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is there any other way to optimize this code. Anyone can come up with better way because this is taking lot of time in main code. Thanks alot;)
HashMap<String, Integer> hmap = new HashMap<String, Integer>();
List<String> dup = new ArrayList<String>();
List<String> nondup = new ArrayList<String>();
for (String num : nums) {
String x= num;
String result = x.toLowerCase();
if (hmap.containsKey(result)) {
hmap.put(result, hmap.get(result) + 1);
}
else {
hmap.put(result,1);
}
}
for(String num:nums){
int count= hmap.get(num.toLowerCase());
if (count == 1){
nondup.add(num);
}
else{
dup.add(num);
}
}
output:
[A/tea, C/SEA.java, C/clock, aep, aeP, C/SEA.java]
Dups: [C/SEA.java, aep, aeP, C/SEA.java]
nondups: [A/tea, C/clock]
How much time is "a lot of time"? Is your input bigger than what you've actually shown us?
You could parallelize this with something like Arrays.parallelStream(nums).collect(Collectors.groupingByConcurrent(k -> k, Collectors.counting()), which would get you a Map<String, Long>, but that would only speed up your code if you have a lot of input, which it doesn't look like you have right now.
You could parallelize the next step, if you liked, like so:
Map<String, Long> counts = Arrays.parallelStream(nums)
.collect(Collectors.groupingByConcurrent(k -> k, Collectors.counting());
Map<Boolean, List<String>> hasDup =
counts.entrySet().parallelStream()
.collect(Collectors.partitioningBy(
entry -> entry.getValue() > 1,
Collectors.mapping(Entry::getKey, Collectors.toList())));
List<String> dup = hasDup.get(true);
List<String> nodup = hasDup.get(false);
The algorithms in the other answers can speed up execution using multiple threads.
This can theoretically reduce the processing time with a factor of M, where M is the maximum number of threads that your system can run concurrently. However, as M is a constant number, this does not change the order of complexity, which therefore remains O(N).
At a glance, I do not see a way to solve your problem in less than O(N), I am afraid.
I would like to convert the string containing abc to a list of characters and a hashset of characters. How can I do that in Java ?
List<Character> charList = new ArrayList<Character>("abc".toCharArray());
In Java8 you can use streams I suppose.
List of Character objects:
List<Character> chars = str.chars()
.mapToObj(e->(char)e).collect(Collectors.toList());
And set could be obtained in a similar way:
Set<Character> charsSet = str.chars()
.mapToObj(e->(char)e).collect(Collectors.toSet());
You will have to either use a loop, or create a collection wrapper like Arrays.asList which works on primitive char arrays (or directly on strings).
List<Character> list = new ArrayList<Character>();
Set<Character> unique = new HashSet<Character>();
for(char c : "abc".toCharArray()) {
list.add(c);
unique.add(c);
}
Here is an Arrays.asList like wrapper for strings:
public List<Character> asList(final String string) {
return new AbstractList<Character>() {
public int size() { return string.length(); }
public Character get(int index) { return string.charAt(index); }
};
}
This one is an immutable list, though. If you want a mutable list, use this with a char[]:
public List<Character> asList(final char[] string) {
return new AbstractList<Character>() {
public int size() { return string.length; }
public Character get(int index) { return string[index]; }
public Character set(int index, Character newVal) {
char old = string[index];
string[index] = newVal;
return old;
}
};
}
Analogous to this you can implement this for the other primitive types.
Note that using this normally is not recommended, since for every access you
would do a boxing and unboxing operation.
The Guava library contains similar List wrapper methods for several primitive array classes, like Chars.asList, and a wrapper for String in Lists.charactersOf(String).
The lack of a good way to convert between a primitive array and a collection of its corresponding wrapper type is solved by some third party libraries. Guava, a very common one, has a convenience method to do the conversion:
List<Character> characterList = Chars.asList("abc".toCharArray());
Set<Character> characterSet = new HashSet<Character>(characterList);
Use a Java 8 Stream.
myString.chars().mapToObj(i -> (char) i).collect(Collectors.toList());
Breakdown:
myString
.chars() // Convert to an IntStream
.mapToObj(i -> (char) i) // Convert int to char, which gets boxed to Character
.collect(Collectors.toList()); // Collect in a List<Character>
(I have absolutely no idea why String#chars() returns an IntStream.)
