30. Substring with Concatenation of All Words

Problem Link

Description

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You are given a string s and an array of strings words. All the strings of words are of the same length.

A concatenated string is a string that exactly contains all the strings of any permutation of words concatenated.

• For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated string because it is not the concatenation of any permutation of words.

Return an array of the starting indices of all the concatenated substrings in s. You can return the answer in any order.

 

Example 1:

Input: s = "barfoothefoobarman", words = ["foo","bar"]

Output: [0,9]

Explanation:

The substring starting at 0 is "barfoo". It is the concatenation of ["bar","foo"] which is a permutation of words.
The substring starting at 9 is "foobar". It is the concatenation of ["foo","bar"] which is a permutation of words.

Example 2:

Input: s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]

Output: []

Explanation:

There is no concatenated substring.

Example 3:

Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]

Output: [6,9,12]

Explanation:

The substring starting at 6 is "foobarthe". It is the concatenation of ["foo","bar","the"].
The substring starting at 9 is "barthefoo". It is the concatenation of ["bar","the","foo"].
The substring starting at 12 is "thefoobar". It is the concatenation of ["the","foo","bar"].

 

Constraints:

• 1 <= s.length <= 104
• 1 <= words.length <= 5000
• 1 <= words[i].length <= 30
• s and words[i] consist of lowercase English letters.

Solution

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C++

class Solution {
public:
    vector<int> findSubstring(string S, vector<string>& L) {
            // travel all the words combinations to maintain a window
    // there are wl(word len) times travel
    // each time, n/wl words, mostly 2 times travel for each word
    // one left side of the window, the other right side of the window
    // so, time complexity O(wl * 2 * N/wl) = O(2N)
        vector<int> ans;
        int n = S.size(), cnt = L.size();
        if (n <= 0 || cnt <= 0) return ans;
        
        // init word occurence
        unordered_map<string, int> dict;
        for (int i = 0; i < cnt; ++i) dict[L[i]]++;
        
        // travel all sub string combinations
        int wl = L[0].size();
        for (int i = 0; i < wl; ++i) {
            int left = i, count = 0;
            unordered_map<string, int> tdict;
            for (int j = i; j <= n - wl; j += wl) {
                string str = S.substr(j, wl);
                // a valid word, accumulate results
                if (dict.count(str)) {
                    tdict[str]++;
                    if (tdict[str] <= dict[str]) 
                        count++;
                    else {
                        // a more word, advance the window left side possiablly
                        while (tdict[str] > dict[str]) {
                            string str1 = S.substr(left, wl);
                            tdict[str1]--;
                            if (tdict[str1] < dict[str1]) count--;
                            left += wl;
                        }
                    }
                    // come to a result
                    if (count == cnt) {
                        ans.push_back(left);
                        // advance one word
                        tdict[S.substr(left, wl)]--;
                        count--;
                        left += wl;
                    }
                }
                // not a valid word, reset all vars
                else {
                    tdict.clear();
                    count = 0;
                    left = j + wl;
                }
            }
        }
        
        return ans;
    }
};