Given is a String word and a String array book that contains some strings. The program should give out the number of possibilities to create word only using elements in book. An element can be used as many times as we want and the program must terminate in under 6 seconds.
For example, input:
String word = "stackoverflow";
String[] book = new String[9];
book[0] = "st";
book[1] = "ck";
book[2] = "CAG";
book[3] = "low";
book[4] = "TC";
book[5] = "rf";
book[6] = "ove";
book[7] = "a";
book[8] = "sta";
The output should be 2, since we can create "stackoverflow" in two ways:
1: "st" + "a" + "ck" + "ove" + "rf" + "low"
2: "sta" + "ck" + "ove" + "rf" + "low"
My implementation of the program only terminates in the required time if word is relatively small (<15 characters). However, as I mentioned before, the running time limit for the program is 6 seconds and it should be able to handle very large word strings (>1000 characters). Here is an example of a large input.
Here is my code:
1) the actual method:
input: a String word and a String[] book
output: the number of ways word can be written only using strings in book
public static int optimal(String word, String[] book){
int count = 0;
List<List<String>> allCombinations = allSubstrings(word);
List<String> empty = new ArrayList<>();
List<String> wordList = Arrays.asList(book);
for (int i = 0; i < allCombinations.size(); i++) {
allCombinations.get(i).retainAll(wordList);
if (!sumUp(allCombinations.get(i), word)) {
allCombinations.remove(i);
allCombinations.add(i, empty);
}
else count++;
}
return count;
}
2) allSubstrings():
input: a String input
output: A list of lists, each containing a combination of substrings that add up to input
static List<List<String>> allSubstrings(String input) {
if (input.length() == 1) return Collections.singletonList(Collections.singletonList(input));
List<List<String>> result = new ArrayList<>();
for (List<String> temp : allSubstrings(input.substring(1))) {
List<String> firstList = new ArrayList<>(temp);
firstList.set(0, input.charAt(0) + firstList.get(0));
if (input.startsWith(firstList.get(0), 0)) result.add(firstList);
List<String> l = new ArrayList<>(temp);
l.add(0, input.substring(0, 1));
if (input.startsWith(l.get(0), 0)) result.add(l);
}
return result;
}
3.) sumup():
input: A String list input and a String expected
output: true if the elements in input add up to expected
public static boolean sumUp (List<String> input, String expected) {
String x = "";
for (int i = 0; i < input.size(); i++) {
x = x + input.get(i);
}
if (expected.equals(x)) return true;
return false;
}
I've figured out what I was doing wrong in my previous answer: I wasn't using memoization, so I was redoing an awful lot of unnecessary work.
Consider a book array {"a", "aa", "aaa"}, and a target word "aaa". There are four ways to construct this target:
"a" + "a" + "a"
"aa" + "a"
"a" + "aa"
"aaa"
My previous attempt would have walk through all four, separately. But instead, one can observe that:
There is 1 way to construct "a"
You can construct "aa" in 2 ways, either "a" + "a" or using "aa" directly.
You can construct "aaa" either by using "aaa" directly (1 way); or "aa" + "a" (2 ways, since there are 2 ways to construct "aa"); or "a" + "aa" (1 way).
Note that the third step here only adds a single additional string to a previously-constructed string, for which we know the number of ways it can be constructed.
This suggests that if we count the number of ways in which a prefix of word can be constructed, we can use that to trivially calculate the number of ways a longer prefix by adding just one more string from book.
I defined a simple trie class, so you can quickly look up prefixes of the book words that match at any given position in word:
class TrieNode {
boolean word;
Map<Character, TrieNode> children = new HashMap<>();
void add(String s, int i) {
if (i == s.length()) {
word = true;
} else {
children.computeIfAbsent(s.charAt(i), k -> new TrieNode()).add(s, i + 1);
}
}
}
For each letter in s, this creates an instance of TrieNode, and stores the TrieNode for the subsequent characters etc.
static long method(String word, String[] book) {
// Construct a trie from all the words in book.
TrieNode t = new TrieNode();
for (String b : book) {
t.add(b, 0);
}
// Construct an array to memoize the number of ways to construct
// prefixes of a given length: result[i] is the number of ways to
// construct a prefix of length i.
long[] result = new long[word.length() + 1];
// There is only 1 way to construct a prefix of length zero.
result[0] = 1;
for (int m = 0; m < word.length(); ++m) {
if (result[m] == 0) {
// If there are no ways to construct a prefix of this length,
// then just skip it.
continue;
}
// Walk the trie, taking the branch which matches the character
// of word at position (n + m).
