LeetCode 518: Coin Change II Solution

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will create a 1D array dp where dp[i] represents the number of combinations to make up amount i. We initialize dp[0] as 1, since there is one way to make up amount 0, which is by using no coins. Then, for each coin, we iterate through all amounts from the coin value to the target amount and update the number of combinations accordingly.

Algorithm:

Initialize a 1D array dp of size amount + 1 with all values set to 0, except dp[0] = 1.
Iterate through each coin in the coins array.
For each coin, iterate through all amounts from the coin value to the target amount.
Update dp[j] = dp[j] + dp[j - coin] to accumulate the number of combinations.

Time Complexity: O(amount * n), where n is the number of coins Space Complexity: O(amount)

Solution Code

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

        for (int coin : coins) {
            for (int j = coin; j <= amount; j++) {
                dp[j] += dp[j - coin];
            }
        }

        return dp[amount];
    }
}

Frequently Asked Questions

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518. Coin Change II