The most straightforward way is to use a for loop to add elements to a new List:
String abc = "abc";
List<Character> charList = new ArrayList<Character>();
for (char c : abc.toCharArray()) {
charList.add(c);
}
Similarly, for a Set:
String abc = "abc";
Set<Character> charSet = new HashSet<Character>();
for (char c : abc.toCharArray()) {
charSet.add(c);
}
List<String> result = Arrays.asList("abc".split(""));
Create an empty list of Character and then make a loop to get every character from the array and put them in the list one by one.
List<Character> characterList = new ArrayList<Character>();
char arrayChar[] = abc.toCharArray();
for (char aChar : arrayChar)
{
characterList.add(aChar); // autoboxing
}
You can do this without boxing if you use Eclipse Collections:
CharAdapter abc = Strings.asChars("abc");
CharList list = abc.toList();
CharSet set = abc.toSet();
CharBag bag = abc.toBag();
Because CharAdapter is an ImmutableCharList, calling collect on it will return an ImmutableList.
ImmutableList<Character> immutableList = abc.collect(Character::valueOf);
If you want to return a boxed List, Set or Bag of Character, the following will work:
LazyIterable<Character> lazyIterable = abc.asLazy().collect(Character::valueOf);
List<Character> list = lazyIterable.toList();
Set<Character> set = lazyIterable.toSet();
Bag<Character> set = lazyIterable.toBag();
Note: I am a committer for Eclipse Collections.
IntStream can be used to access each character and add them to the list.
String str = "abc";
List<Character> charList = new ArrayList<>();
IntStream.range(0,str.length()).forEach(i -> charList.add(str.charAt(i)));
Using Java 8 - Stream Funtion:
Converting A String into Character List:
ArrayList<Character> characterList = givenStringVariable
.chars()
.mapToObj(c-> (char)c)
.collect(collectors.toList());
Converting A Character List into String:
String givenStringVariable = characterList
.stream()
.map(String::valueOf)
.collect(Collectors.joining())
To get a list of Characters / Strings -
List<String> stringsOfCharacters = string.chars().
mapToObj(i -> (char)i).
map(c -> c.toString()).
collect(Collectors.toList());
I have a method as below
private Map<String,List<String>> createTableColumnListMap(List<Map<String,String>> nqColumnMapList){
Map<String, List<String>> targetTableColumnListMap =
nqColumnMapList.stream()
.flatMap(m -> m.entrySet().stream())
.collect(groupingBy(Map.Entry::getKey, mapping(Map.Entry::getValue, toList())));
return targetTableColumnListMap;
}
I want to uppercase the map keys but couldn't find a way to do it. is there a java 8 way to achieve this?
This doesn't require any fancy manipulation of Collectors. Lets say you have this map
Map<String, Integer> imap = new HashMap<>();
imap.put("One", 1);
imap.put("Two", 2);
Just get a stream for the keySet() and collect into a new map where the keys you insert are uppercased:
Map<String, Integer> newMap = imap.keySet().stream()
.collect(Collectors.toMap(key -> key.toUpperCase(), key -> imap.get(key)));
// ONE - 1
// TWO - 2
Edit:
#Holger's comment is correct, it would be better (and cleaner) to just use an entry set, so here is the updated solution
Map<String, Integer> newMap = imap.entrySet().stream()
.collect(Collectors.toMap(entry -> entry.getKey().toUpperCase(), entry -> entry.getValue()));
Answer for your question [which you can copy and paste] :
Map<String, List<String>> targetTableColumnListMap = nqColumnMapList.stream().flatMap(m -> m.entrySet().stream())
.collect(Collectors.groupingBy(e -> e.getKey().toUpperCase(), Collectors.mapping(Map.Entry::getValue, Collectors.toList())));
I'm trying to find a suitable DP algorithm for simplifying a string. For example I have a string a b a b and a list of rules
a b -> b
a b -> c
b a -> a
c c -> b
The purpose is to get all single chars that can be received from the given string using these rules. For this example it will be b, c. The length of the given string can be up to 200 symbols. Could you please prompt an effective algorithm?
Rules always are 2 -> 1. I've got an idea of creating a tree, root is given string and each child is a string after one transform, but I'm not sure if it's the best way.