TrieNode tt = t;
for (int n = 0; tt != null && n + m <= word.length(); ++n) {
if (tt.word) {
// We have reached the end of a word: we can reach a prefix
// of length (n + m) from a prefix of length (m).
// Increment the number of ways to reach (n+m) by the number
// of ways to reach (m).
// (Increment, because there may be other ways).
result[n + m] += result[m];
if (n + m == word.length()) {
break;
}
}
tt = tt.children.get(word.charAt(n + m));
}
}
// The number of ways to reach a prefix of length (word.length())
// is now stored in the last element of the array.
return result[word.length()];
}
For the very long input given by OP, this gives output:
$ time java Ideone
2217093120
real 0m0.126s
user 0m0.146s
sys 0m0.036s
Quite a bit faster than the required 6 seconds - and this includes JVM startup time too.
Edit: in fact, the trie isn't necessary. You can simply replace the "Walk the trie" loop with:
for (String b : book) {
if (word.regionMatches(m, b, 0, b.length())) {
result[m + b.length()] += result[m];
}
}
and it performs slower, but still way faster than 6s:
2217093120
real 0m0.173s
user 0m0.226s
sys 0m0.033s
A few observations:
x = x + input.get(i);
As you are looping, using String+ isn't a good idea. Use a StringBuilder and append to that within the loop, and in the end return builder.toString(). Or you follow the idea from Andy. There is no need to merge strings, you already know the target word. See below.
Then: List implies that adding/removing elements might be costly. So see if you can get rid of that part, and if it would be possible to use maps, sets instead.
Finally: the real point would be to look into your algorithm. I would try to work "backwards". Meaning: first identify those array elements that actually occur in your target word. You can ignore all others right from start.
Then: look at all array entries that **start*+ your search word. In your example you can notice that there are just two array elements that fit. And then work your way from there.
My first observation would be that you don't actually need to build anything: you know what string you are trying to construct (e.g. stackoverflow), so all you really need to keep track of is how much of that string you have matched so far. Call this m.
Next, having matched m characters, provided m < word.length(), you need to choose a next string from book which matches the portion of word from m to m + nextString.length().
You could do this by checking each string in turn:
if (word.matches(m, nextString, 0, nextString.length()) { ...}
But you can do better, by determining strings that can't match in advance: the next string you append will have the following properties:
word.charAt(m) == nextString.charAt(0) (the next characters match)
m + nextString.length() <= word.length() (adding the next string shouldn't make the constructed string longer than word)
So, you can cut down the potential words from book that you might check by constructing a map of letters to words that start with that (point 1); and if you store the words with the same starting letter in increasing length order, you can stop checking that letter as soon as the length gets too big (point 2).
You can construct a map once and reuse:
Map<Character, List<String>> prefixMap =
Arrays.asList(book).stream()
.collect(groupingBy(
s -> s.charAt(0),
collectingAndThen(
toList(),
ss -> {
ss.sort(comparingInt(String::length));
return ss;
})));
You can count the number of ways recursively, without constructing any additional objects (*):
int method(String word, String[] book) {
return method(word, 0, /* construct map as above */);
}
int method(String word, int m, Map<Character, List<String>> prefixMap) {
if (m == word.length()) {
return 1;
}
int result = 0;
for (String nextString : prefixMap.getOrDefault(word.charAt(m), emptyList())) {
if (m + nextString.length() > word.length()) {
break;
}
// Start at m+1, because you already know they match at m.
if (word.regionMatches(m + 1, nextString, 1, nextString.length()-1)) {
// This is a potential match!
// Make a recursive call.
result += method(word, m + nextString.length(), prefixMap);
}
}
return result;
}
(*) This may construct new instances of Character, because of the boxing of the word.charAt(m): cached instances are guaranteed to be used for chars in the range 0-127 only. There are ways to work around this, but they would only clutter the code.
