feat: solve coin change

Author: GangBean

coin-change/GangBean.java

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+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        /**
+        1. understanding
+        - given coins that can be used, find the minimum count of coins sum up to input amount value.
+        - [1,2,5]: 11
+            - 2 * 5 + 1 * 1: 3 -> use high value coin as much as possible if the remain can be sumed up by remain coins.
+        2. strategy
+        - If you search in greedy way, it will takes over O(min(amount/coin) ^ N), given N is the length of coins.
+        - Let dp[k] is the number of coins which are sum up to amount k, in a given coin set.
+        - Then, dp[k] = min(dp[k], dp[k-coin] + 1)
+        3. complexity
+        - time: O(CA), where C is the length of coins, A is amount value
+        - space: O(A), where A is amount value
+        */
+        Arrays.sort(coins);
+
+        int[] dp = new int[amount + 1];
+        for (int i = 1; i <= amount; i++) {
+            dp[i] = amount + 1;
+        }
+
+        for (int coin: coins) { // O(C)
+            for (int k = coin; k <= amount; k++) { // O(A)
+                dp[k] = Math.min(dp[k], dp[k-coin] + 1);
+            }
+        }
+
+        return (dp[amount] >= amount + 1) ? -1 : dp[amount];
+    }
+}
+