week11 mission coin-change

Author: dev-jonghoonpark

coin-change/dev-jonghoonpark.md

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+- 문제: https://leetcode.com/problems/coin-change/
+- 풀이: https://algorithm.jonghoonpark.com/2024/02/26/leetcode-322
+
+## dfs로 풀기
+
+```java
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        if(amount == 0) {
+            return 0;
+        }
+
+        int[] dp = new int[amount + 1];
+
+        List<Integer> sortedCoins = Arrays.stream(coins).boxed()
+                .sorted(Collections.reverseOrder())
+                .toList();
+
+        sortedCoins.forEach(coin -> dfs(dp, sortedCoins, amount, coin));
+
+        return dp[0] == 0 ? -1 : dp[0];
+    }
+
+    void dfs(int[] dp, List<Integer> coins, int amount, int selectedCoin) {
+        int currentPointer = amount - selectedCoin;
+        if (currentPointer < 0) {
+            return;
+        }
+
+        if (dp[currentPointer] == 0 || dp[currentPointer] > dp[amount] + 1) {
+            dp[currentPointer] = dp[amount] + 1;
+            coins.forEach(coin -> dfs(dp, coins, currentPointer, coin));
+        }
+    }
+}
+```
+
+### TS, SC
+
+코인의 수를 n 이라고 했을 때, `O(n * amount ^ 2)` 의 시간복잡도와 `O(amount)` 의 공간복잡도를 가진다.
+
+## dp로 풀기
+
+```java
+public class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int max = amount + 1;
+        int[] dp = new int[amount + 1];
+        Arrays.fill(dp, max);
+        dp[0] = 0;
+
+        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);
+                }
+            }
+        }
+
+        return dp[amount] > amount ? -1 : dp[amount];
+    }
+}
+```
+
+### TS, SC
+
+코인의 수를 n 이라고 했을 때, `O(n * amount)` 의 시간복잡도와 `O(amount)` 의 공간복잡도를 가진다.