I think you are already doing a pretty good job at optimizing your application. In addition to the answer by GhostCat here are a few suggestions of my own:
public static int optimal(String word, String[] book){
int count = 0;
List<List<String>> allCombinations = allSubstrings(word);
List<String> wordList = Arrays.asList(book);
for (int i = 0; i < allCombinations.size(); i++)
{
/*
* allCombinations.get(i).retainAll(wordList);
*
* There is no need to retrieve the list element
* twice, just set it in a local variable
*/
java.util.List<String> combination = allCombinations.get(i);
combination.retainAll(wordList);
/*
* Since we are only interested in the count here
* there is no need to remove and add list elements
*/
if (sumUp(combination, word))
{
/*allCombinations.remove(i);
allCombinations.add(i, empty);*/
count++;
}
/*else count++;*/
}
return count;
}
public static boolean sumUp (List<String> input, String expected) {
String x = "";
for (int i = 0; i < input.size(); i++) {
x = x + input.get(i);
}
// No need for if block here, just return comparison result
/*if (expected.equals(x)) return true;
return false;*/
return expected.equals(x);
}
And since you are interested in seeing the execution time of your method I would recommend implementing a benchmarking system of some sort. Here is a quick mock-up:
private static long benchmarkOptima(int cycles, String word, String[] book) {
long totalTime = 0;
for (int i = 0; i < cycles; i++)
{
long startTime = System.currentTimeMillis();
int a = optimal(word, book);
long executionTime = System.currentTimeMillis() - startTime;
totalTime += executionTime;
}
return totalTime / cycles;
}
public static void main(String[] args)
{
String word = "stackoverflow";
String[] book = new String[] {
"st", "ck", "CAG", "low", "TC",
"rf", "ove", "a", "sta"
};
int result = optimal(word, book);
final int cycles = 50;
long averageTime = benchmarkOptima(cycles, word, book);
System.out.println("Optimal result: " + result);
System.out.println("Average execution time - " + averageTime + " ms");
}
Output
2
Average execution time - 6 ms
Note: The implementation is getting stuck in the test case mentioned by #user1221, working on it.
What I could think of is a Trie based approach that is O(sum of length of words in dict) space. Time is not optimal.
Procedure:
Build a Trie of all the words in the dictionary. This is a pre-processing task that will take O(sum of lengths of all strings in dict).
We try finding the string that you want to make in the trie, with a twist. We start with searching a prefix of the string. If we get a prefix in the trie, we start the search from the top recursively and continue to look for more prefixes.
When we reach the end of out string i.e. stackoverflow, we check if we arrived at the end of any string, if yes, then we reached a valid combination of this string. we count this while going back up the recursion.
eg:
In the above case, we use the dict as {"st", "sta", "a", "ck"}
We construct our trie ($ is the sentinel char, i.e. a char which is not in the dict):
$___s___t.___a.
|___a.
|___c___k.
the . represents that a word in the dict ends at that position.
We try to find the no of constructions of stack.
We start searching stack in the trie.
depth=0
$___s(*)___t.___a.
|___a.
|___c___k.
We see that we are at the end of one word, we start a new search with the remaining string ack from the top.
depth=0
$___s___t(*).___a.
|___a.
|___c___k.
Again we are at the end of one word in the dict. We start a new search for ck.
depth=1
$___s___t.___a.
|___a(*).
|___c___k.
depth=2
$___s___t.___a.
|___a.
|___c(*)___k.
We reach the end of stack and end of a word in the dict, hence we have 1 valid representation of stack.
depth=2
$___s___t.___a.
|___a.
|___c___k(*).
We go back to the caller of depth=2
No next char is available, we return to the caller of depth=1.
depth=1
$___s___t.___a.
|___a(*, 1).
|___c___k.
depth=0
$___s___t(*, 1).___a.
|___a.
|___c___k.
We move to next char. We see that we reached the end of one word in the dict, we launch a new search for ck in the dict.
depth=0
$___s___t.___a(*, 1).
|___a.
|___c___k.
depth=1
$___s___t.___a.
|___a.
|___c(*)___k.
We reach the end of the stack and a work in the dict, so another valid representation. We go back to the caller of depth=1
depth=1
$___s___t.___a.
|___a.
|___c___k(*, 1).
There are no more chars to proceed, we return with the result 2.
depth=0
$___s___t.___a(*, 2).
|___a.
|___c___k.
Note: The implementation is in C++, shouldn't be too hard to convert to Java and this implementation assumes that all chars are lowercase, it's trivial to extend it to both cases.
