsigned int>(amount + 1, 0);
        dp[0] = 1;

        for (auto coin : coins) {
            for (auto j = coin; j < dp.size(); j++) {
                dp[j] += dp[j - coin];
            }
        }

        return dp.back();
    }
};

Java

class Solution {
    public int change(int amount, int[] coins) {
        var dp = new int[amount + 1];
        dp[0] = 1;

        for (var coin : coins) {
            for (var j = coin; j < dp.length; j++) {
                dp[j] += dp[j - coin];
            }
        }

        return dp[dp.length - 1];
    }
}

JavaScript

var change = function (amount, coins) {
   const dp = Array(amount + 1).fill(0)
   dp[0] = 1

   for (const coin of coins) {
      for (let j = coin; j < dp.length; j++) {
         dp[j] += dp[j - coin]
      }
   }

   return dp.at(-1);
};

Go

func change(amount int, coins []int) int {
    dp := make([]int, amount + 1)
    dp[0] = 1

    for _, coin := range coins {
        for j := coin; j < len(dp); j++ {
            dp[j] += dp[j - coin]
        }
    }

    return dp[len(dp) - 1]
}

Ruby

def change(amount, coins)
  dp = Array.new(amount + 1, 0)
  dp[0] = 1

  coins.each do |coin|
    (coin...dp.size).each do |j|
      dp[j] += dp[j - coin]
    end
  end

  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!