New Problem Solution - "Sum of Beauty of All Substrings"

Author: haoel

README.md

@@ -12,6 +12,7 @@ LeetCode
 |1786|[Number of Restricted Paths From First to Last Node](https://leetcode.com/problems/number-of-restricted-paths-from-first-to-last-node/) | [C++](./algorithms/cpp/numberOfRestrictedPathsFromFirstToLastNode/NumberOfRestrictedPathsFromFirstToLastNode.cpp)|Medium|
 |1785|[Minimum Elements to Add to Form a Given Sum](https://leetcode.com/problems/minimum-elements-to-add-to-form-a-given-sum/) | [C++](./algorithms/cpp/minimumElementsToAddToFormAGivenSum/MinimumElementsToAddToFormAGivenSum.cpp)|Medium|
 |1784|[Check if Binary String Has at Most One Segment of Ones](https://leetcode.com/problems/check-if-binary-string-has-at-most-one-segment-of-ones/) | [C++](./algorithms/cpp/checkIfBinaryStringHasAtMostOneSegmentOfOnes/CheckIfBinaryStringHasAtMostOneSegmentOfOnes.cpp)|Easy|
+|1781|[Sum of Beauty of All Substrings](https://leetcode.com/problems/sum-of-beauty-of-all-substrings/) | [C++](./algorithms/cpp/sumOfBeautyOfAllSubstrings/SumOfBeautyOfAllSubstrings.cpp)|Medium|
 |1780|[Check if Number is a Sum of Powers of Three](https://leetcode.com/problems/check-if-number-is-a-sum-of-powers-of-three/) | [C++](./algorithms/cpp/checkIfNumberIsASumOfPowersOfThree/CheckIfNumberIsASumOfPowersOfThree.cpp)|Medium|
 |1779|[Find Nearest Point That Has the Same X or Y Coordinate](https://leetcode.com/problems/find-nearest-point-that-has-the-same-x-or-y-coordinate/) | [C++](./algorithms/cpp/findNearestPointThatHasTheSameXOrYCoordinate/FindNearestPointThatHasTheSameXOrYCoordinate.cpp)|Easy|
 |1763|[Longest Nice Substring](https://leetcode.com/problems/longest-nice-substring/) | [C++](./algorithms/cpp/longestNiceSubstring/LongestNiceSubstring.cpp)|Easy|

algorithms/cpp/sumOfBeautyOfAllSubstrings/SumOfBeautyOfAllSubstrings.cpp

@@ -0,0 +1,82 @@
+// Source : https://leetcode.com/problems/sum-of-beauty-of-all-substrings/
+// Author : Hao Chen
+// Date   : 2021-03-13
+
+/***************************************************************************************************** 
+ *
+ * The beauty of a string is the difference in frequencies between the most frequent and least 
+ * frequent characters.
+ * 
+ * 	For example, the beauty of "abaacc" is 3 - 1 = 2.
+ * 
+ * Given a string s, return the sum of beauty of all of its substrings.
+ * 
+ * Example 1:
+ * 
+ * Input: s = "aabcb"
+ * Output: 5
+ * Explanation: The substrings with non-zero beauty are ["aab","aabc","aabcb","abcb","bcb"], each with 
+ * beauty equal to 1.
+ * 
+ * Example 2:
+ * 
+ * Input: s = "aabcbaa"
+ * Output: 17
+ * 
+ * Constraints:
+ * 
+ * 	1 <= s.length <= 500
+ * 	s consists of only lowercase English letters.
+ ******************************************************************************************************/
+
+class Solution {
+private:
+    int beauty(string& s, int start, int end) {
+        int stat[26] = {0};
+        for (int i=start; i<=end; i++){
+            stat[s[i]-'a']++;
+        }
+        int max = INT_MIN, min = INT_MAX; 
+        for (auto s: stat) {
+            if (s == 0 ) continue;
+            max = s > max ? s : max;
+            min = s < min ? s : min;
+        }
+        return max - min;
+    }
+public:
+    int beautySum(string s) {
+        return beautySum02(s);
+        return beautySum01(s);
+    }
+    
+    int beautySum01(string& s) {
+        int sum = 0; 
+        for (int i=0; i<s.size()-1; i++) {
+            for (int j=i+1; j<s.size(); j++) {
+                sum += beauty(s, i, j);
+            }
+        }
+        return sum;
+    }
+    
+    //same as beautySum01(), but optimazed slightly
+    int beautySum02(string& s) {
+        int sum = 0; 
+        for (int i=0; i<s.size()-1; i++) {
+            int stat[26] = {0};
+            for (int j=i; j<s.size(); j++) {
+                stat[s[j]-'a']++;
+                int max = INT_MIN, min = INT_MAX; 
+                for (auto& n: stat) {
+                    if (n <= 0 ) continue; 
+                    max = n > max ? n : max;
+                    min = n < min ? n : min;
+                }
+                //cout << s.substr(i, j-i+1)  << " --> "<< max << ":" << min << endl;
+                sum += (max - min);
+            }
+        }
+        return sum;
+    }
+};