Sample code (full version):
/**
Node *base: head of the trie
Node *h : current node in the trie
string s : string to search
int idx : the current position in the string
*/
int count(Node *base, Node *h, string s, int idx) {
// step 3: found a valid combination.
if (idx == s.size()) return h->end;
int res = 0;
// step 2: we recursively start a new search.
if (h->end) {
res += count(base, base, s, idx);
}
// move ahead in the trie.
if (h->next[s[idx] - 'a'] != NULL) {
res += count(base, h->next[s[idx] - 'a'], s, idx + 1);
}
return res;
}
def cancons(target,wordbank, memo={}):
if target in memo:
return memo[target]
if target =='':
return 1
total_count =0
for word in wordbank:
if target.startswith(word):
l= len(word)
number_of_way=cancons(target[l:],wordbank,memo)
total_count += number_of_way
memo[target]= total_count
return total_count
if __name__ == '__main__':
word = "stackoverflow";
String= ["st", "ck","CAG","low","TC","rf","ove","a","sta"]
b=cancons(word,String,memo={})
print(b)
For instance, take the following list of Strings, disregarding the inverted commas:
"Hello"
"Hello!"
"I'm saying Hello!"
"I haven't said hello yet, but I will."
Now let's say I'd like to perform a certain operation on the characters of each word β for instance, say I'd like to reverse the characters, but keep the positions of the punctuation. So the result would be:
"olleH"
"olleH!"
"m'I gniyas olleH!"
"I tneva'h dias olleh tey, tub I lliw."
Ideally I'd like my code to be independent of the operation performed on the string (another example would be a random shuffling of letters), and independent of all punctuationβso hyphens, apostrophes, commas, full stops, en/em dashes, etc. all remain in their original positions after the operation is performed. This probably requires some form of regular expressions.
For this, I was thinking that I should save the indices and characters of all punctuation in a given word, perform the operation, and then re-insert all punctuation at their correct positions. However, I can't think of a way to do this, or a class to use.
I have a first attempt, but this unfortunately does not work with punctuation, which is the key:
jshell> String str = "I haven't said hello yet, but I will."
str ==> "I haven't said hello yet, but I will."
jshell> Arrays.stream(str.split("\\s+")).map(x -> (new StringBuilder(x)).reverse().toString()).reduce((x, y) -> x + " " + y).get()
$2 ==> "I t'nevah dias olleh ,tey tub I .lliw"
Has anyone got an idea how I might fix this? Thanks very much. There's no need for full working codeβmaybe just a signpost to an appropriate class I could use to perform this operation.
No need to use regex for this, and you certainly shouldn't use split("\\s+"), since you'd lose consecutive spaces, and the type of whitespace characters, i.e. the spaces of the result could be incorrect.
You also shouldn't use charAt() or anything like it, since that would not support letters from the Unicode Supplemental Planes, i.e. Unicode characters that are stored in Java strings as surrogate pairs.
Basic logic:
Locate start of word, i.e. start of string or first character following whitespace.
Locate end of word, i.e. last character preceding whitespace or end of string.
Iterating from beginning and end in parallel:
Skip characters that are not letters.
Swap the letters.
As Java code, with full Unicode support:
public static String reverseLettersOfWords(String input) {
int[] codePoints = input.codePoints().toArray();
for (int i = 0, start = 0; i <= codePoints.length; i++) {
if (i == codePoints.length || Character.isWhitespace(codePoints[i])) {
for (int end = i - 1; ; start++, end--) {
while (start < end && ! Character.isLetter(codePoints[start]))
start++;
while (start < end && ! Character.isLetter(codePoints[end]))
end--;
if (start >= end)
break;
int tmp = codePoints[start];
codePoints[start] = codePoints[end];
codePoints[end] = tmp;
}
start = i + 1;
}
}
return new String(codePoints, 0, codePoints.length);
}
Test
System.out.println(reverseLettersOfWords("Hello"));
System.out.println(reverseLettersOfWords("Hello!"));
System.out.println(reverseLettersOfWords("I'm saying Hello!"));
System.out.println(reverseLettersOfWords("I haven't said hello yet, but I will."));
System.out.println(reverseLettersOfWords("Works with surrogate pairs: πππ+π "));
Output
olleH
olleH!
m'I gniyas olleH!
I tneva'h dias olleh tey, tub I lliw.
skroW htiw etagorrus sriap: πππ+π
Note that the special letters at the end are the first 4 shown here in column "Script (or Calligraphy)", "Bold", e.g. the π is Unicode Character 'MATHEMATICAL BOLD SCRIPT CAPITAL A' (U+1D4D0), which in Java is two characters "\uD835\uDCD0".
