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Substring with Concatenation of All Words
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/** | |
* You are given a string, S, and a list of words, L, that are all of the same length. | |
* Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters. | |
* For example, given: | |
* S: "barfoothefoobarman" | |
* L: ["foo", "bar"] | |
* You should return the indices: [0,9]. | |
* (order does not matter). | |
*/ | |
public class Solution { | |
public ArrayList<Integer> findSubstring(String S, String[] L) { | |
ArrayList<Integer> indices = new ArrayList<Integer>(); | |
if (L.length == 0) | |
return indices; | |
// notice: all the same length | |
int total = L.length, wordLen = L[0].length(); | |
// Store the words and frequences in a hash table for L array | |
HashMap<String, Integer> words = new HashMap<String, Integer>(); | |
for (String s : L) { | |
if(words.containsKey(s)) | |
words.put(s, words.get(s) + 1); | |
else | |
words.put(s, 1); | |
} | |
// find the concatenations | |
for (int i = 0; i <= S.length() - total * wordLen; i++) { | |
// check if it is a concatenation | |
HashMap<String, Integer> target = new HashMap<String, Integer>(words); | |
for (int j = i; j <= S.length() - wordLen && !target.isEmpty(); j += wordLen) { | |
String sub = S.substring(j, j + wordLen); | |
if(!target.containsKey(sub)) | |
break; | |
else if (target.get(sub) > 1) { | |
// reduce the frequency | |
target.put(sub, target.get(sub) - 1); | |
} else { | |
// remove the word if only one left | |
target.remove(sub); | |
} | |
} | |
if (target.isEmpty()) { | |
indices.add(i); | |
} | |
} | |
return indices; | |
} | |
public List<Integer> findSubstring(String s, String[] words) { | |
List<Integer> res = new ArrayList<Integer>(); | |
if(words.length == 0) return res; | |
int count = words.length, wordLen = words[0].length(); | |
HashMap<String, Integer> map = new HashMap<String, Integer>(); | |
for(String word : words) { | |
map.put(word, map.getOrDefault(word, 0) + 1); | |
} | |
HashMap<String, Integer> found = new HashMap<String, Integer>(); | |
for(int i = 0; i <= s.length() - count * wordLen; i++) { | |
found.clear(); | |
for(int j = 0; j < count; j++) { | |
int k = i + j * wordLen; | |
String str = s.substring(k, k + wordLen); | |
if(!map.containsKey(str)) break; | |
found.put(str, found.getOrDefault(str, 0) + 1); | |
if(found.get(str) > map.get(str)) break; | |
if(j == count - 1) res.add(i); | |
} | |
} | |
return res; | |
} | |
} |
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