In this post, we are going to solve the 518. Coin Change 2 problem of Leetcode. This problem 518. Coin Change 2 is a Leetcode medium level problem. Let’s see the code, 518. Coin Change 2 – Leetcode Solution.

Problem

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

Now, let’s see the code of 518. Coin Change 2 – Leetcode Solution.

Coin Change 2 – Leetcode Solution

518. Coin Change 2 – Solution in Java

class Solution {
    public int solveTab(int amount, int[] coins){
        int[] dp = new int[amount+1];
        dp[0] = 1;
       
        for(int e:coins){
            for(int i=1;i<=amount;i++){
                if(i-e >= 0)
                    dp[i] += dp[i-e];
            }
        }

        return dp[amount];
    }
    public int change(int amount, int[] coins) {
        return solveTab(amount,coins);
    }
}

518. Coin Change 2 – Solution in C++

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        vector<int> dp(amount+1, 0);
        dp[0] = 1;
        for(int x: coins){
            for(int i=x; i<=amount; i++) dp[i] += dp[i-x];
        }
        return dp[amount];
    }
};

518. Coin Change 2 – Solution in Python

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

Note: This problem 518. Coin Change 2 is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.