UPDATE
The above implementation is optimized for reversing the letters of the word. To apply an arbitrary operation to mangle the letters of the word, use the following implementation:
public static String mangleLettersOfWords(String input) {
int[] codePoints = input.codePoints().toArray();
for (int i = 0, start = 0; i <= codePoints.length; i++) {
if (i == codePoints.length || Character.isWhitespace(codePoints[i])) {
int wordCodePointLen = 0;
for (int j = start; j < i; j++)
if (Character.isLetter(codePoints[j]))
wordCodePointLen++;
if (wordCodePointLen != 0) {
int[] wordCodePoints = new int[wordCodePointLen];
for (int j = start, k = 0; j < i; j++)
if (Character.isLetter(codePoints[j]))
wordCodePoints[k++] = codePoints[j];
int[] mangledCodePoints = mangleWord(wordCodePoints.clone());
if (mangledCodePoints.length != wordCodePointLen)
throw new IllegalStateException("Mangled word is wrong length: '" + new String(wordCodePoints, 0, wordCodePoints.length) + "' (" + wordCodePointLen + " code points)" +
" vs mangled '" + new String(mangledCodePoints, 0, mangledCodePoints.length) + "' (" + mangledCodePoints.length + " code points)");
for (int j = start, k = 0; j < i; j++)
if (Character.isLetter(codePoints[j]))
codePoints[j] = mangledCodePoints[k++];
}
start = i + 1;
}
}
return new String(codePoints, 0, codePoints.length);
}
private static int[] mangleWord(int[] codePoints) {
return mangleWord(new String(codePoints, 0, codePoints.length)).codePoints().toArray();
}
private static CharSequence mangleWord(String word) {
return new StringBuilder(word).reverse();
}
You can of course replace the hardcoded call to the either mangleWord method with a call to a passed-in Function<int[], int[]> or Function<String, ? extends CharSequence> parameter, if needed.
The result with that implementation of the mangleWord method(s) is the same as the original implementation, but you can now easily implement a different mangling algorithm.
E.g. to randomize the letters, simply shuffle the codePoints array:
private static int[] mangleWord(int[] codePoints) {
Random rnd = new Random();
for (int i = codePoints.length - 1; i > 0; i--) {
int j = rnd.nextInt(i + 1);
int tmp = codePoints[j];
codePoints[j] = codePoints[i];
codePoints[i] = tmp;
}
return codePoints;
}
Sample Output
Hlelo
Hlleo!
m'I nsayig oHlel!
I athen'v siad eohll yte, btu I illw.
srWok twih rueoatrsg rpasi: πππ+π
I suspect there's a more efficient solution but here's a naive one:
Split sentence into words on spaces (note - if you have multiple spaces my implementation will have problems)
Strip punctuation
Reverse each word
Go through each letter, and insert character from reversed word AND insert punctuation from original word if necessary
public class Reverser {
public String reverseSentence(String sentence) {
String[] words = sentence.split(" ");
return Arrays.stream(words).map(this::reverseWord).collect(Collectors.joining(" "));
}
private String reverseWord(String word) {
String noPunctuation = word.replaceAll("\\W", "");
String reversed = new StringBuilder(noPunctuation).reverse().toString();
StringBuilder result = new StringBuilder();
for (int i = 0; i < word.length(); ++i) {
char ch = word.charAt(i);
if (!Character.isAlphabetic(ch) && !Character.isDigit(ch)) {
result.append(ch);
}
if (i < reversed.length()) {
result.append(reversed.charAt(i));
}
}
return result.toString();
}
}
Trying to search for patterns of letters in a file, the pattern is entered by a user and comes out as a String, so far I've got it to find the first letter by unsure how to make it test to see if the next letter also matches the pattern.
This is the loop I currently have. any help would be appreciated
public void exactSearch(){
if (pattern==null){UI.println("No pattern");return;}
UI.println("===================\nExact searching for "+patternString);
int j = 0 ;
for(int i=0; i<data.size(); i++){
if(patternString.charAt(i) == data.get(i) )
j++;
UI.println( "found at " + j) ;
}
}
You need to iterate over the first string until you find the first character of the other string. From there, you can create an inner loop and iterate on both simultaneously, like you did.
