New Problem Solution"Longest Substring with At Least K Repeating Characters"

haoel · haoel · commit 2f3a77de45a6 · 2016-09-08T01:48:16.000+08:00
diff --git a/README.md b/README.md
@@ -8,6 +8,8 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|395|[Longest Substring with At Least K Repeating Characters](https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/) | [C++](./algorithms/cpp/longestSubstringWithAtLeastKRepeatingCharacters/LongestSubstringWithAtLeastKRepeatingCharacters.cpp)|Medium|
+|394|[Decode String](https://leetcode.com/problems/decode-string/) | [C++](./algorithms/cpp/decodeString/DecodeString.cpp)|Medium|
 |393|[UTF-8 Validation](https://leetcode.com/problems/utf-8-validation/) | [C++](./algorithms/cpp/UTF8Validation/UTF8Validation.cpp)|Medium|
 |392|[Is Subsequence](https://leetcode.com/problems/is-subsequence/) | [C++](./algorithms/cpp/isSubsequence/IsSubsequence.cpp)|Medium|
 |391|[Perfect Rectangle](https://leetcode.com/problems/perfect-rectangle/) | [C++](./algorithms/cpp/perfectRectangle/PerfectRectangle.cpp)|Hard|
diff --git a/algorithms/cpp/decodeString/DecodeString.cpp b/algorithms/cpp/decodeString/DecodeString.cpp
@@ -0,0 +1,105 @@
+// Source : https://leetcode.com/problems/decode-string/
+// Author : Hao Chen
+// Date   : 2016-09-08
+
+/*************************************************************************************** 
+ *
+ * Given an encoded string, return it's decoded string.
+ * 
+ * The encoding rule is: k[encoded_string], where the encoded_string inside the square 
+ * brackets is being repeated exactly k times. Note that k is guaranteed to be a 
+ * positive integer.
+ * 
+ * You may assume that the input string is always valid; No extra white spaces, square 
+ * brackets are well-formed, etc.
+ * 
+ * Furthermore, you may assume that the original data does not contain any digits and 
+ * that digits are only for those repeat numbers, k. For example, there won't be input 
+ * like 3a or 2[4].
+ * 
+ * Examples:
+ * 
+ * s = "3[a]2[bc]", return "aaabcbc".
+ * s = "3[a2[c]]", return "accaccacc".
+ * s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
+ ***************************************************************************************/
+
+class Solution {
+public:
+    string decodeString(string s) {
+        if (!isValid(s)) return "";
+        
+        stack<string> _stack;
+        stack<int> _nstack;
+        
+        string result;
+        string tmp;
+        int n=0;
+        for (int i=0; i<s.size(); i++) {
+            
+            if ( isNum(s[i]) ) {
+                n = 0;
+                for(; isNum(s[i]); i++ ) {
+                   n = n*10 + s[i] - '0'; 
+                }
+            }
+            
+            if (s[i] == '[') {
+                tmp="";
+                _stack.push(tmp);
+                _nstack.push(n);
+            } else if (s[i] == ']') {
+                n = _nstack.top(); 
+                tmp="";
+                for (; n>0; n--) {
+                    tmp += _stack.top();
+                }
+                _stack.pop();
+                _nstack.pop();
+                if ( ! _stack.empty() ) {
+                    _stack.top() += tmp;
+                }else {
+                    result += tmp;
+                }
+            } else {
+                if ( ! _stack.empty() ) {
+                    _stack.top() += s[i];
+                } else {
+                    result += s[i];
+                }
+                
+            }
+        }
+        
+        return result;
+    }
+    
+private:
+
+    //only check the following rules:
+    // 1) the number must be followed by '['
+    // 2) the '[' and ']' must be matched.
+    bool isValid(string& s) {
+        stack<char> _stack;
+        for (int i=0; i<s.size(); i++) {
+            if ( isNum(s[i]) ) {
+                for(; isNum(s[i]); i++);
+                if (s[i] != '[') {
+                    return false;
+                } 
+                _stack.push('[');
+                continue;
+            } else if (s[i] == ']' ) {
+                if ( _stack.top() != '[' ) return false;
+                _stack.pop();
+            }
+        }
+        
+        return (_stack.size() == 0);
+    }
+    
+    bool isNum(char ch) {
+        return (ch>='0' && ch<='9');
+    }
+};
+
diff --git a/algorithms/cpp/longestSubstringWithAtLeastKRepeatingCharacters/LongestSubstringWithAtLeastKRepeatingCharacters.cpp b/algorithms/cpp/longestSubstringWithAtLeastKRepeatingCharacters/LongestSubstringWithAtLeastKRepeatingCharacters.cpp
@@ -0,0 +1,109 @@
+// Source : https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/
+// Author : Hao Chen
+// Date   : 2016-09-08
+
+/*************************************************************************************** 
+ *
+ * Find the length of the longest substring T of a given string (consists of lowercase 
+ * letters only) such that every character in T appears no less than k times.
+ * 
+ * Example 1:
+ * 
+ * Input:
+ * s = "aaabb", k = 3
+ * 
+ * Output:
+ * 3
+ * 
+ * The longest substring is "aaa", as 'a' is repeated 3 times.
+ * 
+ * Example 2:
+ * 
+ * Input:
+ * s = "ababbc", k = 2
+ * 
+ * Output:
+ * 5
+ * 
+ * The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 
+ * times.
+ ***************************************************************************************/
+
+const int NO_OF_CHARS = 256;
+
+/* if every character appears at least k times, the whole string is ok. 
+ * Otherwise split by a least frequent character.
+ * 
+ * Because it will always be too infrequent and thus can't be part of any ok substring
+ * and make the most out of the splits.
+ */
+
+
+class Solution {
+public:
+    int longestSubstring(string s, int k) {
+        
+        //deal with edge cases
+        if (s.size() == 0 || s.size() < k) return 0;
+        if (k==1) return s.size();
+
+        //declare a map for every char's counter
+        int count[NO_OF_CHARS];
+        memset(count , 0, sizeof(count));
+        
+        //counting every char
+        for (char ch : s) {
+            count[ch]++;
+        }
+        
+        int i=0;
+        for ( i=0; i<NO_OF_CHARS; i++) {
+            if (count[i] !=0 && count[i] < k) break;
+        }
+        //all of the chars meet the requirement
+        if ( i >= NO_OF_CHARS ) return s.size();
+        
+        // find the most infrequent char
+        char least = 0;
+        for (int c = 0; c < NO_OF_CHARS; c++) {
+            if (count[c] == 0) continue;
+            if (least == 0) {
+                least = c;
+            } else if ( count[c] < count[least]) {
+                least = c;
+            }
+        }
+        
+        //split the string and run them recursively
+        vector<string> subs;
+        split(s, least, subs);
+        
+        int res = 0;
+        for (string str: subs) {
+            res = max(res, longestSubstring(str, k));
+        }
+        return res;
+        return 0;
+    }
+    
+private:
+    
+    inline int max(int x, int y) { return x>y? x:y; }
+    
+    inline void split(const string &s, char delim, vector<string> &elems) {
+        stringstream ss;
+        ss.str(s);
+        string item;
+        while (getline(ss, item, delim)) {
+            cout << item << endl;
+            elems.push_back(item);
+        }
+    }
+
+
+    inline vector<string> split(const string &s, char delim) {
+        vector<string> elems;
+        split(s, delim, elems);
+        return elems;
+    }
+};
