30. Substring with Concatenation of All Words && 76. Minimum Window Substring

30. Substring with Concatenation of All Words

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s:

"barfoothefoobarman"

words:

["foo", "bar"]

You should return the indices:

[0,9]

.
(order does not matter).

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Hash Table Two Pointers String

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(H) Minimum Window Substring

Brute-force solution: (There are better solutions.)

public class Solution {
public static List<Integer> findSubstring(String s, String[] words) {
List<Integer> res = new ArrayList<>();
if (s == null || words == null || words.length == 0)
return res;
int len = words[0].length(); //all of words are of the same length

Map<String, Integer> wordCountMap = new HashMap<>();
for (String w : words)
wordCountMap.put(w, wordCountMap.containsKey(w) ? wordCountMap.get(w) + 1 : 1);

for (int i = 0; i <= s.length() - len * words.length; ++i) {
//the copy of this map keeps the remaining of words to be used.
Map<String, Integer> copy = new HashMap<>(wordCountMap);

for (int j = 0; j < words.length; ++j) {
int start = i + j * len;
String word = s.substring(start, start + len); // next word
Integer count = copy.get(word);
if (count == null)
break; //the word not in map

if (count.equals(1))
copy.remove(word);
else
copy.put(word, count - 1);

//check the map after updates
if (copy.isEmpty()) {
res.add(i);
break;
}
}
}
return res;
}
}

76. Minimum Window Substring

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.

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Hash Table Two Pointers String

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(H) Substring with Concatenation of All Words (M) Minimum Size Subarray Sum (H) Sliding Window Maximum

This solution could serve as a template for this type of substring problems.

public class Solution {
public String minWindow(String s, String t) {
Map<Character, Integer> tcCount = new HashMap<>(t.length());
for (int i = 0; i < t.length(); ++i) {
Character c = t.charAt(i);
Integer count = tcCount.get(c);
if (count == null)
tcCount.put(c, 1);
else
tcCount.put(c, count + 1);
}

int counter = t.length(); // indicate how many more need to be matched.
int begin = 0, end = 0; //two pointers, one points to head; one tail.
int d = Integer.MAX_VALUE; //the length of substring

int head = 0;

while (end < s.length()) {
char tailChar = s.charAt(end++);
Integer tailCharCount = tcCount.get(tailChar);
if (tailCharCount == null)
continue;
if(tailCharCount > 0) // found a matching character.
--counter;
tcCount.put(tailChar, tailCharCount - 1); //must allow it to go negative.

while (counter == 0) { //Valid. All character counts in the map == 0.
if (end - begin < d) {
head = begin; //cache the updated substring.
d = end - begin;
}
char exitingChar = s.charAt(begin++);
Integer exitingCharCount = tcCount.get(exitingChar);

if (exitingCharCount == null)
continue;
if(exitingCharCount == 0) //the substring no longer matches the condition.
++counter;
tcCount.put(exitingChar, exitingCharCount + 1);
}
}
return d == Integer.MAX_VALUE ? "" : s.substring(head, head + d);
}
}