New Problem Solution -"Longest Nice Substring"

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|
+|1763|[Longest Nice Substring](https://leetcode.com/problems/longest-nice-substring/) | [C++](./algorithms/cpp/longestNiceSubstring/LongestNiceSubstring.cpp)|Easy|
 |1761|[Minimum Degree of a Connected Trio in a Graph](https://leetcode.com/problems/minimum-degree-of-a-connected-trio-in-a-graph/) | [C++](./algorithms/cpp/minimumDegreeOfAConnectedTrioInAGraph/MinimumDegreeOfAConnectedTrioInAGraph.cpp)|Hard|
 |1760|[Minimum Limit of Balls in a Bag](https://leetcode.com/problems/minimum-limit-of-balls-in-a-bag/) | [C++](./algorithms/cpp/minimumLimitOfBallsInABag/MinimumLimitOfBallsInABag.cpp)|Medium|
 |1759|[Count Number of Homogenous Substrings](https://leetcode.com/problems/count-number-of-homogenous-substrings/) | [C++](./algorithms/cpp/countNumberOfHomogenousSubstrings/CountNumberOfHomogenousSubstrings.cpp)|Medium|

algorithms/cpp/longestNiceSubstring/LongestNiceSubstring.cpp

@@ -0,0 +1,79 @@
+// Source : https://leetcode.com/problems/longest-nice-substring/
+// Author : Hao Chen
+// Date   : 2021-03-08
+
+/***************************************************************************************************** 
+ *
+ * A string s is nice if, for every letter of the alphabet that s contains, it appears both in 
+ * uppercase and lowercase. For example, "abABB" is nice because 'A' and 'a' appear, and 'B' and 'b' 
+ * appear. However, "abA" is not because 'b' appears, but 'B' does not.
+ * 
+ * Given a string s, return the longest substring of s that is nice. If there are multiple, return the 
+ * substring of the earliest occurrence. If there are none, return an empty string.
+ * 
+ * Example 1:
+ * 
+ * Input: s = "YazaAay"
+ * Output: "aAa"
+ * Explanation: "aAa" is a nice string because 'A/a' is the only letter of the alphabet in s, and both 
+ * 'A' and 'a' appear.
+ * "aAa" is the longest nice substring.
+ * 
+ * Example 2:
+ * 
+ * Input: s = "Bb"
+ * Output: "Bb"
+ * Explanation: "Bb" is a nice string because both 'B' and 'b' appear. The whole string is a substring.
+ * 
+ * Example 3:
+ * 
+ * Input: s = "c"
+ * Output: ""
+ * Explanation: There are no nice substrings.
+ * 
+ * Example 4:
+ * 
+ * Input: s = "dDzeE"
+ * Output: "dD"
+ * Explanation: Both "dD" and "eE" are the longest nice substrings.
+ * As there are multiple longest nice substrings, return "dD" since it occurs earlier.
+ * 
+ * Constraints:
+ * 
+ * 	1 <= s.length <= 100
+ * 	s consists of uppercase and lowercase English letters.
+ ******************************************************************************************************/
+
+class Solution {
+    inline int getCharIndex(char c) {
+        return  (c >='A' && c <='Z') ? c - 'A' : c - 'a';
+    }
+    inline int getCaseIndex(char c) {
+         return (c >='A' && c <='Z') ? 1 : 0;
+    }
+public:
+    string longestNiceSubstring(string s) {
+        vector<bitset<26>> check(2);
+        int start = 0, len = 0;
+        for (int i = 0; i < s.size() -1; i++){
+            for (int j = i+1; j < s.size(); j++) {
+                
+                check[0] = check[1] = 0;
+                for (int x=i; x<=j; x++){
+                    int i = getCaseIndex(s[x]);
+                    int j = getCharIndex(s[x]);
+                    check[i][j] = true;
+                }
+                
+                if ( (check[0] ^ check[1]) == 0 ) {
+                    if ( j - i + 1 > len ){
+                        start = i;
+                        len = j-i+1;
+                    }
+                }
+            }
+        }
+        
+        return s.substr(start, len);
+    }
+};