Substring with Concatenation of All Words
2014.2.27 00:46
You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.
For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]
You should return the indices: [0,9].
(order does not matter).
Solution:
My solution to this problem is basically brute-force, starting from every position and check if a complete concatenation is possible.
<unordered_map> will provide efficient hashing.
Total time complexity is O(n^3). Space complexity is O(n).
Accepted code:
1 #include <string> 2 #include <unordered_map> 3 #include <vector> 4 using namespace std; 5 6 class Solution { 7 public: 8 vector<int> findSubstring(string S, vector<string> &L) { 9 vector<int> result_set; 10 vector<int> vi; 11 vector<int> vic; 12 unordered_map<string, int> um; 13 unordered_map<string, int>::iterator uit; 14 int wl; 15 int nc; 16 int n = (int)L.size(); 17 int slen = (int)S.length(); 18 19 if (slen == 0 || n == 0) { 20 return result_set; 21 } 22 wl = (int)L[0].length(); 23 if (wl == 0) { 24 return result_set; 25 } 26 if (slen < n * wl) { 27 return result_set; 28 } 29 30 um.clear(); 31 vi.clear(); 32 nc = 0; 33 34 int i, j; 35 for (i = 0; i < n; ++i) { 36 uit = um.find(L[i]); 37 if (uit == um.end()) { 38 um[L[i]] = nc; 39 vi.push_back(1); 40 ++nc; 41 } else { 42 ++vi[um[L[i]]]; 43 } 44 } 45 vic.resize(nc); 46 47 int cc, ll, rr; 48 int idx; 49 string str; 50 51 for (i = 0; i < wl; ++i) { 52 for (j = 0; j < nc; ++j) { 53 vic[j] = vi[j]; 54 } 55 cc = 0; 56 57 ll = i; 58 rr = ll; 59 while (rr + wl <= slen) { 60 str = S.substr(rr, wl); 61 uit = um.find(str); 62 if (uit != um.end()) { 63 idx = uit->second; 64 if (vic[idx] == 0) { 65 while (true) { 66 str = S.substr(ll, wl); 67 if (um[str] != idx) { 68 ll += wl; 69 ++vic[um[str]]; 70 --cc; 71 } else { 72 ll += wl; 73 ++vic[um[str]]; 74 --cc; 75 break; 76 } 77 } 78 } else { 79 --vic[idx]; 80 ++cc; 81 rr += wl; 82 } 83 } else { 84 ll = rr + wl; 85 for (j = 0; j < nc; ++j) { 86 vic[j] = vi[j]; 87 } 88 rr = ll; 89 cc = 0; 90 } 91 if (cc == n) { 92 result_set.push_back(ll); 93 str = S.substr(ll, wl); 94 idx = um[str]; 95 ++vic[idx]; 96 --cc; 97 ll += wl; 98 } 99 } 100 } 101 vic.clear(); 102 103 return result_set; 104 } 105 };