Formatted question description: https://leetcode.ca/all/1048.html

1048. Longest String Chain (Medium)

Given a list of words, each word consists of English lowercase letters.

Let's say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2.  For example, "abc" is a predecessor of "abac".

A word chain is a sequence of words [word_1, word_2, ..., word_k] with k >= 1, where word_1 is a predecessor of word_2, word_2 is a predecessor of word_3, and so on.

Return the longest possible length of a word chain with words chosen from the given list of words.

 

Example 1:

Input: ["a","b","ba","bca","bda","bdca"]
Output: 4
Explanation: one of the longest word chain is "a","ba","bda","bdca".

 

Note:

  1. 1 <= words.length <= 1000
  2. 1 <= words[i].length <= 16
  3. words[i] only consists of English lowercase letters.

 

Related Topics:
Hash Table, Dynamic Programming

Solution 1.

// OJ: https://leetcode.com/problems/longest-string-chain/

// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int longestStrChain(vector<string>& A) {
        int N = A.size(), ans = 1;
        vector<int> id(N), len(N, 1);
        iota(begin(id), end(id), 0);
        sort(begin(id), end(id), [&](int a, int b) { return A[a].size() < A[b].size(); });
        vector<array<int, 26>> cnts(N);
        map<int, vector<int>> m;
        for (int i = 0; i < N; ++i) {
            for (char c : A[i]) cnts[i][c - 'a']++;
            m[A[i].size()].push_back(i);
        }
        for (auto &[cnt, ids] : m) {
            if (m.count(cnt - 1) == 0) continue;
            for (int i : ids) {
                for (int j : m[cnt - 1]) {
                    int diff = 0;
                    for (int k = 0; k < 26; ++k) {
                        diff += abs(cnts[i][k] - cnts[j][k]);
                        if (diff > 1) break;
                    }
                    if (diff == 1) len[i] = max(len[i], len[j] + 1);
                }
                ans = max(ans, len[i]);
            }
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/longest-string-chain/

// Time: O(NSS)
// Space: O(NS)
// Ref: https://leetcode.com/problems/longest-string-chain/discuss/294890/C%2B%2BJavaPython-DP-Solution
class Solution {
public:
    int longestStrChain(vector<string>& A) {
        auto cmp = [](string &a, string &b) { return a.size() < b.size(); };
        sort(begin(A), end(A), cmp);
        unordered_map<string, int> dp;
        int ans = 1;
        for (auto &s : A) {
            for (int i = 0; i < s.size(); ++i) {
                dp[s] = max(dp[s], dp[s.substr(0, i) + s.substr(i + 1)] + 1);
                ans = max(ans, dp[s]);
            }
        }
        return ans;
    }
};

Java

class Solution {
    public int longestStrChain(String[] words) {
        Arrays.sort(words, new Comparator<String>() {
            public int compare(String str1, String str2) {
                if (str1.length() != str2.length())
                    return str1.length() - str2.length();
                else
                    return str1.compareTo(str2);
            }
        });
        int smallestLength = words[0].length();
        int nextLength = smallestLength + 1;
        int length = words.length;
        int[] dp = new int[length];
        int dpStartIndex = -1;
        for (int i = 0; i < length; i++) {
            dp[i] = 1;
            if (words[i].length() == nextLength && dpStartIndex < 0)
                dpStartIndex = i;
        }
        if (dpStartIndex < 0)
            return 1;
        for (int i = dpStartIndex; i < length; i++) {
            String curWord = words[i];
            for (int j = i - 1; j >= 0; j--) {
                String prevWord = words[j];
                if (prevWord.length() == curWord.length())
                    continue;
                else if (prevWord.length() < curWord.length() - 1)
                    break;
                else {
                    if (isSubsequence(prevWord, curWord))
                        dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
        }
        int maxLength = 0;
        for (int num : dp)
            maxLength = Math.max(maxLength, num);
        return maxLength;
    }

    public boolean isSubsequence(String str1, String str2) {
        int length1 = str1.length(), length2 = str2.length();
        if (length1 > length2)
            return false;
        int index1 = 0, index2 = 0;
        while (index1 < length1 && index2 < length2) {
            char c1 = str1.charAt(index1), c2 = str2.charAt(index2);
            if (c1 == c2)
                index1++;
            index2++;
        }
        return index1 == length1;
    }
}

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