Hint: be sure to look watch for boundaries as the strings might not be of the same size.
You can try this :-
String a1 = "foo-bar-baz-bar-";
String pattern = "bar";
int foundIndex = 0;
while(foundIndex != -1) {
foundIndex = a1.indexOf(pattern,foundIndex);
if(foundIndex != -1)
{
System.out.println(foundIndex);
foundIndex += 1;
}
}
indexOf - first parameter is the pattern string,
second parameter is starting index from where we have to search.
If pattern is found, it will return the starting index from where the pattern matched.
If pattern is not found, indexOf will return -1.
String data = "foo-bar-baz-bar-";
String pattern = "bar";
int foundIndex = data.indexOf(pattern);
while (foundIndex > -1) {
System.out.println("Match found at: " + foundIndex);
foundIndex = data.indexOf(pattern, foundIndex + pattern.length());
}
Based on your request, you can use this algorithm to search for your positions:
1) We check if we reach at the end of the string, to avoid the invalidIndex error, we verify if the remaining substring's size is smaller than the pattern's length.
2) We calculate the substring at each iteration and we verify the string with the pattern.
List<Integer> positionList = new LinkedList<>();
String inputString = "AAACABCCCABC";
String pattern = "ABC";
for (int i = 0 ; i < inputString.length(); i++) {
if (inputString.length() - i < pattern.length()){
break;
}
String currentSubString = inputString.substring(i, i + pattern.length());
if (currentSubString.equals(pattern)){
positionList.add(i);
}
}
for (Integer pos : positionList) {
System.out.println(pos); // Positions : 4 and 9
}
EDIT :
Maybe it can be optimized, not to use a Collection for this simple task, but I used a LinkedList to write a quicker approach.
I facing some issue with write logic for below problem.
I have two ArrayLists of strings:
List1: contains 5 million strings
List2: will create on users input and contains some strings/characters(Ex. a,b,c,g,l,pd,sp,mta)
Now I have to split list1 into multiple Lists according to startsWith strings in list2 like in above case. I need to create 8 lists as starts with 'a', 'b','c', 'g', 'l','pd', 'sp' and 'mta'
But the condition for above is I have to iterate List1 or List2 only once. i.e. worst complexity for algorithm should be size of List1 (5 million).
It is allowed to use collections.sort() method
Code I have tried
// Create List for search strings.
List<String> CharList = new ArrayList<String>();
CharList.add("a");
CharList.add("b");
CharList.add("e");
CharList.add("z");
CharList.add("4");
CharList.add("1");
CharList.add("zi");
List<String> recordList = new ArrayList<String>();
// Creating dummy data with 100 character in live environment it can be
// around 50 lakhs strings
for (int i = 0; i < 100; i++) {
char[] chars = "abcdefghijklmnopqrstuvwxyzABCGDKL0123456789".toCharArray();
StringBuilder sb = new StringBuilder();
Random random = new Random();
for (int i1 = 0; i1 < 6; i1++) {
char c = chars[random.nextInt(chars.length)];
sb.append(c);
}
String output = sb.toString();
recordList.add(output);
}
// Adding some data mannually
recordList.add("zink");
recordList.add("zebra");
recordList.add("zzzzzz");
Collections.sort(CharList, String.CASE_INSENSITIVE_ORDER);
Collections.sort(recordList, String.CASE_INSENSITIVE_ORDER);
System.out.println("RECORDLIST ===>" + recordList);
System.out.println("***************************************************");
System.out.println("Charlist ===>" + CharList);
System.out.println("***************************************************");
List<List> lists = new ArrayList<List>();
int startIndex = 0, charPointer = 0;
while (startIndex < recordList.size() && charPointer < CharList.size()) {
List<String> temp = new ArrayList<String>();
boolean isHit = false;
String currentRecord = recordList.get(startIndex);
String partitionSattement = CharList.get(charPointer);
while (currentRecord.startsWith(partitionSattement.toUpperCase())
|| currentRecord.startsWith(partitionSattement.toLowerCase())) {
temp.add(recordList.get(startIndex));
isHit = true;
startIndex++;
}
if (!isHit) {
startIndex++;
}
if (!temp.isEmpty()) {
lists.add(temp);
System.out.println(CharList.get(charPointer) + "====>" + temp);
}
charPointer++;
}
Just using the String startsWith method won't work in this case. Consider what happens if the first pattern does not match any input - you'll loop through all strings in the input list without finding a match, even though subsequent pattern matches do exist.
