if my string is lets say "Alfa1234Beta"
how can I convert all the number in to "_"
for example "Alfa1234Beta"
will be "Alfa____Beta"
Going with the Regex approach pointed out by others is possibly OK for your scenario. Mind you however, that Regex sometimes tend to be overused. A hand rolled approach could be like this:
static string ReplaceDigits(string str)
{
StringBuilder sb = null;
for (int i = 0; i < str.Length; i++)
{
if (Char.IsDigit(str[i]))
{
if (sb == null)
{
// Seen a digit, allocate StringBuilder, copy non-digits we might have skipped over so far.
sb = new StringBuilder();
if (i > 0)
{
sb.Append(str, 0, i);
}
}
// Replace current character (a digit)
sb.Append('_');
}
else
{
if (sb != null)
{
// Seen some digits (being replaced) already. Collect non-digits as well.
sb.Append(str[i]);
}
}
}
if (sb != null)
{
return sb.ToString();
}
return str;
}
It is more light weight than Regex and only allocates when there is actually something to do (replace). So, go ahead use the Regex version if you like. If you figure out during profiling that is too heavy weight, you can use something like the above. YMMV
You can run for loop on the string and then use the following method to replace numbers with _
if (!System.Text.RegularExpressions.Regex.IsMatch(i, "^[0-9]*$"))
Here variable i is the character in the for loop .
You can use this:
var s = "Alfa1234Beta";
var s2 = System.Text.RegularExpressions.Regex.Replace(s, "[0-9]", "_");
s2 now contains "Alfa____Beta".
Explanation: the regex [0-9] matches any digit from 0 to 9 (inclusive). The Regex.Replace then replaces all matched characters with an "_".
EDIT
And if you want it a bit shorter AND also match non-latin digits, use \d as a regex:
var s = "Alfa1234Beta๓"; // ๓ is "Thai digit three"
var s2 = System.Text.RegularExpressions.Regex.Replace(s, #"\d", "_");
s2 now contains "Alfa____Beta_".
Related
Let's assume we have a multiline string, like
var s:String = "my first line\nmy second line\nmy third line\nand so on!";
What is the best way to get (only) the first line of this string in Haxe? I know I can do something like:
static function getFirstLine(s:String):String {
var t:String = s.split("\n")[0];
if(t.charAt(t.length - 1) == "\r") {
t = t.substring(0, t.length - 1);
}
return t;
}
However I'm wondering if there is any easier (predefined) method for this ...
Caveat that #Gama11's answer works well and is more elegant than this.
If your string is long, split will iterate over the whole thing and allocate an array containing every line in your string, both of which are unnecessary here. Another option would be indexOf:
static function getFirstLine(s:String):String {
var i = s.indexOf("\n");
if (i == -1) return s;
if (i > 0 && s.charAt(i - 1) == "\r") --i;
return s.substr(0, i);
}
There's no built-in utility in the standard library for this that I know of, but you make it a bit more elegant and avoid the substring() handling for \r by splitting on a regex:
static function getFirstLine(s:String):String {
return ~/\r?\n/.split(s)[0];
}
The regex \r?\n optionally matches a carriage return followed by a line feed character.
Using std C++, I would like to split a string delimited by commas but ignore commas in strings that are surrounded by single quotes. For example:
1,'2,3',4,5,'6,7',8
when split becomes
1
'2,3'
4
5
'6,7'
8
I think this might be best handled with regex, but I'm not sure how to construct the pattern. Any solution without regex is welcome, too. Thanks.
I'm not sure what the C++ syntax would be, but here's some pseudocode:
vector<string> split(const string& value)
{
bool is_escaped = false;
vector<char> current;
vector<string> result;
for (char c : value)
{
if (c == '\'')
{
is_escaped = !is_escaped;
}
if (c == ',' && !is_escaped)
{
result.push_back(string(current.begin(), current.end());
current.clear();
}
else
{
current.push_back(c);
}
}
result.push_back(string(current.begin(), current.end());
return result;
}
Obviously you'll need to adjust it to be valid C++ but it should do the trick.
