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Find longest common sub sequence in two strings.
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| #include <string> | |
| std::string LongestSubSeq(std::string s1, std::string s2) | |
| { | |
| // space complexity O(2 *LengthOfLongestString) | |
| // time Complexity O(s1Length*s2Length) | |
| size_t Len = s1.length() > s2.length() ? s1.length() : s2.length(); | |
| std::string result,temp; | |
| result.reserve(Len); | |
| temp.reserve(Len); | |
| for (size_t s1idx{ 0 }; s1idx < s1.length(); ++s1idx) | |
| { | |
| const char currCh = s1[s1idx]; | |
| size_t s2firstPos = s2.find(currCh, 0); | |
| if (s2firstPos == s2.npos) continue; | |
| temp.clear(); | |
| temp.push_back(currCh); | |
| size_t s2nextPos = s2firstPos + 1; | |
| for (size_t s1subIdx{ s1idx + 1 }; s1subIdx < s1.length(); ++s1subIdx) | |
| { | |
| const char nextCh = s1[s1subIdx]; | |
| size_t s2subPos = s2.find(nextCh, s2nextPos); | |
| if (s2subPos != s2.npos) | |
| { | |
| temp.push_back(nextCh); | |
| s2nextPos = s2subPos + 1; | |
| } | |
| } | |
| if (temp.length() > result.length()) | |
| { | |
| result = temp; | |
| } | |
| } | |
| return result; | |
| } |
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