missed coommit the source files

algorithms/cpp/minimumCostForTickets/MinimumCostForTickets.cpp

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+// Source : https://leetcode.com/problems/minimum-cost-for-tickets/
+// Author : Hao Chen
+// Date   : 2019-01-29
+
+/***************************************************************************************************** 
+ *
+ * In a country popular for train travel, you have planned some train travelling one year in advance.  
+ * The days of the year that you will travel is given as an array days.  Each day is an integer from 1 
+ * to 365.
+ * 
+ * Train tickets are sold in 3 different ways:
+ * 
+ * 	a 1-day pass is sold for costs[0] dollars;
+ * 	a 7-day pass is sold for costs[1] dollars;
+ * 	a 30-day pass is sold for costs[2] dollars.
+ * 
+ * The passes allow that many days of consecutive travel.  For example, if we get a 7-day pass on day 
+ * 2, then we can travel for 7 days: day 2, 3, 4, 5, 6, 7, and 8.
+ * 
+ * Return the minimum number of dollars you need to travel every day in the given list of days.
+ * 
+ * Example 1:
+ * 
+ * Input: days = [1,4,6,7,8,20], costs = [2,7,15]
+ * Output: 11
+ * Explanation: 
+ * For example, here is one way to buy passes that lets you travel your travel plan:
+ * On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
+ * On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
+ * On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
+ * In total you spent $11 and covered all the days of your travel.
+ * 
+ * Example 2:
+ * 
+ * Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
+ * Output: 17
+ * Explanation: 
+ * For example, here is one way to buy passes that lets you travel your travel plan:
+ * On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30.
+ * On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
+ * In total you spent $17 and covered all the days of your travel.
+ * 
+ * Note:
+ * 
+ * 	1 <= days.length <= 365
+ * 	1 <= days[i] <= 365
+ * 	days is in strictly increasing order.
+ * 	costs.length == 3
+ * 	1 <= costs[i] <= 1000
+ * 
+ ******************************************************************************************************/
+
+class Solution {
+private:
+    int min(int x, int y){
+        return x < y ? x : y;
+    }
+    int min(int x, int y, int z) {
+        return min(min(x, y), z);
+    }
+public:
+    int mincostTickets(vector<int>& days, vector<int>& costs) {
+
+        // Dynamic Programming
+        vector<int> dp(days.size(), INT_MAX);
+
+        // dp[i] is the minimal cost from Days[0] to Days[i]
+        dp[0] = costs[0];
+
+        for (int i = 1; i< days.size(); i ++) {
+
+            // the currnet day need at least 1-day pass cost
+            int OneDayPass = dp[i-1] + costs[0];
+
+            // Seprating the array to two parts.
+            //     days[0] -> days[j] -> day[i]
+            //
+            // calculate the day[i] - day[j] whether can use 7-day pass or 30-day pass
+            //
+            // Traking the minimal costs, then can have dp[i] minimal cost
+
+            int SevenDayPass = INT_MAX, ThrityDayPass = INT_MAX;
+            for (int j=i-1; j>=0; j--){
+                if (days[i] - days[j] < 7 )  {
+                    SevenDayPass  = dp[j-1] + costs[1];
+                } else if (days[i] - days[j] < 30 ) {
+                    ThrityDayPass = dp[j-1] + costs[2];
+                } else {
+                    break;
+                }
+                int m = min(OneDayPass, SevenDayPass, ThrityDayPass);
+                if ( dp[i] > m )  dp[i] = m;
+            }
+
+        }
+
+        return dp[dp.size()-1];
+    }
+};

algorithms/cpp/stringWithoutAAAOrBBB/StringWithoutAAAOrBBB.cpp

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+// Source : https://leetcode.com/problems/string-without-aaa-or-bbb/
+// Author : Hao Chen
+// Date   : 2019-01-29
+
+/***************************************************************************************************** 
+ *
+ * Given two integers A and B, return any string S such that:
+ * 
+ * 	S has length A + B and contains exactly A 'a' letters, and exactly B 'b' letters;
+ * 	The substring 'aaa' does not occur in S;
+ * 	The substring 'bbb' does not occur in S.
+ * 
+ * Example 1:
+ * 
+ * Input: A = 1, B = 2
+ * Output: "abb"
+ * Explanation: "abb", "bab" and "bba" are all correct answers.
+ * 
+ * Example 2:
+ * 
+ * Input: A = 4, B = 1
+ * Output: "aabaa"
+ * 
+ * Note:
+ * 
+ * 	0 <= A <= 100
+ * 	0 <= B <= 100
+ * 	It is guaranteed such an S exists for the given A and B.
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    string strWithout3a3b(int A, int B) {
+
+        string result;
+        while (true){
+
+            // if A == B, then just simpley repeat "ab" pattern;
+            if (A == B) {
+                for ( int i=0; i<A; i++ ) {
+                    result += "ab";
+                }
+                break;
+            }
+
+            // if A+B less then 3 and A != B, which means A=1,2 && B=0 or A=0 && B=1,2
+            if (A+B <3) {
+                while  ( A-- > 0 ) result += 'a';
+                while  ( B-- > 0 ) result += 'b';
+                break;
+            }
+
+            // if A+B >=3 and A !=B
+
+            // if A > B, then we need consume 'a' more than 'b'
+            // So, only "aab" can be used.
+            if ( A > B ) {
+                result += "aab";
+                A -= 2;
+                B--;
+                continue;
+            }
+
+            // if A > B, then we need consume 'b' more than 'a'
+            // So, only "bba" can be used.
+            if (B > A ){
+                result += "bba";
+                B-=2;
+                A--;
+                continue;
+            }
+
+        }
+
+        return result;
+    }
+};