If you read those rules from right to left, they look exactly like the rules of a context free grammar, and have basically the same meaning. You could apply a bottom-up parsing algorithm like the Earley algorithm to your data, along with a suitable starting rule; something like
start <- start a
| start b
| start c
and then just examine the parse forest for the shortest chain of starts. The worst case remains O(n^3) of course, but Earley is fairly effective, these days.
You can also produce parse forests when parsing with derivatives. You might be able to efficiently check them for short chains of starts.
For a DP problem, you always need to understand how you can construct the answer for a big problem in terms of smaller sub-problems. Assume you have your function simplify which is called with an input of length n. There are n-1 ways to split the input in a first and a last part. For each of these splits, you should recursively call your simplify function on both the first part and the last part. The final answer for the input of length n is the set of all possible combinations of answers for the first and for the last part, which are allowed by the rules.
In Python, this can be implemented like so:
rules = {'ab': set('bc'), 'ba': set('a'), 'cc': set('b')}
all_chars = set(c for cc in rules.values() for c in cc)
# memoize
def simplify(s):
if len(s) == 1: # base case to end recursion
return set(s)
possible_chars = set()
# iterate over all the possible splits of s
for i in range(1, len(s)):
head = s[:i]
tail = s[i:]
# check all possible combinations of answers of sub-problems
for c1 in simplify(head):
for c2 in simplify(tail):
possible_chars.update(rules.get(c1+c2, set()))
# speed hack
if possible_chars == all_chars: # won't get any bigger
return all_chars
return possible_chars
Quick check:
In [53]: simplify('abab')
Out[53]: {'b', 'c'}
To make this fast enough for large strings (to avoiding exponential behavior), you should use a memoize decorator. This is a critical step in solving DP problems, otherwise you are just doing a brute-force calculation. A further tiny speedup can be obtained by returning from the function as soon as possible_chars == set('abc'), since at that point, you are already sure that you can generate all possible outcomes.
Analysis of running time: for an input of length n, there are 2 substrings of length n-1, 3 substrings of length n-2, ... n substrings of length 1, for a total of O(n^2) subproblems. Due to the memoization, the function is called at most once for every subproblem. Maximum running time for a single sub-problem is O(n) due to the for i in range(len(s)), so the overall running time is at most O(n^3).
Let N - length of given string and R - number of rules.
Expanding a tree in a top down manner yields computational complexity O(NR^N) in the worst case (input string of type aaa... and rules aa -> a).
Proof:
Root of the tree has (N-1)R children, which have (N-1)R^2 children, ..., which have (N-1)R^N children (leafs). So, the total complexity is O((N-1)R + (N-1)R^2 + ... (N-1)R^N) = O(N(1 + R^2 + ... + R^N)) = (using binomial theorem) = O(N(R+1)^N) = O(NR^N).
Recursive Java implementation of this naive approach:
public static void main(String[] args) {
Map<String, Character[]> rules = new HashMap<String, Character[]>() {{
put("ab", new Character[]{'b', 'c'});
put("ba", new Character[]{'a'});
put("cc", new Character[]{'b'});
}};
System.out.println(simplify("abab", rules));
}
public static Set<String> simplify(String in, Map<String, Character[]> rules) {
Set<String> result = new HashSet<String>();
simplify(in, rules, result);
return result;
}
private static void simplify(String in, Map<String, Character[]> rules, Set<String> result) {
if (in.length() == 1) {
result.add(in);
}
for (int i = 0; i < in.length() - 1; i++) {
String two = in.substring(i, i + 2);
Character[] rep = rules.get(two);
if (rep != null) {
for (Character c : rep) {
simplify(in.substring(0, i) + c + in.substring(i + 2, in.length()), rules, result);
}
}
}
}
Bas Swinckels's O(RN^3) Java implementation (with HashMap as a memoization cache):
public static Set<String> simplify2(final String in, Map<String, Character[]> rules) {
Map<String, Set<String>> cache = new HashMap<String, Set<String>>();
return simplify2(in, rules, cache);
}
private static Set<String> simplify2(final String in, Map<String, Character[]> rules, Map<String, Set<String>> cache) {
final Set<String> cached = cache.get(in);
if (cached != null) {
return cached;
}
Set<String> ret = new HashSet<String>();
if (in.length() == 1) {
ret.add(in);
return ret;
}
for (int i = 1; i < in.length(); i++) {
String head = in.substring(0, i);
String tail = in.substring(i, in.length());
for (String c1 : simplify2(head, rules)) {
for (String c2 : simplify2(tail, rules, cache)) {
Character[] rep = rules.get(c1 + c2);
if (rep != null) {
for (Character c : rep) {
ret.add(c.toString());
}
}
}
}
}
cache.put(in, ret);
return ret;
}
Output in both approaches:
[b, c]
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I saw this in an interview question ,
Given a sorting order string, you are asked to sort the input string based on the given sorting order string.
