0322. Coin Change

https://leetcode.com/problems/coin-change

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 fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

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

Example 1:

**Input:** coins = [1,2,5], amount = 11
**Output:** 3
**Explanation:** 11 = 5 + 5 + 1

Example 2:

**Input:** coins = [2], amount = 3
**Output:** -1

Example 3:

**Input:** coins = [1], amount = 0
**Output:** 0

Example 4:

**Input:** coins = [1], amount = 1
**Output:** 1

Example 5:

**Input:** coins = [1], amount = 2
**Output:** 2

Constraints:

• 1 <= coins.length <= 12

• 1 <= coins[i] <= 231 - 1

• 0 <= amount <= 104

ac: dp

class Solution {
    public int coinChange(int[] coins, int amount) {
        // edge cases
        if (coins == null || coins.length == 0 || amount <= 0) return 0;

        // DP
        int[] dp = new int[amount+1];
            // init
        for (int i : coins) {
            if (i <= amount) dp[i] = 1;
        }
            // iterate
        for (int i = 1; i <= amount; i++) {
            if (dp[i] != 0) continue;
            int min = Integer.MAX_VALUE;
            for (int j = 0; j < coins.length; j++) {
                int prev = i - coins[j];
                if (prev <= 0 || dp[prev] == -1) continue;
                min = Math.min(min, dp[prev] + 1);
            }
            dp[i] = min == Integer.MAX_VALUE ? -1 : min;
        }


        // res
        return dp[amount];
    }
}

/*
state: dp[i] means fewest coins for i amount
func: dp[i], iterate coins, i - coin, get minimun
init: dp[coin] = 1
result: last one

*/

This one is a little bit concise

class Solution {
    public int coinChange(int[] coins, int amount) {
        // edge cases
        if (coins == null || coins.length == 0 || amount <= 0) return 0;

        int[] dp = new int[amount+1];
        // init
        Arrays.fill(dp, amount + 1);
        dp[0] = 0;

        // iterate
        for (int i = 1; i <= amount; i++) {
            for (int j = 0; j < coins.length; j++) {
                if (coins[j] <= i) {
                    dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }

        // res
        return dp[amount] > amount ? -1 : dp[amount];
    }
}