792. Number of Matching Subsequences

/**
 * Pre Processing Time : O(words.length)
 *
 * Time Complexity: O(S.length() + Summation of all chars in all words)
 *
 * Space Complexity: O(words.length)
 */
class Solution {
    class Word {
        String word;
        int index;

        Word(String w) {
            word = w;
            index = 0;
        }
    }

    public int numMatchingSubseq(String S, String[] words) {
        if (S == null || words == null || S.length() == 0 || words.length == 0) {
            return 0;
        }

        Map<Character, List<Word>> map = new HashMap();
        for (String word : words) {
            if (word.length() > S.length()) {
                continue;
            }
            char c = word.charAt(0);
            if (!map.containsKey(c)) {
                map.put(c, new ArrayList<Word>());
            }
            map.get(c).add(new Word(word));
        }

        int result = 0;
        for (char c : S.toCharArray()) {
            if (!map.containsKey(c)) {
                continue;
            }
            List<Word> curBucket = map.get(c);
            map.put(c, new ArrayList<Word>());
            // Overall this for loop will run for summation of all chars in all words.
            for (Word w : curBucket) {
                w.index++;
                if (w.word.length() == w.index) {
                    result++;
                } else {
                    char nextChar = w.word.charAt(w.index);
                    if (!map.containsKey(nextChar)) {
                        map.put(nextChar, new ArrayList<Word>());
                    }
                    map.get(nextChar).add(w);
                }
            }
            curBucket.clear();
        }
        return result;
    }
}

/*
 * Using Solution of https://leetcode.com/problems/is-subsequence
 *
 * Pre Processing Time = O(S.length())
 *
 * Time Complexity: O(Sum of all word lengths * log S)
 *
 * Space Complexity: O(S)
 */
class Solution {
    public int numMatchingSubseq(String S, String[] words) {
        if (S == null || words == null || S.length() == 0 || words.length == 0) {
            return 0;
        }

        Map<Character, List<Integer>> map = processStr(S);
        int result = 0;
        for (String word : words) {
            if (word.length() > S.length()) {
                continue;
            }
            if (isSubsequence(map, word)) {
                result++;
            }
        }
        return result;
    }

    private Map<Character, List<Integer>> processStr(String s) {
        Map<Character, List<Integer>> map = new HashMap();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (!map.containsKey(c)) {
                map.put(c, new ArrayList<Integer>());
            }
            map.get(c).add(i);
        }
        return map;
    }

    private boolean isSubsequence(Map<Character, List<Integer>> map, String s) {
        int prevIdx = -1;
        for (char c : s.toCharArray()) {
            if (!map.containsKey(c)) {
                return false;
            }
            int nextIdx = binarySearchInList(map.get(c), prevIdx);
            if (nextIdx == -1) {
                return false;
            }
            prevIdx = nextIdx;
        }
        return true;
    }

    private int binarySearchInList(List<Integer> list, int prev) {
        int start = 0;
        int end = list.size() - 1;
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (list.get(mid) <= prev) {
                start = mid + 1;
            } else {
                end = mid;
            }
        }
        return list.get(start) > prev ? list.get(start) : -1;
    }
}