From c33b849863d8eac306b6520f195c8e8a97aa4607 Mon Sep 17 00:00:00 2001
From: Hao Chen <haoel@hotmail.com>
Date: Mon, 18 Mar 2019 17:22:55 +0800
Subject: [PATCH] New Problem Solution "Coin Change 2"

---
 README.md                                 |  1 +
 algorithms/cpp/coinChange/CoinChange2.cpp | 84 +++++++++++++++++++++++
 2 files changed, 85 insertions(+)
 create mode 100644 algorithms/cpp/coinChange/CoinChange2.cpp

diff --git a/README.md b/README.md
index e368948..547671b 100644
--- a/README.md
+++ b/README.md
@@ -61,6 +61,7 @@ LeetCode
 |538|[Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/) | [Python](./algorithms/python/ConvertBSTtoGreaterTree/convertBST.py)|Easy|
 |532|[K-diff Pairs in an Array](https://leetcode.com/problems/k-diff-pairs-in-an-array/) | [Python](./algorithms/python/K-diffPairsInAnArray/findPairs.py)|Easy|
 |520|[Detect Capital](https://leetcode.com/problems/detect-capital/) | [C++](./algorithms/cpp/detectCapital/DetectCapital.cpp)|Easy|
+|518|[Coin Change 2](https://leetcode.com/problems/coin-change-2/) | [C++](./algorithms/cpp/coinChange/CoinChange2.cpp)|Medium|
 |509|[Fibonacci Number](https://leetcode.com/problems/fibonacci-number/) | [Python](./algorithms/python/FibonacciNumber/fib.py)|Easy|
 |477|[Total Hamming Distance](https://leetcode.com/problems/total-hamming-distance/) | [C++](./algorithms/cpp/totalHammingDistance/totalHammingDistance.cpp)|Medium|
 |463|[Island Perimeter](https://leetcode.com/problems/island-perimeter/) | [C++](./algorithms/cpp/islandPerimeter/IslandPerimeter.cpp)|Easy|
diff --git a/algorithms/cpp/coinChange/CoinChange2.cpp b/algorithms/cpp/coinChange/CoinChange2.cpp
new file mode 100644
index 0000000..2bcb992
--- /dev/null
+++ b/algorithms/cpp/coinChange/CoinChange2.cpp
@@ -0,0 +1,84 @@
+// Source : https://leetcode.com/problems/coin-change-2/
+// Author : Hao Chen
+// Date   : 2019-03-18
+
+/***************************************************************************************************** 
+ *
+ * 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
+ * 
+ ******************************************************************************************************/
+class Solution {
+public:
+    int change(int amount, vector<int>& coins) {
+        return change_dp(amount, coins);
+        return change_recursive(amount, coins); // Time Limit Error
+    }
+    
+    
+    int change_recursive(int amount, vector<int>& coins) {
+        int result = 0;
+        change_recursive_helper(amount, coins, 0, result);
+        return result;
+    }
+    
+    // the `idx` is used for remove the duplicated solutions.
+    void change_recursive_helper(int amount, vector<int>& coins, int idx, int& result) {
+        if (amount == 0) { 
+            result++; 
+            return;
+        }
+    
+        for ( int i = idx; i < coins.size(); i++ ) {
+            if (amount < coins[i]) continue;
+            change_recursive_helper(amount - coins[i], coins, i, result);
+        }
+        return;
+    }
+    
+    int change_dp(int amount, vector<int>& coins) {
+        vector<int> dp(amount+1, 0);
+        dp[0] = 1;
+        for (int i=0; i<coins.size(); i++) {
+            for(int n=1; n<=amount; n++) {
+                if (n >= coins[i]) {
+                    dp[n] += dp[n-coins[i]];
+                }
+            }
+        }
+
+        return dp[amount];
+    }
+};
+