for example if the sorting order string is dfbcae
and the Input string is abcdeeabc
the output should be dbbccaaee.
any ideas on how to do this , in an efficient way ?
The Counting Sort option is pretty cool, and fast when the string to be sorted is long compared to the sort order string.
create an array where each index corresponds to a letter in the alphabet, this is the count array
for each letter in the sort target, increment the index in the count array which corresponds to that letter
for each letter in the sort order string
add that letter to the end of the output string a number of times equal to it's count in the count array
Algorithmic complexity is O(n) where n is the length of the string to be sorted. As the Wikipedia article explains we're able to beat the lower bound on standard comparison based sorting because this isn't a comparison based sort.
Here's some pseudocode.
char[26] countArray;
foreach(char c in sortTarget)
{
countArray[c - 'a']++;
}
int head = 0;
foreach(char c in sortOrder)
{
while(countArray[c - 'a'] > 0)
{
sortTarget[head] = c;
head++;
countArray[c - 'a']--;
}
}
Note: this implementation requires that both strings contain only lowercase characters.
Here's a nice easy to understand algorithm that has decent algorithmic complexity.
For each character in the sort order string
scan string to be sorted, starting at first non-ordered character (you can keep track of this character with an index or pointer)
when you find an occurrence of the specified character, swap it with the first non-ordered character
increment the index for the first non-ordered character
This is O(n*m), where n is the length of the string to be sorted and m is the length of the sort order string. We're able to beat the lower bound on comparison based sorting because this algorithm doesn't really use comparisons. Like Counting Sort it relies on the fact that you have a predefined finite external ordering set.
Here's some psuedocode:
int head = 0;
foreach(char c in sortOrder)
{
for(int i = head; i < sortTarget.length; i++)
{
if(sortTarget[i] == c)
{
// swap i with head
char temp = sortTarget[head];
sortTarget[head] = sortTarget[i];
sortTarget[i] = temp;
head++;
}
}
}
In Python, you can just create an index and use that in a comparison expression:
order = 'dfbcae'
input = 'abcdeeabc'
index = dict([ (y,x) for (x,y) in enumerate(order) ])
output = sorted(input, cmp=lambda x,y: index[x] - index[y])
print 'input=',''.join(input)
print 'output=',''.join(output)
gives this output:
input= abcdeeabc
output= dbbccaaee
Use binary search to find all the "split points" between different letters, then use the length of each segment directly. This will be asymptotically faster then naive counting sort, but will be harder to implement:
Use an array of size 26*2 to store the begin and end of each letter;
Inspect the middle element, see if it is different from the element left to it. If so, then this is the begin for the middle element and end for the element before it;
Throw away the segment with identical begin and end (if there are any), recursively apply this algorithm.
Since there are at most 25 "split"s, you won't have to do the search for more than 25 segemnts, and for each segment it is O(logn). Since this is constant * O(logn), the algorithm is O(nlogn).
And of course, just use counting sort will be easier to implement:
Use an array of size 26 to record the number of different letters;
Scan the input string;
Output the string in the given sorting order.
This is O(n), n being the length of the string.
Interview questions are generally about thought process and don't usually care too much about language features, but I couldn't resist posting a VB.Net 4.0 version anyway.
"Efficient" can mean two different things. The first is "what's the fastest way to make a computer execute a task" and the second is "what's the fastest that we can get a task done". They might sound the same but the first can mean micro-optimizations like int vs short, running timers to compare execution times and spending a week tweaking every millisecond out of an algorithm. The second definition is about how much human time would it take to create the code that does the task (hopefully in a reasonable amount of time). If code A runs 20 times faster than code B but code B took 1/20th of the time to write, depending on the granularity of the timer (1ms vs 20ms, 1 week vs 20 weeks), each version could be considered "efficient".
