Create CoinChangeII.cpp

Coin-Change-II.cpp

@@ -0,0 +1,79 @@
+/***************************************************************************************************** 
+ *
+ * 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];
+    }
+};

Find-All-Anagrams-in-a-String.cpp

@@ -0,0 +1,47 @@
+/***************************************************************************************************** 
+* Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
+* Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
+* The order of output does not matter.
+* Example 1:
+* Input:
+   s: "cbaebabacd" p: "abc"
+* Output:
+  [0, 6]
+* Explanation:
+  The substring with start index = 0 is "cba", which is an anagram of "abc".
+  The substring with start index = 6 is "bac", which is an anagram of "abc".
+*Example 2:
+* Input:
+   s: "abab" p: "ab"
+* Output:
+   [0, 1, 2]
+* Explanation:
+* The substring with start index = 0 is "ab", which is an anagram of "ab".
+* The substring with start index = 1 is "ba", which is an anagram of "ab".
+* The substring with start index = 2 is "ab", which is an anagram of "ab".
+ ******************************************************************************************************/
+
+
+class Solution {
+    public:
+        vector<int> findAnagrams(string s, string p) {
+            vector<int>result;
+            vector<int>v1(26);
+            if(s.size()<p.size()){
+                return result;
+            }
+            for(int i=0;i<p.size();i++){
+                v1[p[i]-97]++;
+            }
+            for(int i=0;i<=(s.size()-p.size());i++){
+                vector<int>v2(26);
+                for(int index=i;index<(i+p.size());index++){
+                    v2[s[index]-97]++;
+                }
+                if(v1==v2){
+                    result.push_back(i);
+                }        
+            }
+            return result;
+        }
+};