Related
So, I am trying to generate a word and then check if it is a real word using word index. Can someone help me figure out how to solve this, it is constantly giving me a never-ending loop. I just simply wanted to create random characters, and after each character see if it exists in the hashmap. First by checking to find the key which is the first letter of the word, and then by checking the substring of each word in that key selection.
public ArrayList<String> randomSize(int length) {
ArrayList<String> randomWords = new ArrayList<>(length);
String letters = "abcdefghijklmnopqrstuvwxyz";
// for loop add a new slot into randomWords
for (int eachSlot = 0; eachSlot < length; eachSlot++) {
// for each slot generate a random length for the word
int randWordLength = (int) (Math.random() * (10 - 0) + 0);
// every slot generate a random firstLetter
int slotFound = 0;
String firstConfirmedLetter = "";
// while the first letter is not found in WordIndex
while (slotFound == 0) {
int randNumer = (int) (Math.random() * (24 - 0) + 0);
String firstLetter = letters.substring(randNumer, randNumer + 1);
if (wordIndex.containsKey(firstLetter) == true)
{
firstConfirmedLetter = firstLetter;
randomWords.add(firstConfirmedLetter);
System.out.println(firstLetter);// working
randWordLength--;
slotFound = 1;
// if it is found end the while loop
}
}
// we found the first letter, now we need to find the rest of the letters
for (int eachLetter = 0; eachLetter < randWordLength; eachLetter++){
int isFound = 0;
// while letter is not found loop through until it is found with combimation to the previous letter
while (isFound == 0){
// gerate a random letter
int randLetter = (int) (Math.random() * (24 - 0) + 0);
String nextLetter = letters.substring(randLetter, randLetter + 1);
//create curr word
String currWord = randomWords.get(eachSlot) + nextLetter;// works until here
// loop through each word in wordIdex to find match
System.out.println(wordIndex.get(firstConfirmedLetter).size());
for(int i = 0; i< wordIndex.get(firstConfirmedLetter).size(); i++){
String test = wordIndex.get(firstConfirmedLetter).get(i);
if(test.length() > eachLetter+2){
System.out.println(test.substring(0,eachLetter+2));
if(test.substring(0,eachLetter+2).equals(currWord)){
String currState = randomWords.get(eachSlot);
randomWords.set(eachSlot,currWord);
isFound =1;
}
}
}
}
}
}
return randomWords;
}
The odds of generating a valid word randomly are low, so this approach is inefficient. Instead, randomly select a valid word from your dictionary:
private final List<String> words = wordIndex.values().stream()
.flatMap(List::stream)
.collect(Collectors.toList());
public List<String> randomSize(int length) {
Collections.shuffle(words);
return new ArrayList<>(words.subList(0, length));
}
Your code is very difficult to read and therefore difficult to fix, but here are some bugs in it:
The randWordLength chosen can be zero, but this will still result in a one character word being selected. I presume you intended to select words of length 1–10, inclusive.
You only randomly choose from the first 24 letters of your 26-letter set. Presumably, your dictionary contains words with all 26 letters, but these can never be found.
Most importantly, you are testing prefixes of words of any length, but you keep looping until you find a word of the specified length. As a simplified example, consider a dictionary of two words, "ab" and "acc". If a length of three is required, but the loop chooses the prefix "ab", it will loop forever, trying to find a three-letter word that starts with "ab".
You must match prefixes only against words of the required length.
You underestimate the universe of existing words.
There are: 24^10 possibilities, that is 3.6520347e16 possible words. It's not infinite if you let it running long enough, is just a terrible strategy.
You could increase the probability for early positives by combining random weighted generation like here https://stackoverflow.com/a/6409791/1803810 , with a distribution based on the frequency of the appearance of letters in english http://pi.math.cornell.edu/~mec/2003-2004/cryptography/subs/frequencies.html.
Given n string of max length m. How can we find the longest common prefix shared by at least two strings among them?
Example: ['flower', 'flow', 'hello', 'fleet']
Answer: fl
I was thinking of building a Trie for all the string and then checking the deepest node (satisfies longest) that branches out to two/more substrings (satisfies commonality). This takes O(n*m) time and space. Is there a better way to do this
Why to use trie(which takes O(mn) time and O(mn) space, just use the basic brute force way. first loop, find the shortest string as minStr, which takes o(n) time, second loop, compare one by one with this minStr, and keep an variable which indicates the rightmost index of minStr, this loop takes O(mn) where m is the shortest length of all strings. The code is like below,
public String longestCommonPrefix(String[] strs) {
if(strs.length==0) return "";
String minStr=strs[0];
for(int i=1;i<strs.length;i++){
if(strs[i].length()<minStr.length())
minStr=strs[i];
}
int end=minStr.length();
for(int i=0;i<strs.length;i++){
int j;
for( j=0;j<end;j++){
if(minStr.charAt(j)!=strs[i].charAt(j))
break;
}
if(j<end)
end=j;
}
return minStr.substring(0,end);
}
there is an O(|S|*n) solution to this problem, using a trie. [n is the number of strings, S is the longest string]
(1) put all strings in a trie
(2) do a DFS in the trie, until you find the first vertex with more than 1 "edge".
(3) the path from the root to the node you found at (2) is the longest common prefix.
There is no possible faster solution then it [in terms of big O notation], at the worst case, all your strings are identical - and you need to read all of them to know it.
I would sort them, which you can do in n lg n time. Then any strings with common prefixes will be right next to eachother. In fact you should be able to keep a pointer of which index you're currently looking at and work your way down for a pretty speedy computation.
As a completely different answer from my other answer...
You can, with one pass, bucket every string based on its first letter.
With another pass you can sort each bucket based on its second later. (This is known as radix sort, which is O(n*m), and O(n) with each pass.) This gives you a baseline prefix of 2.
You can safely remove from your dataset any elements that do not have a prefix of 2.