What we need to do instead is compare each pattern against the initial characters of each input string and process accordingly. Let's say we have an input string str and a pattern pat. Let subStr be the first pat.length() characters of str. Now we can compare subStr and pat using the String compareToIgnoreCase method. There are three cases to consider:
subStr < pat Move to the next input string.
subStr == pat Add str to output for pat and move to the next input string.
subStr > pat Move to the next pattern.
Here's some code to illustrate (I've kept your variable names where possible).
List<List<String>> output = new ArrayList<>();
for(int i=0; i<CharList.size(); i++) output.add(new ArrayList<String>());
int startIndex=0;
int charPointer=0;
while(startIndex < recordList.size() && charPointer < CharList.size())
{
String charStr = CharList.get(charPointer);
String recStr = recordList.get(startIndex);
int cmp;
if(recStr.length() < charStr.length())
{
cmp = -1;
}
else
{
String recSubStr = recStr.substring(0, charStr.length());
cmp = recSubStr.compareToIgnoreCase(charStr);
}
if(cmp <= 0)
{
if(cmp == 0) output.get(charPointer).add(recStr);
startIndex++;
}
else
{
charPointer++;
}
}
for(int i=0; i<CharList.size(); i++)
{
System.out.println(CharList.get(i) + " : " + output.get(i));
}
Also, you should note that when you include a pattern that itself starts with another pattern (e.g. "zi" and "z") the longer pattern will never be matched, since the shorter one will capture all inputs.
I can see two problems in your code:
You should remove the following segment:
if (!isHit) {
startIndex++;
}
Actually you don't need that isHit variable at all. If a string doesn't match with a pattern then you still have to compare it with the next pattern.
You should sort the arrays in descending order. As SirRaffleBuffle noted in the other answer you should compare the strings with the longer pattern first. Sorting the strings and patterns in descending order will automatically solve this problem.
Having problems trying to show every combination of a character of array without repeating letters.
public static String[] getAllLists(String[] elements, int lengthOfList)
{
//initialize our returned list with the number of elements calculated above
String[] allLists = new String[(int)Math.pow(elements.length, lengthOfList)];
//lists of length 1 are just the original elements
if(lengthOfList == 1) return elements;
else
{
//the recursion--get all lists of length 3, length 2, all the way up to 1
String[] allSublists = getAllLists(elements, lengthOfList - 1);
//append the sublists to each element
int arrayIndex = 0;
for(int i = 0; i < elements.length; i++)
{
for(int j = 0; j < allSublists.length; j++)
{
//add the newly appended combination to the list
allLists[arrayIndex] = elements[i] + allSublists[j];
arrayIndex++;
}
}
return allLists;
}
}
The above code works perfect but use's each letter more than once which cant be done in this case.
And i am stuck how to do this now.
Here's an example implementation. Essentially it takes a String and iterates over every character, putting that character at the front. It then recurses on the remaining characters. That structure removes your issue of repeated letters, because the input to the recursion has removed the character you've already used.
I also stored results in a set to remove semantic equivalences. The input 'aab' can switch char 0 and char 1 but still be 'aab.' I used a TreeSet to preserve ordering for easier verification of the output, but HashSet would be faster.
public static Set<String> permute(String chars)
{
// Use sets to eliminate semantic duplicates (aab is still aab even if you switch the two 'a's)
// Switch to HashSet for better performance
Set<String> set = new TreeSet<String>();
// Termination condition: only 1 permutation for a string of length 1
if (chars.length() == 1)
{
set.add(chars);
}
else
{
// Give each character a chance to be the first in the permuted string
for (int i=0; i<chars.length(); i++)
{
// Remove the character at index i from the string
String pre = chars.substring(0, i);
String post = chars.substring(i+1);
String remaining = pre+post;
// Recurse to find all the permutations of the remaining chars
for (String permutation : permute(remaining))
{
// Concatenate the first character with the permutations of the remaining chars
set.add(chars.charAt(i) + permutation);
}
}
}
return set;
}
Example run:
public static void main(String[] args)
{
for (String s : CharPermuter.permute("abca"))
{
System.out.println(s);
}
}
Generates:
aabc
aacb
abac
abca
acab
acba
baac
baca
bcaa
caab
caba
cbaa