Leetcode 518: Coin Change 2

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

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

Note:

You can assume that

• 0 <= amount <= 5000
• 1 <= coin <= 5000
• the number of coins is less than 500
• the answer is guaranteed to fit into signed 32-bit integer

Code (Java):

class Solution {
    public int change(int amount, int[] coins) {
        if (amount < 0 || coins == null || coins.length == 0) {
            return amount == 0 ? 1 : 0;
        }

        int[][] dp = new int[coins.length + 1][amount + 1];
        for (int i = 0; i <= coins.length; i++) {
            dp[i][0] = 1;
        }

        for (int i = 1; i <= coins.length; i++) {
            for (int j = 1; j <= amount; j++) {
                dp[i][j] = dp[i - 1][j];

                if (j - coins[i - 1] >= 0) {
                    dp[i][j] += dp[i][j - coins[i - 1]];
                }
            }
        }

        return dp[coins.length][amount];

    }
}

Space optimization:

class Solution {
    public int change(int amount, int[] coins) {
        if (amount < 0 || coins == null || coins.length == 0) {
            return amount == 0 ? 1 : 0;
        }

        int[][] dp = new int[2][amount + 1];
        for (int i = 0; i <= coins.length; i++) {
            dp[0][0] = 1;
        }

        int old = 0;
        int cur = 0;

        for (int i = 1; i <= coins.length; i++) {
            old = cur;
            cur = 1 - cur;
            dp[cur][0] = 1;
            for (int j = 1; j <= amount; j++) {
                dp[cur][j] = dp[old][j];

                if (j - coins[i - 1] >= 0) {
                    dp[cur][j] += dp[cur][j - coins[i - 1]];
                }
            }
        }

        return dp[cur][amount];
    }
}

Space Optimization 2:

class Solution {
    public int change(int amount, int[] coins) {
        if (amount < 0 || coins == null || coins.length == 0) {
            return amount == 0 ? 1 : 0;
        }

        int[] dp = new int[amount + 1];
        for (int i = 0; i <= coins.length; i++) {
            dp[0] = 1;
        }

        for (int i = 1; i <= coins.length; i++) {
            dp[0] = 1;
            for (int j = 1; j <= amount; j++) {
                if (j - coins[i - 1] >= 0) {
                    dp[j] += dp[j - coins[i - 1]];
                }
            }
        }

        return dp[amount];
    }
}