Runtime: 1433 ms (Top 23.41%) | Memory: 179.2 MB (Top 26.07%)

Author: AnasImloul

scripts/algorithms/L/Longest Substring of One Repeating Character/Longest Substring of One Repeating Character.cpp

@@ -1,20 +1,21 @@
+// Runtime: 1433 ms (Top 23.41%) | Memory: 179.2 MB (Top 26.07%)
 class Solution {
 public:
     vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
         multiset<int> ls; // ordered set of lengths of all intervals at present
         set<int> ins; // ordered set representing the left-ends of intervals
         int n = s.length();
         ins.insert(n); // allow us to get length of current interval by *next(it) - *it for all cases
-        
+
         // initialize the ordered sets with the original string
         for (int i = 0, j = 1; i < n;) {
             while (j < n && s[j] == s[j-1]) ++j;
             ins.insert(i);
             ls.insert(j-i);
             i = j++;
         }
-        
-        // update the string and track the length of the longest substring 
+
+        // update the string and track the length of the longest substring
         // by merging/splitting intervals
         vector<int> ans;
         for (int i = 0; i < queryIndices.size(); ++i) {
@@ -24,41 +25,41 @@ class Solution {
                 ans.push_back(*ls.rbegin());
                 continue;
             }
-            
+
             // update the ordered sets
             // 1. split the involved interval to parts: (left), self, (right)
-            //    ins = 0     3     6
-            //    s   = b b b c c c
-            //  query :       b
-            // => ins = 0     3 4   6  
-            // => s   = b b b b c c
-            
+            // ins = 0 3 6
+            // s = b b b c c c
+            // query : b
+            // => ins = 0 3 4 6
+            // => s = b b b b c c
+
             // find the left-end of the target interval
             auto it = prev(ins.upper_bound(queryIndices[i])); // old/left interval
             int oldl = *next(it) - *it; // old interval length
             ls.erase(ls.find(oldl));
-             
-            ins.insert(queryIndices[i]); // self interval 
+
+            ins.insert(queryIndices[i]); // self interval
             ins.insert(queryIndices[i] + 1); // right interval
             if (*it != queryIndices[i]) ls.insert(queryIndices[i] - *it); // left interval length
             ls.insert(1); // self interval length
             if (oldl - (queryIndices[i] - *it + 1)) ls.insert(oldl - (queryIndices[i] - *it + 1) ); //right interval length
-            
+
             s[queryIndices[i]] = queryCharacters[i];
 
             // 2. merge the adjacent intervals if the characters are identical
-            //    ins = 0     3 4   6  
-            //    s   = b b b b c c
-            // => ins = 0       4   6
-            
+            // ins = 0 3 4 6
+            // s = b b b b c c
+            // => ins = 0 4 6
+
             it = ins.find(queryIndices[i]); // self interval
             // merge self and right
             if (queryIndices[i] + 1 < n && s[queryIndices[i]] == s[queryIndices[i] + 1]) {
                 ls.erase(ls.find(*next(it, 2) - *next(it)));
                 ls.erase(ls.find(1));
                 ins.erase(next(it));
                 ls.insert(*next(it) - *it);
-            } 
+            }
             // merge left and self
             if (it != ins.begin() && s[queryIndices[i]] == s[queryIndices[i] - 1]) {
                 ls.erase(ls.find(*next(it) - *it) );
@@ -71,4 +72,4 @@ class Solution {
         }
         return ans;
     }
-};
+};