New Solution "1071 - Greatest Common Divisor of Strings"

README.md

@@ -10,6 +10,7 @@ LeetCode
 |---| ----- | -------- | ---------- |
 |1207|[Unique Number of Occurrences](https://leetcode.com/problems/unique-number-of-occurrences/) | [C++](./algorithms/cpp/uniqueNumberOfOccurrences/Unique-Number-of-Occurrences.cpp)|Easy|
 |1170|[Compare Strings by Frequency of the Smallest Character](https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/) | [C++](./algorithms/cpp/compareStringsByFrequencyOfTheSmallestCharacter/CompareStringsByFrequencyOfTheSmallestCharacter.cpp)|Easy|
+|1071|[Greatest Common Divisor of Strings](https://leetcode.com/problems/greatest-common-divisor-of-strings/) | [C++](./algorithms/cpp/greatestCommonDivisorOfStrings/GreatestCommonDivisorOfStrings.cpp)|Easy|
 |1030|[Matrix Cells in Distance Order](https://leetcode.com/problems/matrix-cells-in-distance-order/) | [C++](./algorithms/cpp/matrixCellsInDistanceOrder/MatrixCellsInDistanceOrder.cpp)|Easy|
 |1029|[Two City Scheduling](https://leetcode.com/problems/two-city-scheduling/) | [C++](./algorithms/cpp/twoCityScheduling/TwoCityScheduling.cpp)|Easy|
 |1028|[Recover a Tree From Preorder Traversal](https://leetcode.com/problems/recover-a-tree-from-preorder-traversal/) | [C++](./algorithms/cpp/recoverATreeFromPreorderTraversal/recoverATreeFromPreorderTraversal.cpp)|Hard|

algorithms/cpp/greatestCommonDivisorOfStrings/GreatestCommonDivisorOfStrings.cpp

@@ -0,0 +1,87 @@
+// Source : https://leetcode.com/problems/greatest-common-divisor-of-strings/
+// Author : Hao Chen
+// Date   : 2020-07-19
+
+/***************************************************************************************************** 
+ *
+ * For strings S and T, we say "T divides S" if and only if S = T + ... + T  (T concatenated with 
+ * itself 1 or more times)
+ * 
+ * Return the largest string X such that X divides str1 and X divides str2.
+ * 
+ * Example 1:
+ * 
+ * Input: str1 = "ABCABC", str2 = "ABC"
+ * Output: "ABC"
+ * 
+ * Example 2:
+ * 
+ * Input: str1 = "ABABAB", str2 = "ABAB"
+ * Output: "AB"
+ * 
+ * Example 3:
+ * 
+ * Input: str1 = "LEET", str2 = "CODE"
+ * Output: ""
+ * 
+ * Note:
+ * 
+ * 	1 <= str1.length <= 1000
+ * 	1 <= str2.length <= 1000
+ * 	str1[i] and str2[i] are English uppercase letters.
+ ******************************************************************************************************/
+
+class Solution {
+private:
+
+
+    // Euclidean algorithm
+    // https://en.wikipedia.org/wiki/Euclidean_algorithm
+    // recursive way
+    int findGCD_r(int a, int b) {
+        if (a == 0)
+            return b;
+        return gcd(b % a, a);
+    }
+    // non-recursive way
+    int findGCD(int a, int b) {
+        int t = 1;
+        while(t != 0) {
+            t = a % b;
+            a = b;
+            b = t;
+        }
+        return a;
+    }
+
+    bool isStrRepeatByLen(string& s, int len) {
+        if (s.size() == len) return true;
+        if (s.size() % len != 0 ) return false;
+        for (int l=0; l<len; l++) {
+            for (int i=1; i<s.size()/len; i++) {
+                if (s[l] != s[i*len+l]) return false;
+            }
+        }
+        return true;
+    }
+
+    bool strPrefixComp(string& s1, string &s2, int len){
+        for(int i=0; i<len; i++) {
+            if (s1[i] != s2[i]) return false;
+        }
+        return true;
+    }
+public:
+    string gcdOfStrings(string s1, string s2) {
+
+        int gcd = findGCD(s1.size(), s2.size());
+
+        if (strPrefixComp(s1, s2, gcd) &&
+            isStrRepeatByLen(s1, gcd)&&
+            isStrRepeatByLen(s2, gcd) ){
+            return s2.substr(0,gcd);
+        }
+
+        return "";
+    }
+};