You can continue the radix sort, removing elements without a shared prefix of p, as p approaches m.
This will give you the same O(n*m) time that the trie approach does, but will always be faster than the trie since the trie must look at every character in every string (as it enters the structure), while this approach is only guaranteed to look at 2 characters per string, at which point it culls much of the dataset.
The worst case is still that every string is identical, which is why it shares the same big O notation, but will be faster in all cases as is guaranteed to use less comparisons since on any "non-worst-case" there are characters that never need to be visited.
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0)
return "";
char[] c_list = strs[0].toCharArray();
int len = c_list.length;
int j = 0;
for (int i = 1; i < strs.length; i++) {
for (j = 0; j < len && j < strs[i].length(); j++)
if (c_list[j] != strs[i].charAt(j))
break;
len = j;
}
return new String(c_list).substring(0, len);
}
It happens that the bucket sort (radix sort) described by corsiKa can be extended such that all strings are eventually placed alone in a bucket, and at that point, the LCP for such a lonely string is known. Further, the shustring of each string is also known; it is one longer than is the LCP. The bucket sort is defacto the construction of a suffix array but, only partially so. Those comparisons that are not performed (as described by corsiKa) indeed represent those portions of the suffix strings that are not added to the suffix array. Finally, this method allows for determination of not just the LCP and shustrings, but also one may easily find those subsequences that are not present within the string.
Since the world is obviously begging for an answer in Swift, here's mine ;)
func longestCommonPrefix(strings:[String]) -> String {
var commonPrefix = ""
var indices = strings.map { $0.startIndex}
outerLoop:
while true {
var toMatch: Character = "_"
for (whichString, f) in strings.enumerate() {
let cursor = indices[whichString]
if cursor == f.endIndex { break outerLoop }
indices[whichString] = cursor.successor()
if whichString == 0 { toMatch = f[cursor] }
if toMatch != f[cursor] { break outerLoop }
}
commonPrefix.append(toMatch)
}
return commonPrefix
}
Swift 3 Update:
func longestCommonPrefix(strings:[String]) -> String {
var commonPrefix = ""
var indices = strings.map { $0.startIndex}
outerLoop:
while true {
var toMatch: Character = "_"
for (whichString, f) in strings.enumerated() {
let cursor = indices[whichString]
if cursor == f.endIndex { break outerLoop }
indices[whichString] = f.characters.index(after: cursor)
if whichString == 0 { toMatch = f[cursor] }
if toMatch != f[cursor] { break outerLoop }
}
commonPrefix.append(toMatch)
}
return commonPrefix
}
What's interesting to note:
this runs in O^2, or O(n x m) where n is the number of strings and m
is the length of the shortest one.
this uses the String.Index data type and thus deals with Grapheme Clusters which the Character type represents.
And given the function I needed to write in the first place:
/// Takes an array of Strings representing file system objects absolute
/// paths and turn it into a new array with the minimum number of common
/// ancestors, possibly pushing the root of the tree as many level downwards
/// as necessary
///
/// In other words, we compute the longest common prefix and remove it
func reify(fullPaths:[String]) -> [String] {
let lcp = longestCommonPrefix(fullPaths)
return fullPaths.map {
return $0[lcp.endIndex ..< $0.endIndex]
}
}
here is a minimal unit test:
func testReifySimple() {
let samplePaths:[String] = [
"/root/some/file"
, "/root/some/other/file"
, "/root/another/file"
, "/root/direct.file"
]
let expectedPaths:[String] = [
"some/file"
, "some/other/file"
, "another/file"
, "direct.file"
]
let reified = PathUtilities().reify(samplePaths)
for (index, expected) in expectedPaths.enumerate(){
XCTAssert(expected == reified[index], "failed match, \(expected) != \(reified[index])")
}
}
Perhaps a more intuitive solution. Channel the already found prefix out of earlier iteration as input string to the remaining or next string input. [[[w1, w2], w3], w4]... so on], where [] is supposedly the LCP of two strings.
public String findPrefixBetweenTwo(String A, String B){
String ans = "";
for (int i = 0, j = 0; i < A.length() && j < B.length(); i++, j++){
if (A.charAt(i) != B.charAt(j)){
return i > 0 ? A.substring(0, i) : "";
}
}
// Either of the string is prefix of another one OR they are same.
return (A.length() > B.length()) ? B.substring(0, B.length()) : A.substring(0, A.length());
}
public String longestCommonPrefix(ArrayList<String> A) {
if (A.size() == 1) return A.get(0);
String prefix = A.get(0);
for (int i = 1; i < A.size(); i++){
prefix = findPrefixBetweenTwo(prefix, A.get(i)); // chain the earlier prefix
}
return prefix;
}
What is the quickest way to find the first character which only appears once in a string?
It has to be at least O(n) because you don't know if a character will be repeated until you've read all characters.
So you can iterate over the characters and append each character to a list the first time you see it, and separately keep a count of how many times you've seen it (in fact the only values that matter for the count is "0", "1" or "more than 1").
When you reach the end of the string you just have to find the first character in the list that has a count of exactly one.
Example code in Python:
def first_non_repeated_character(s):
counts = defaultdict(int)
l = []
for c in s:
counts[c] += 1
if counts[c] == 1:
l.append(c)
for c in l:
if counts[c] == 1:
return c
return None
This runs in O(n).
I see that people have posted some delightful answers below, so I'd like to offer something more in-depth.
An idiomatic solution in Ruby
We can find the first un-repeated character in a string like so:
def first_unrepeated_char string
string.each_char.tally.find { |_, n| n == 1 }.first
end
How does Ruby accomplish this?
Reading Ruby's source
Let's break down the solution and consider what algorithms Ruby uses for each step.
First we call each_char on the string. This creates an enumerator which allows us to visit the string one character at a time. This is complicated by the fact that Ruby handles Unicode characters, so each value we get from the enumerator can be a variable number of bytes. If we know our input is ASCII or similar, we could use each_byte instead.
