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76 Minimum Window Substring

Problem

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example, S = "ADOBECODEBANC" T = "ABC" Minimum window is "BANC".

Note: If there is no such window in S that covers all characters in T, return the empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

Solutions

class Solution {
    public String minWindow(String s, String t) {
        HashMap<Character, Integer> miss = new HashMap<Character, Integer>();
        HashSet<Character> chars = new HashSet<Character>();
        HashMap<Character, Integer> count = new HashMap<Character, Integer>();
        for (int i = 0; i < t.length(); i ++) {
            char c = t.charAt(i);
            chars.add(c);
            if (!miss.containsKey(c)) {
                miss.put(c, 0);
            }
            if (!count.containsKey(c)) {
                count.put(c, 0);
            }
            miss.put(c, miss.get(c) + 1);
        }
        int left = -1, right = 0;
        int min = Integer.MAX_VALUE;
        String res = "";
        int target = t.length();
        int matched = 0;
        while (right <= s.length()) {
            if (matched == target) {
                String cand = s.substring(left, right);
                if (cand.length() < min) {
                    min = cand.length();
                    res = cand;
                }
                char toremove = s.charAt(left);
                count.put(toremove, count.get(toremove) - 1);
                if (count.get(toremove) < miss.get(toremove)) {
                    matched --;
                }
                left ++;
                while (left < right && !chars.contains(s.charAt(left))) {
                    left ++;
                }
                continue;
            }
            if (right == s.length()) {
                break;
            }
            char c = s.charAt(right);
            if (chars.contains(c)) {
                if (matched == 0) {
                    left = right;
                }
                count.put(c, count.get(c) + 1);
                if (count.get(c) <= miss.get(c)) {
                    matched ++;
                }
            }
            right ++;
        }
        return res;
    }
}

class Solution {
    public String minWindow(String s, String t) {
        HashMap<Character, Integer> count = new HashMap<>();
        HashMap<Character, Integer> has = new HashMap<>();
        int miss = t.length();
        for (int i = 0; i < t.length(); i ++) {
            char c = t.charAt(i);
            if (!count.containsKey(c)) {
                count.put(c, 0);
            }
            count.put(c, count.get(c) + 1);
        }
        String min = "";
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i ++) {
            char c = s.charAt(i);
            sb.append(c);
            if (!count.containsKey(c)) {
                continue;
            }
            if (!has.containsKey(c)) {
                has.put(c, 0);
            }
            has.put(c, has.get(c) + 1);
            if (has.get(c) <= count.get(c)) {
                miss --;
            }
            while (miss == 0) {
                if (min.equals("") || min.length() > sb.length()) {
                    min = sb.toString();
                }
                char remove = sb.charAt(0);
                sb.deleteCharAt(0);
                if (has.containsKey(remove)) {
                    has.put(remove, has.get(remove) - 1);
                    if (has.get(remove) < count.get(remove)) {
                        miss ++;
                    }
                }
            }
        }

        return min;
    }
}