Link

Medium DP

Amazon

Bloomberg

2020-06-07

518. Coin Change 2

Question:

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

Solution:

Using DP to solve the problem, DP[i][j] means select ith coin with total sum to j.

Time: O(n * m)
Space: O(n * m)
m means amount

class Solution {
    public int change(int amount, int[] coins) {
        int[][] dp = new int[coins.length + 1][amount + 1];
        dp[0][0] = 1;
        
        for(int i = 1; i <= coins.length; i++) {
            dp[i][0] = 1;
            int curr = coins[i - 1];
            for (int j = 1; j <= amount; j++) {
                if (j >= curr) {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - curr];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[coins.length][amount];
    }
}

It can be reduced to 1-D array.

Time: O(n * m)
Space: O(m)

class Solution {
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        
        for(int i = 1; i <= coins.length; i++) {
            int curr = coins[i - 1];
            for (int j = 1; j <= amount; j++) {
                if (j >= curr) {
                    dp[j] += dp[j - curr];
                } 
            }
        }
        return dp[amount];
    }
}