The each_char method is implemented like so:
rb_str_each_char(VALUE str)
{
RETURN_SIZED_ENUMERATOR(str, 0, 0, rb_str_each_char_size);
return rb_str_enumerate_chars(str, 0);
}
In turn, rb_string_enumerate_chars is implemented as:
rb_str_enumerate_chars(VALUE str, VALUE ary)
{
VALUE orig = str;
long i, len, n;
const char *ptr;
rb_encoding *enc;
str = rb_str_new_frozen(str);
ptr = RSTRING_PTR(str);
len = RSTRING_LEN(str);
enc = rb_enc_get(str);
if (ENC_CODERANGE_CLEAN_P(ENC_CODERANGE(str))) {
for (i = 0; i < len; i += n) {
n = rb_enc_fast_mbclen(ptr + i, ptr + len, enc);
ENUM_ELEM(ary, rb_str_subseq(str, i, n));
}
}
else {
for (i = 0; i < len; i += n) {
n = rb_enc_mbclen(ptr + i, ptr + len, enc);
ENUM_ELEM(ary, rb_str_subseq(str, i, n));
}
}
RB_GC_GUARD(str);
if (ary)
return ary;
else
return orig;
}
From this we can see that it calls rb_enc_mbclen (or its fast version) to get the length (in bytes) of the next character in the string so that it can iterate the next step. By lazily iterating over a string, reading just one character at a time, we end up doing just one full pass over the input string as tally consumes the iterator.
Tally is then implemented like so:
static void
tally_up(VALUE hash, VALUE group)
{
VALUE tally = rb_hash_aref(hash, group);
if (NIL_P(tally)) {
tally = INT2FIX(1);
}
else if (FIXNUM_P(tally) && tally < INT2FIX(FIXNUM_MAX)) {
tally += INT2FIX(1) & ~FIXNUM_FLAG;
}
else {
tally = rb_big_plus(tally, INT2FIX(1));
}
rb_hash_aset(hash, group, tally);
}
static VALUE
tally_i(RB_BLOCK_CALL_FUNC_ARGLIST(i, hash))
{
ENUM_WANT_SVALUE();
tally_up(hash, i);
return Qnil;
}
Here, tally_i uses RB_BLOCK_CALL_FUNC_ARGLIST to call repeatedly to tally_up, which updates the tally hash on every iteration.
Rough time & memory analysis
The each_char method doesn't allocate an array to eagerly hold the characters of the string, so it has a small constant memory overhead. When we tally the characters, we allocate a hash and put our tally data into it which in the worst case scenario can take up as much memory as the input string times some constant factor.
Time-wise, tally does a full scan of the string, and calling find to locate the first non-repeated character will scan the hash again, each of which carry O(n) worst-case complexity.
However, tally also updates a hash on every iteration. Updating the hash on every character can be as slow as O(n) again, so the worst case complexity of this Ruby solution is perhaps O(n^2).
However, under reasonable assumptions, updating a hash has an O(1) complexity, so we can expect the average case amortized to look like O(n).
My old accepted answer in Python
You can't know that the character is un-repeated until you've processed the whole string, so my suggestion would be this:
def first_non_repeated_character(string):
chars = []
repeated = []
for character in string:
if character in chars:
chars.remove(character)
repeated.append(character)
else:
if not character in repeated:
chars.append(character)
if len(chars):
return chars[0]
else:
return False
Edit: originally posted code was bad, but this latest snippet is Certified To Work On Ryan's Computer™.
Why not use a heap based data structure such as a minimum priority queue. As you read each character from the string, add it to the queue with a priority based on the location in the string and the number of occurrences so far. You could modify the queue to add priorities on collision so that the priority of a character is the sum of the number appearances of that character. At the end of the loop, the first element in the queue will be the least frequent character in the string and if there are multiple characters with a count == 1, the first element was the first unique character added to the queue.
Here is another fun way to do it. Counter requires Python2.7 or Python3.1
>>> from collections import Counter
>>> def first_non_repeated_character(s):
... return min((k for k,v in Counter(s).items() if v<2), key=s.index)
...
>>> first_non_repeated_character("aaabbbcddd")
'c'
>>> first_non_repeated_character("aaaebbbcddd")
'e'
Lots of answers are attempting O(n) but are forgetting the actual costs of inserting and removing from the lists/associative arrays/sets they're using to track.
If you can assume that a char is a single byte, then you use a simple array indexed by the char and keep a count in it. This is truly O(n) because the array accesses are guaranteed O(1), and the final pass over the array to find the first element with 1 is constant time (because the array has a small, fixed size).
If you can't assume that a char is a single byte, then I would propose sorting the string and then doing a single pass checking adjacent values. This would be O(n log n) for the sort plus O(n) for the final pass. So it's effectively O(n log n), which is better than O(n^2). Also, it has virtually no space overhead, which is another problem with many of the answers that are attempting O(n).
Counter requires Python2.7 or Python3.1
>>> from collections import Counter
>>> def first_non_repeated_character(s):
... counts = Counter(s)
... for c in s:
... if counts[c]==1:
... return c
... return None
...
>>> first_non_repeated_character("aaabbbcddd")
'c'
>>> first_non_repeated_character("aaaebbbcddd")
'e'
Refactoring a solution proposed earlier (not having to use extra list/memory). This goes over the string twice. So this takes O(n) too like the original solution.
def first_non_repeated_character(s):
counts = defaultdict(int)
for c in s:
counts[c] += 1
for c in s:
if counts[c] == 1:
return c
return None
The following is a Ruby implementation of finding the first nonrepeated character of a string:
def first_non_repeated_character(string)
string1 = string.split('')
string2 = string.split('')
string1.each do |let1|
counter = 0
string2.each do |let2|
if let1 == let2
counter+=1
end
end
if counter == 1
return let1
break
end
end
end
p first_non_repeated_character('dont doddle in the forest')
And here is a JavaScript implementation of the same style function:
var first_non_repeated_character = function (string) {
var string1 = string.split('');
var string2 = string.split('');
var single_letters = [];
for (var i = 0; i < string1.length; i++) {
var count = 0;
for (var x = 0; x < string2.length; x++) {
if (string1[i] == string2[x]) {
count++
}
}
if (count == 1) {
return string1[i];
}
}
}
console.log(first_non_repeated_character('dont doddle in the forest'));
console.log(first_non_repeated_character('how are you today really?'));
In both cases I used a counter knowing that if the letter is not matched anywhere in the string, it will only occur in the string once so I just count it's occurrence.
