245. Shortest Word Distance III

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

word1 and word2 may be the same and they represent two individual words in the list.

Example: Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Input: word1 = “makes”, word2 = “coding”
Output: 1

Input: word1 = "makes", word2 = "makes"
Output: 3

Note: You may assume word1 and word2 are both in the list.

// Brute Force
int shortestWordDistance(vector<string>& words, string word1, string word2) { // time: O(n^2); space: O(1)
    if (words.empty()) return 0;
    int n = words.size(), res = INT_MAX;
    for (int i = 0; i < n; ++i) {
        if (words[i] != word1) continue;
        for (int j = 0; j < n; ++j) {
            if (words[j] != word2 || j == i) continue;
            res = min(res, abs(i - j));
        }
    }
    return res;
}

// Optimized method
int shortestWordDistance(vector<string>& words, string word1, string word2) { // time: O(n); space: O(1)
    int idx1 = -1, idx2 = -1, res = words.size();
    for (int i = 0; i < words.size(); ++i) {
        int t = idx1; // for duplicates
        if (words[i] == word1) {
            idx1 = i;
        }
        if (words[i] == word2) {
            idx2 = i;
        }
        if (idx1 != -1 && idx2 != -1) {
            if (word1 == word2 && t != -1 && t != idx1) {
                res = min(res, abs(idx1 - t));
            } else if (idx1 != idx2) {
                res = min(res, abs(idx1 - idx2));
            }
        }
    }
    return res;
}

// Optimized method
int shortestWordDistance(vector<string>& words, string word1, string word2) { // time: O(n); space: O(1)
    int idx1 = words.size(), idx2 = -words.size(), res = INT_MAX;
    for (int i = 0; i < words.size(); ++i) {
        if (words[i] == word1) idx1 = word1 == word2 ? idx2 : i;
        if (words[i] == word2) idx2 = i;
        res = min(res, abs(idx1 - idx2));
    }
    return res;
}

// Optimized method
int shortestWordDistance(vector<string>& words, string word1, string word2) { // time: O(n); space: O(1)
    int idx = -1, res = words.size();
    for (int i = 0; i < words.size(); ++i) {
        if (words[i] == word1 || words[i] == word2) {
            if (idx != -1 && (word1 == word2 || words[idx] != words[i])) {
                res = min(res, i - idx);
            }
            idx = i;
        }
    }
    return res;
}