Dim input = "abcdeeabc"
Dim sort = "dfbcae"
Dim SortChars = sort.ToList()
Dim output = New String((From c In input.ToList() Select c Order By SortChars.IndexOf(c)).ToArray())
Trace.WriteLine(output)
Here is my solution to the question
import java.util.*;
import java.io.*;
class SortString
{
public static void main(String arg[])throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// System.out.println("Enter 1st String :");
// System.out.println("Enter 1st String :");
// String s1=br.readLine();
// System.out.println("Enter 2nd String :");
// String s2=br.readLine();
String s1="tracctor";
String s2="car";
String com="";
String uncom="";
for(int i=0;i<s2.length();i++)
{
if(s1.contains(""+s2.charAt(i)))
{
com=com+s2.charAt(i);
}
}
System.out.println("Com :"+com);
for(int i=0;i<s1.length();i++)
if(!com.contains(""+s1.charAt(i)))
uncom=uncom+s1.charAt(i);
System.out.println("Uncom "+uncom);
System.out.println("Combined "+(com+uncom));
HashMap<String,Integer> h1=new HashMap<String,Integer>();
for(int i=0;i<s1.length();i++)
{
String m=""+s1.charAt(i);
if(h1.containsKey(m))
{
int val=(int)h1.get(m);
val=val+1;
h1.put(m,val);
}
else
{
h1.put(m,new Integer(1));
}
}
StringBuilder x=new StringBuilder();
for(int i=0;i<com.length();i++)
{
if(h1.containsKey(""+com.charAt(i)))
{
int count=(int)h1.get(""+com.charAt(i));
while(count!=0)
{x.append(""+com.charAt(i));count--;}
}
}
x.append(uncom);
System.out.println("Sort "+x);
}
}
Here is my version which is O(n) in time. Instead of unordered_map, I could have just used a char array of constant size. i.,e. char char_count[256] (and done ++char_count[ch - 'a'] ) assuming the input strings has all ASCII small characters.
string SortOrder(const string& input, const string& sort_order) {
unordered_map<char, int> char_count;
for (auto ch : input) {
++char_count[ch];
}
string res = "";
for (auto ch : sort_order) {
unordered_map<char, int>::iterator it = char_count.find(ch);
if (it != char_count.end()) {
string s(it->second, it->first);
res += s;
}
}
return res;
}
private static String sort(String target, String reference) {
final Map<Character, Integer> referencesMap = new HashMap<Character, Integer>();
for (int i = 0; i < reference.length(); i++) {
char key = reference.charAt(i);
if (!referencesMap.containsKey(key)) {
referencesMap.put(key, i);
}
}
List<Character> chars = new ArrayList<Character>(target.length());
for (int i = 0; i < target.length(); i++) {
chars.add(target.charAt(i));
}
Collections.sort(chars, new Comparator<Character>() {
#Override
public int compare(Character o1, Character o2) {
return referencesMap.get(o1).compareTo(referencesMap.get(o2));
}
});
StringBuilder sb = new StringBuilder();
for (Character c : chars) {
sb.append(c);
}
return sb.toString();
}
In C# I would just use the IComparer Interface and leave it to Array.Sort
void Main()
{
// we defin the IComparer class to define Sort Order
var sortOrder = new SortOrder("dfbcae");
var testOrder = "abcdeeabc".ToCharArray();
// sort the array using Array.Sort
Array.Sort(testOrder, sortOrder);
Console.WriteLine(testOrder.ToString());
}
public class SortOrder : IComparer
{
string sortOrder;
public SortOrder(string sortOrder)
{
this.sortOrder = sortOrder;
}
public int Compare(object obj1, object obj2)
{
var obj1Index = sortOrder.IndexOf((char)obj1);
var obj2Index = sortOrder.IndexOf((char)obj2);
if(obj1Index == -1 || obj2Index == -1)
{
throw new Exception("character not found");
}
if(obj1Index > obj2Index)
{
return 1;
}
else if (obj1Index == obj2Index)
{
return 0;
}
else
{
return -1;
}
}
}