I think this should do it in C. This operates in O(n) time with no ambiguity about order of insertion and deletion operators. This is a counting sort (simplest form of a bucket sort, which itself is the simple form of a radix sort).
unsigned char find_first_unique(unsigned char *string)
{
int chars[256];
int i=0;
memset(chars, 0, sizeof(chars));
while (string[i++])
{
chars[string[i]]++;
}
i = 0;
while (string[i++])
{
if (chars[string[i]] == 1) return string[i];
}
return 0;
}
In Ruby:
(Original Credit: Andrew A. Smith)
x = "a huge string in which some characters repeat"
def first_unique_character(s)
s.each_char.detect { |c| s.count(c) == 1 }
end
first_unique_character(x)
=> "u"
def first_non_repeated_character(string):
chars = []
repeated = []
for character in string:
if character in repeated:
... discard it.
else if character in chars:
chars.remove(character)
repeated.append(character)
else:
if not character in repeated:
chars.append(character)
if len(chars):
return chars[0]
else:
return False
Other JavaScript solutions are quite c-style solutions here is a more JavaScript-style solution.
var arr = string.split("");
var occurences = {};
var tmp;
var lowestindex = string.length+1;
arr.forEach( function(c){
tmp = c;
if( typeof occurences[tmp] == "undefined")
occurences[tmp] = tmp;
else
occurences[tmp] += tmp;
});
for(var p in occurences) {
if(occurences[p].length == 1)
lowestindex = Math.min(lowestindex, string.indexOf(p));
}
if(lowestindex > string.length)
return null;
return string[lowestindex];
}
in C, this is almost Shlemiel the Painter's Algorithm (not quite O(n!) but more than 0(n2)).
But will outperform "better" algorithms for reasonably sized strings because O is so small. This can also easily tell you the location of the first non-repeating string.
char FirstNonRepeatedChar(char * psz)
{
for (int ii = 0; psz[ii] != 0; ++ii)
{
for (int jj = ii+1; ; ++jj)
{
// if we hit the end of string, then we found a non-repeat character.
//
if (psz[jj] == 0)
return psz[ii]; // this character doesn't repeat
// if we found a repeat character, we can stop looking.
//
if (psz[ii] == psz[jj])
break;
}
}
return 0; // there were no non-repeating characters.
}
edit: this code is assuming you don't mean consecutive repeating characters.
Here's an implementation in Perl (version >=5.10) that doesn't care whether the repeated characters are consecutive or not:
use strict;
use warnings;
foreach my $word(#ARGV)
{
my #distinct_chars;
my %char_counts;
my #chars=split(//,$word);
foreach (#chars)
{
push #distinct_chars,$_ unless $_~~#distinct_chars;
$char_counts{$_}++;
}
my $first_non_repeated="";
foreach(#distinct_chars)
{
if($char_counts{$_}==1)
{
$first_non_repeated=$_;
last;
}
}
if(length($first_non_repeated))
{
print "For \"$word\", the first non-repeated character is '$first_non_repeated'.\n";
}
else
{
print "All characters in \"$word\" are repeated.\n";
}
}
Storing this code in a script (which I named non_repeated.pl) and running it on a few inputs produces:
jmaney> perl non_repeated.pl aabccd "a huge string in which some characters repeat" abcabc
For "aabccd", the first non-repeated character is 'b'.
For "a huge string in which some characters repeat", the first non-repeated character is 'u'.
All characters in "abcabc" are repeated.
Here's a possible solution in ruby without using Array#detect (as in this answer). Using Array#detect makes it too easy, I think.
ALPHABET = %w(a b c d e f g h i j k l m n o p q r s t u v w x y z)
def fnr(s)
unseen_chars = ALPHABET.dup
seen_once_chars = []
s.each_char do |c|
if unseen_chars.include?(c)
unseen_chars.delete(c)
seen_once_chars << c
elsif seen_once_chars.include?(c)
seen_once_chars.delete(c)
end
end
seen_once_chars.first
end
Seems to work for some simple examples:
fnr "abcdabcegghh"
# => "d"
fnr "abababababababaqababa"
=> "q"
Suggestions and corrections are very much appreciated!
Try this code:
public static String findFirstUnique(String str)
{
String unique = "";
foreach (char ch in str)
{
if (unique.Contains(ch)) unique=unique.Replace(ch.ToString(), "");
else unique += ch.ToString();
}
return unique[0].ToString();
}
In Mathematica one might write this:
string = "conservationist deliberately treasures analytical";
Cases[Gather # Characters # string, {_}, 1, 1][[1]]
{"v"}
This snippet code in JavaScript
var string = "tooth";
var hash = [];
for(var i=0; j=string.length, i<j; i++){
if(hash[string[i]] !== undefined){
hash[string[i]] = hash[string[i]] + 1;
}else{
hash[string[i]] = 1;
}
}
for(i=0; j=string.length, i<j; i++){
if(hash[string[i]] === 1){
console.info( string[i] );
return false;
}
}
// prints "h"
Different approach here.
scan each element in the string and create a count array which stores the repetition count of each element.