I'm stuck on a task of trying to print words that contain only lowercase letters a-z. I have already stripped out an inputted string if it contains any number 0-9 and if it contains an Uppercase letter:
String[] textParts;
textParts = text.Split(delimChars);
for (int i = 0; i < textParts.Length; i++) //adds s to words list and checks for capitals
{
String s = textParts[i];
bool valid = true;
foreach (char c in textParts[i])
{
if (char.IsUpper(c))
{
valid = false;
break;
}
if (c >= '0' && c <= '9')
{
valid = false;
break;
}
if (char.IsPunctuation(c))
{
valid = false;
break;
}
}
if (valid) pageIn.words.Add(s);
This is my code so far. The last part I'm trying to check to see if a word contains any punctuation (it's not working) is there an easier way I could do this and how could I get the last part of my code to work?
P.S. I'm not that comfortable with using Regex.
Many Thanks,
Ellie
Without regex, you can use LINQ (might be less performant)
bool isOnlyLower = s.Count(c => Char.IsLower(c)) == s.Length;
Count will retrieve the number of char that are in lower in the following string. If it match the string's length, string is composed only of lowercase letters.
An alternative would be to check if there's any UpperCase :
bool isOnlyLower = !s.Any(c => Char.IsUpper(c));
var regex = new Regex("^[a-z]+$");
if (!regex.IsMatch(input))
{
// is't not only lower case letters, remove input
}
I'm not sure whether I get your question right, but shouldn't the following work?
for (int i = 0; i < textParts.Length; i++) //adds s to words list and checks for capitals
{
String s = textParts[i];
if(s.Equals(s.ToLower()))
{
// string is all lower
}
}
Closed. This question needs to be more focused. It is not currently accepting answers.
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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;
}
}
}
I asked this question in a few interviews. I want to know from the Stackoverflow readers as to what should be the answer to this question.
Such a seemingly simple question, but has been interpreted quite a few different ways.
if your definition of a "word" is a series of non-whitespace characters surrounded by a whitespace character, then in 5 second pseudocode you do:
var words = split(inputString, " ")
var reverse = new array
var count = words.count -1
var i = 0
while count != 0
reverse[i] = words[count]
count--
i++
return reverse
If you want to take into consideration also spaces, you can do it like that:
string word = "hello my name is";
string result="";
int k=word.size();
for (int j=word.size()-1; j>=0; j--)
{
while(word[j]!= ' ' && j>=0)
j--;
int end=k;
k=j+1;
int count=0;
if (j>=0)
{
int temp=j;
while (word[temp]==' '){
count++;
temp--;
}
j-=count;
}
else j=j+1;
result+=word.substr(k,end-k);
k-=count;
while(count!=0)
{
result+=' ';
count--;
}
}
It will print out for you "is name my hello"
Taken from something called "Hacking a Google Interview" that was somewhere on my computer ... don't know from where I got it but I remember I saw this exact question inside ... here is the answer:
Reverse the string by swapping the
first character with the last
character, the second with the
second-to-last character, and so on.
Then, go through the string looking
for spaces, so that you find where
each of the words is. Reverse each of
the words you encounter by again
swapping the first character with the
last character, the second character
with the second-to-last character, and
so on.
This came up in LessThanDot Programmer Puzzles
#include<stdio.h>
void reverse_word(char *,int,int);
int main()
{
char s[80],temp;
int l,i,k;
int lower,upper;
printf("Enter the ssentence\n");
gets(s);
l=strlen(s);
printf("%d\n",l);
k=l;
for(i=0;i<l;i++)
{
if(k<=i)
{temp=s[i];
s[i]=s[l-1-i];
s[l-1-i]=temp;}
k--;
}
printf("%s\n",s);
lower=0;
upper=0;
for(i=0;;i++)
{
if(s[i]==' '||s[i]=='\0')
{upper=i-1;
reverse_word(s,lower,upper);
lower=i+1;
}
if(s[i]=='\0')
break;
}
printf("%s",s);
return 0;
}
void reverse_word(char *s,int lower,int upper)
{
char temp;
//int i;
while(upper>lower)
{
temp=s[lower];
s[lower]=s[upper];
s[upper]=temp;
upper=upper-1;
lower=lower+1;
}
}
The following code (C++) will convert a string this is a test to test a is this:
string reverseWords(string str)
{
string result = "";
vector<string> strs;
stringstream S(str);
string s;
while (S>>s)
strs.push_back(s);
reverse(strs.begin(), strs.end());
if (strs.size() > 0)
result = strs[0];
for(int i=1; i<strs.size(); i++)
result += " " + strs[i];
return result;
}
PS: it's actually a google code jam question, more info can be found here.