[LeetCode] Substring with Concatenation of All Words
I think the following code is self-explanatory enough. We use an unordered_map<string, int> counts to record the expected times of each word and another unordered_map<string, int> seento record the times we have seen. Then we check for every possible position of i. Once we meet an unexpected word or the times of some word is larger than its expected times, we stop the check. If we finish the check successfully, push i to the result indexes.
1 class Solution { 2 public: 3 vector<int> findSubstring(string s, vector<string>& words) { 4 unordered_map<string, int> counts; 5 for (string word : words) 6 counts[word]++; 7 int n = s.length(), num = words.size(), len = words[0].length(); 8 vector<int> indexes; 9 for (int i = 0; i < n - num * len + 1; i++) { 10 unordered_map<string, int> seen; 11 int j = 0; 12 for (; j < num; j++) { 13 string word = s.substr(i + j * len, len); 14 if (counts.find(word) != counts.end()) { 15 seen[word]++; 16 if (seen[word] > counts[word]) 17 break; 18 } 19 else break; 20 } 21 if (j == num) indexes.push_back(i); 22 } 23 return indexes; 24 } 25 };
The following is a more sophisticated solution taken from this link. You may need to spend some time understanding how it works.
1 class Solution { 2 public: 3 vector<int> findSubstring(string s, vector<string>& words) { 4 int n = s.length(), len = words[0].length(), num = words.size(); 5 unordered_map<string, int> counts; 6 for (string word : words) 7 counts[word]++; 8 vector<int> indexes; 9 for (int i = 0; i < len; i++) { 10 int left = i, valid = 0; 11 unordered_map<string, int> seen; 12 for (int j = i; j <= n - len; j += len) { 13 string word = s.substr(j, len); 14 if (counts.find(word) != counts.end()) { 15 seen[word]++; 16 if (seen[word] <= counts[word]) 17 valid++; 18 else { 19 while (seen[word] > counts[word]) { 20 string wd = s.substr(left, len); 21 seen[wd]--; 22 if (seen[wd] < counts[wd]) valid--; 23 left += len; 24 } 25 } 26 if (valid == num) { 27 indexes.push_back(left); 28 seen[s.substr(left, len)]--; 29 valid--; 30 left += len; 31 } 32 } 33 else { 34 seen.clear(); 35 valid = 0; 36 left = j + len; 37 } 38 } 39 } 40 return indexes; 41 } 42 };