Next time again start from first element in the array and print the first occurrence of element with count = 1
C code
-----
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
char t_c;
char *t_p = argv[1] ;
char count[128]={'\0'};
char ch;
for(t_c = *(argv[1]); t_c != '\0'; t_c = *(++t_p))
count[t_c]++;
t_p = argv[1];
for(t_c = *t_p; t_c != '\0'; t_c = *(++t_p))
{
if(count[t_c] == 1)
{
printf("Element is %c\n",t_c);
break;
}
}
return 0;
}
input is = aabbcddeef output is = c
char FindUniqueChar(char *a)
{
int i=0;
bool repeat=false;
while(a[i] != '\0')
{
if (a[i] == a[i+1])
{
repeat = true;
}
else
{
if(!repeat)
{
cout<<a[i];
return a[i];
}
repeat=false;
}
i++;
}
return a[i];
}
Here is another approach...we could have a array which will store the count and the index of the first occurrence of the character. After filling up the array we could jst traverse the array and find the MINIMUM index whose count is 1 then return str[index]
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <climits>
using namespace std;
#define No_of_chars 256
//store the count and the index where the char first appear
typedef struct countarray
{
int count;
int index;
}countarray;
//returns the count array
countarray *getcountarray(char *str)
{
countarray *count;
count=new countarray[No_of_chars];
for(int i=0;i<No_of_chars;i++)
{
count[i].count=0;
count[i].index=-1;
}
for(int i=0;*(str+i);i++)
{
(count[*(str+i)].count)++;
if(count[*(str+i)].count==1) //if count==1 then update the index
count[*(str+i)].index=i;
}
return count;
}
char firstnonrepeatingchar(char *str)
{
countarray *array;
array = getcountarray(str);
int result = INT_MAX;
for(int i=0;i<No_of_chars;i++)
{
if(array[i].count==1 && result > array[i].index)
result = array[i].index;
}
delete[] (array);
return (str[result]);
}
int main()
{
char str[] = "geeksforgeeks";
cout<<"First non repeating character is "<<firstnonrepeatingchar(str)<<endl;
return 0;
}
Function:
This c# function uses a HashTable (Dictionary) and have a performance O(2n) worstcase.
private static string FirstNoRepeatingCharacter(string aword)
{
Dictionary<string, int> dic = new Dictionary<string, int>();
for (int i = 0; i < aword.Length; i++)
{
if (!dic.ContainsKey(aword.Substring(i, 1)))
dic.Add(aword.Substring(i, 1), 1);
else
dic[aword.Substring(i, 1)]++;
}
foreach (var item in dic)
{
if (item.Value == 1) return item.Key;
}
return string.Empty;
}
Example:
string aword = "TEETER";
Console.WriteLine(FirstNoRepeatingCharacter(aword)); //print: R
I have two strings i.e. 'unique' and 'repeated'. Every character appearing for the first time, gets added to 'unique'. If it is repeated for the second time, it gets removed from 'unique' and added to 'repeated'. This way, we will always have a string of unique characters in 'unique'.
Complexity big O(n)
public void firstUniqueChar(String str){
String unique= "";
String repeated = "";
str = str.toLowerCase();
for(int i=0; i<str.length();i++){
char ch = str.charAt(i);
if(!(repeated.contains(str.subSequence(i, i+1))))
if(unique.contains(str.subSequence(i, i+1))){
unique = unique.replaceAll(Character.toString(ch), "");
repeated = repeated+ch;
}
else
unique = unique+ch;
}
System.out.println(unique.charAt(0));
}
The following code is in C# with complexity of n.
using System;
using System.Linq;
using System.Text;
namespace SomethingDigital
{
class FirstNonRepeatingChar
{
public static void Main()
{
String input = "geeksforgeeksandgeeksquizfor";
char[] str = input.ToCharArray();
bool[] b = new bool[256];
String unique1 = "";
String unique2 = "";
foreach (char ch in str)
{
if (!unique1.Contains(ch))
{
unique1 = unique1 + ch;
unique2 = unique2 + ch;
}
else
{
unique2 = unique2.Replace(ch.ToString(), "");
}
}
if (unique2 != "")
{
Console.WriteLine(unique2[0].ToString());
Console.ReadLine();
}
else
{
Console.WriteLine("No non repeated string");
Console.ReadLine();
}
}
}
}
The following solution is an elegant way to find the first unique character within a string using the new features which have been introduced as part as Java 8. This solution uses the approach of first creating a map to count the number of occurrences of each character. It then uses this map to find the first character which occurs only once. This runs in O(N) time.
import static java.util.stream.Collectors.counting;
import static java.util.stream.Collectors.groupingBy;
import java.util.Arrays;
import java.util.List;
import java.util.Map;
// Runs in O(N) time and uses lambdas and the stream API from Java 8
// Also, it is only three lines of code!
private static String findFirstUniqueCharacterPerformantWithLambda(String inputString) {
// convert the input string into a list of characters
final List<String> inputCharacters = Arrays.asList(inputString.split(""));
// first, construct a map to count the number of occurrences of each character
final Map<Object, Long> characterCounts = inputCharacters
.stream()
.collect(groupingBy(s -> s, counting()));
// then, find the first unique character by consulting the count map
return inputCharacters
.stream()
.filter(s -> characterCounts.get(s) == 1)
.findFirst()
.orElse(null);
}
Here is one more solution with o(n) time complexity.
public void findUnique(String string) {
ArrayList<Character> uniqueList = new ArrayList<>();
int[] chatArr = new int[128];
for (int i = 0; i < string.length(); i++) {
Character ch = string.charAt(i);
if (chatArr[ch] != -1) {
chatArr[ch] = -1;
uniqueList.add(ch);
} else {
uniqueList.remove(ch);
}
}
if (uniqueList.size() == 0) {
System.out.println("No unique character found!");
} else {
System.out.println("First unique character is :" + uniqueList.get(0));
}
}
I read through the answers, but did not see any like mine, I think this answer is very simple and fast, am I wrong?
def first_unique(s):
repeated = []
while s:
if s[0] not in s[1:] and s[0] not in repeated:
return s[0]
else:
repeated.append(s[0])
s = s[1:]
return None
test
(first_unique('abdcab') == 'd', first_unique('aabbccdad') == None, first_unique('') == None, first_unique('a') == 'a')
Question : First Unique Character of a String
This is the simplest solution.
public class Test4 {
public static void main(String[] args) {
String a = "GiniGinaProtijayi";
firstUniqCharindex(a);
}
public static void firstUniqCharindex(String a) {
int[] count = new int[256];
for (int i = 0; i < a.length(); i++) {
count[a.charAt(i)]++;
}
int index = -1;
for (int i = 0; i < a.length(); i++) {
if (count[a.charAt(i)] == 1) {
index = i;
break;
} // if
}
System.out.println(index);// output => 8
System.out.println(a.charAt(index)); //output => P
}// end1
}
IN Python :
def firstUniqChar(a):
count = [0] * 256
for i in a: count[ord(i)] += 1
element = ""
for items in a:
if(count[ord(items) ] == 1):
element = items ;
break
return element
a = "GiniGinaProtijayi";
print(firstUniqChar(a)) # output is P
Using Java 8 :
public class Test2 {
public static void main(String[] args) {
String a = "GiniGinaProtijayi";
Map<Character, Long> map = a.chars()
.mapToObj(
ch -> Character.valueOf((char) ch)
).collect(
Collectors.groupingBy(
Function.identity(),
LinkedHashMap::new,
Collectors.counting()));
System.out.println("MAP => " + map);
// {G=2, i=5, n=2, a=2, P=1, r=1, o=1, t=1, j=1, y=1}
Character chh = map
.entrySet()
.stream()
.filter(entry -> entry.getValue() == 1L)
.map(entry -> entry.getKey())
.findFirst()
.get();
System.out.println("First Non Repeating Character => " + chh);// P
}// main
}
how about using a suffix tree for this case... the first unrepeated character will be first character of longest suffix string with least depth in tree..
Create Two list -
unique list - having only unique character .. UL
non-unique list - having only repeated character -NUL
for(char c in str) {
if(nul.contains(c)){
//do nothing
}else if(ul.contains(c)){
ul.remove(c);
nul.add(c);
}else{
nul.add(c);
}
I recently found a contest problem that asks you to compute the minimum number of characters that must be inserted (anywhere) in a string to turn it into a palindrome.
For example, given the string: "abcbd" we can turn it into a palindrome by inserting just two characters: one after "a" and another after "d": "adbcbda".
This seems to be a generalization of a similar problem that asks for the same thing, except characters can only be added at the end - this has a pretty simple solution in O(N) using hash tables.
I have been trying to modify the Levenshtein distance algorithm to solve this problem, but haven't been successful. Any help on how to solve this (it doesn't necessarily have to be efficient, I'm just interested in any DP solution) would be appreciated.
Note: This is just a curiosity. Dav proposed an algorithm which can be modified to DP algorithm to run in O(n^2) time and O(n^2) space easily (and perhaps O(n) with better bookkeeping).
Of course, this 'naive' algorithm might actually come in handy if you decide to change the allowed operations.
Here is a 'naive'ish algorithm, which can probably be made faster with clever bookkeeping.
Given a string, we guess the middle of the resulting palindrome and then try to compute the number of inserts required to make the string a palindrome around that middle.
If the string is of length n, there are 2n+1 possible middles (Each character, between two characters, just before and just after the string).
Suppose we consider a middle which gives us two strings L and R (one to left and one to right).
If we are using inserts, I believe the Longest Common Subsequence algorithm (which is a DP algorithm) can now be used the create a 'super' string which contains both L and reverse of R, see Shortest common supersequence.
Pick the middle which gives you the smallest number inserts.
This is O(n^3) I believe. (Note: I haven't tried proving that it is true).
My C# solution looks for repeated characters in a string and uses them to reduce the number of insertions. In a word like program, I use the 'r' characters as a boundary. Inside of the 'r's, I make that a palindrome (recursively). Outside of the 'r's, I mirror the characters on the left and the right.
Some inputs have more than one shortest output: output can be toutptuot or outuputuo. My solution only selects one of the possibilities.
Some example runs:
radar -> radar, 0 insertions
esystem -> metsystem, 2 insertions
message -> megassagem, 3 insertions
stackexchange -> stegnahckexekchangets, 8 insertions
First I need to check if an input is already a palindrome:
public static bool IsPalindrome(string str)
{
for (int left = 0, right = str.Length - 1; left < right; left++, right--)
{
if (str[left] != str[right])
return false;
}
return true;
}
Then I need to find any repeated characters in the input. There may be more than one. The word message has two most-repeated characters ('e' and 's'):
private static bool TryFindMostRepeatedChar(string str, out List<char> chs)
{
chs = new List<char>();
int maxCount = 1;
var dict = new Dictionary<char, int>();
foreach (var item in str)
{
int temp;
if (dict.TryGetValue(item, out temp))
{
dict[item] = temp + 1;
maxCount = temp + 1;
}
else
dict.Add(item, 1);
}
foreach (var item in dict)
{
if (item.Value == maxCount)
chs.Add(item.Key);
}
return maxCount > 1;
}
My algorithm is here:
public static string MakePalindrome(string str)
{
List<char> repeatedList;
if (string.IsNullOrWhiteSpace(str) || IsPalindrome(str))
{
return str;
}
//If an input has repeated characters,
// use them to reduce the number of insertions
else if (TryFindMostRepeatedChar(str, out repeatedList))
{
string shortestResult = null;
foreach (var ch in repeatedList) //"program" -> { 'r' }
{
//find boundaries
int iLeft = str.IndexOf(ch); // "program" -> 1
int iRight = str.LastIndexOf(ch); // "program" -> 4
//make a palindrome of the inside chars
string inside = str.Substring(iLeft + 1, iRight - iLeft - 1); // "program" -> "og"
string insidePal = MakePalindrome(inside); // "og" -> "ogo"
string right = str.Substring(iRight + 1); // "program" -> "am"
string rightRev = Reverse(right); // "program" -> "ma"
string left = str.Substring(0, iLeft); // "program" -> "p"
string leftRev = Reverse(left); // "p" -> "p"
//Shave off extra chars in rightRev and leftRev
// When input = "message", this loop converts "meegassageem" to "megassagem",
// ("ee" to "e"), as long as the extra 'e' is an inserted char
while (left.Length > 0 && rightRev.Length > 0 &&
left[left.Length - 1] == rightRev[0])
{
rightRev = rightRev.Substring(1);
leftRev = leftRev.Substring(1);
}
//piece together the result
string result = left + rightRev + ch + insidePal + ch + right + leftRev;
//find the shortest result for inputs that have multiple repeated characters
if (shortestResult == null || result.Length < shortestResult.Length)
shortestResult = result;
}
return shortestResult;
}
else
{
//For inputs that have no repeated characters,
// just mirror the characters using the last character as the pivot.
for (int i = str.Length - 2; i >= 0; i--)
{
str += str[i];
}
return str;
}
}
Note that you need a Reverse function:
public static string Reverse(string str)
{
string result = "";
for (int i = str.Length - 1; i >= 0; i--)
{
result += str[i];
}
return result;
}
C# Recursive solution adding to the end of the string:
There are 2 base cases. When length is 1 or 2. Recursive case: If the extremes are equal, then
make palindrome the inner string without the extremes and return that with the extremes.
If the extremes are not equal, then add the first character to the end and make palindrome the
inner string including the previous last character. return that.
public static string ConvertToPalindrome(string str) // By only adding characters at the end
{
if (str.Length == 1) return str; // base case 1
if (str.Length == 2 && str[0] == str[1]) return str; // base case 2
else
{
if (str[0] == str[str.Length - 1]) // keep the extremes and call
return str[0] + ConvertToPalindrome(str.Substring(1, str.Length - 2)) + str[str.Length - 1];
else //Add the first character at the end and call
return str[0] + ConvertToPalindrome(str.Substring(1, str.Length - 1)) + str[0];
}
}
Given a set of words, we need to find the anagram words and display each category alone using the best algorithm.
input:
man car kile arc none like
output:
man
car arc
kile like
none
The best solution I am developing now is based on an hashtable, but I am thinking about equation to convert anagram word into integer value.
Example: man => 'm'+'a'+'n' but this will not give unique values.
Any suggestion?
See following code in C#:
string line = Console.ReadLine();
string []words=line.Split(' ');
int[] numbers = GetUniqueInts(words);
for (int i = 0; i < words.Length; i++)
{
if (table.ContainsKey(numbers[i]))
{
table[numbers[i]] = table[numbers[i]].Append(words[i]);
}
else
{
table.Add(numbers[i],new StringBuilder(words[i]));
}
}
The problem is how to develop GetUniqueInts(string []) method.
Don't bother with a custom hash function at all. Use the normal string hash function on whatever your platform is. The important thing is to make the key for your hash table the idea of a "sorted word" - where the word is sorted by letter, so "car" => "acr". All anagrams will have the same "sorted word".
Just have a hash from "sorted word" to "list of words for that sorted word". In LINQ this is incredibly easy:
using System;
using System.Collections.Generic;
using System.Linq;
class FindAnagrams
{
static void Main(string[] args)
{
var lookup = args.ToLookup(word => SortLetters(word));
foreach (var entry in lookup)
{
foreach (var word in entry)
{
Console.Write(word);
Console.Write(" ");
}
Console.WriteLine();
}
}
static string SortLetters(string original)
{
char[] letters = original.ToCharArray();
Array.Sort(letters);
return new string(letters);
}
}
Sample use:
c:\Users\Jon\Test>FindAnagrams.exe man car kile arc none like
man
car arc
kile like
none
I used a Godel-inspired scheme:
Assign the primes P_1 to P_26 to the letters (in any order, but to obtain smallish hash values best to give common letters small primes).
Built a histogram of the letters in the word.
Then the hash value is the product of each letter's associated prime raised to the power of its frequency. This gives a unique value to every anagram.
Python code:
primes = [2, 41, 37, 47, 3, 67, 71, 23, 5, 101, 61, 17, 19, 13, 31, 43, 97, 29, 11, 7, 73, 83, 79, 89, 59, 53]
def get_frequency_map(word):
map = {}
for letter in word:
map[letter] = map.get(letter, 0) + 1
return map
def hash(word):
map = get_frequency_map(word)
product = 1
for letter in map.iterkeys():
product = product * primes[ord(letter)-97] ** map.get(letter, 0)
return product
This cleverly transforms the tricky problem of finding subanagrams into the (also known to be tricky) problem of factoring large numbers...
A Python version for giggles:
from collections import defaultdict
res = defaultdict(list)
L = "car, acr, bat, tab, get, cat".split(", ")
for w in L:
res["".join(sorted(w))].append(w)
print(res.values())
I don't think you'll find anything better than a hash table with a custom hash function (that would sort the letters of he word before hashing it).
Sum of the letters will never work, because you can't really make 'ac' and 'bb' different.
You will need large integers (or a bit vector actually) but the following might work
the first occurrence of each letter get's assigned the bit number for that letter, the second occurence gets the bit number for that letter + 26.
For example
a #1 = 1
b #1 = 2
c #1 = 4
a #2 = 2^26
b #2 = 2 ^ 27
You can then sum these together, to get a unique value for the word based on it's letters.
Your storage requirements for the word values will be:
n * 26 bits
where n is the maximum number of occurrences of any repeated letter.
I wouldn't use hashing since it adds additional complexity for look-up and adds. Hashing, sorting and multiplications are all going to be slower than a simple array-based histogram solution with tracking uniques. Worst case is O(2n):
// structured for clarity
static bool isAnagram(String s1, String s2)
{
int[] histogram = new int[256];
int uniques = 0;
// scan first string
foreach (int c in s1)
{
// count occurrence
int count = ++histogram[c];
// count uniques
if (count == 1)
{
++uniques;
}
}
// scan second string
foreach (int c in s2)
{
// reverse count occurrence
int count = --histogram[c];
// reverse count uniques
if (count == 0)
{
--uniques;
}
else if (count < 0) // trivial reject of longer strings or more occurrences
{
return false;
}
}
// final histogram unique count should be 0
return (uniques == 0);
}
I have implemented this before with a simple array of letter counts, e.g.:
unsigned char letter_frequency[26];
Then store that in a database table together with each word. Words that have the same letter frequency 'signature' are anagrams, and a simple SQL query then returns all anagrams of a word directly.
With some experimentation with a very large dictionary, I found no word that exceeded a frequency count of 9 for any letter, so the 'signature' can be represented as a string of numbers 0..9 (The size could be easily halved by packing into bytes as hex, and further reduced by binary encoding the number, but I didn't bother with any of this so far).
Here is a ruby function to compute the signature of a given word and store it into a Hash, while discarding duplicates. From the Hash I later build a SQL table:
def processword(word, downcase)
word.chomp!
word.squeeze!(" ")
word.chomp!(" ")
if (downcase)
word.downcase!
end
if ($dict[word]==nil)
stdword=word.downcase
signature=$letters.collect {|letter| stdword.count(letter)}
signature.each do |cnt|
if (cnt>9)
puts "Signature overflow:#{word}|#{signature}|#{cnt}"
end
end
$dict[word]=[$wordid,signature]
$wordid=$wordid+1
end
end
Assign a unique prime number to the letters a-z
Iterate your word array, creating a product of primes based on the letters in each word.
Store that product in your word list, with the corresponding word.
Sort the array, ascending by the product.
Iterate the array, doing a control break at every product change.
In C, I just implemented the following hash which basically does a 26-bit bitmask on whether the word in the dictionary has a particular letter in it. So, all anagrams have the same hash. The hash doesn't take into account repeated letters, so there will be some additional overloading, but it still manages to be faster than my perl implementation.
#define BUCKETS 49999
struct bucket {
char *word;
struct bucket *next;
};
static struct bucket hash_table[BUCKETS];
static unsigned int hash_word(char *word)
{
char *p = word;
unsigned int hash = 0;
while (*p) {
if (*p < 97 || *p > 122) {
return 0;
}
hash |= 2 << (*p - 97);
*p++;
}
return hash % BUCKETS;
}
Overloaded buckets created and added as linked list, etc. Then just write a function that makes sure that the words that match the hash value are the same length and that the letters in each are 1 to 1 and return that as a match.
I will generate the hasmap based on the sample word and the rest of the alphabets I won't care.
For example if the word is "car"
my hash table will be like this:
a,0
b,MAX
c,1
d,MAX
e,MAX
...
..
r,2
.
As a result any has greater than 3 will consider as not matching
(more tuning...)
And my comparison method will compare the hash total within the hash calculation itself. It won't continue once it can identify the word is not equal.
public static HashMap<String, Integer> getHashMap(String word) {
HashMap<String, Integer> map = new HashMap<String, Integer>();
String[] chars = word.split("");
int index = 0;
for (String c : chars) {
map.put(c, index);
index++;
}
return map;
}
public static int alphaHash(String word, int base,
HashMap<String, Integer> map) {
String[] chars = word.split("");
int result = 0;
for (String c : chars) {
if (c.length() <= 0 || c.equals(null)) {
continue;
}
int index = 0;
if (map.containsKey(c)) {
index = map.get(c);
} else {
index = Integer.MAX_VALUE;
}
result += index;
if (result > base) {
return result;
}
}
return result;
}
Main method
HashMap<String, Integer> map = getHashMap(sample);
int sampleHash = alphaHash(sample, Integer.MAX_VALUE, map);
for (String s : args) {
if (sampleHash == alphaHash(s, sampleHash, map)) {
System.out.print(s + " ");
}
}
Anagrams can be found in following way:
Length of word should match.
Perform addition of each character in terms of integer value. This sum will match if you perform same on anagram.
Perform multiplication of each character in terms of integer value. Evaluated value will match if you perform same on anagram.
So I thought through above three validations, we can find anagrams. Correct me if I'm wrong.
Example: abc cba
Length of both words is 3.
Sum of individual characters for both words is 294.
Prod of individual characters for both words is 941094.
Just want to add simple python solution in addition to the other useful answers:
def check_permutation_group(word_list):
result = {}
for word in word_list:
hash_arr_for_word = [0] * 128 # assuming standard ascii
for char in word:
char_int = ord(char)
hash_arr_for_word[char_int] += 1
hash_for_word = ''.join(str(item) for item in hash_arr_for_word)
if not result.get(hash_for_word, None):
result[str(hash_for_word)] = [word]
else:
result[str(hash_for_word)] += [word]
return list(result.values())
python code:
line = "man car kile arc none like"
hmap = {}
for w in line.split():
ws = ''.join(sorted(w))
try:
hmap[ws].append(w)
except KeyError:
hmap[ws] = [w]
for i in hmap:
print hmap[i]
output:
['car', 'arc']
['kile', 'like']
['none']
['man']
JavaScript version. using hashing.
Time Complexity: 0(nm) , where n is number of words, m is length of word
var words = 'cat act mac tac ten cam net'.split(' '),
hashMap = {};
words.forEach(function(w){
w = w.split('').sort().join('');
hashMap[w] = (hashMap[w]|0) + 1;
});
function print(obj,key){
console.log(key, obj[key]);
}
Object.keys(hashMap).forEach(print.bind(null,hashMap))