There is a strings as below:
2016,07,20,19,20,25
How can I transfer this strings to such as this format string:
2016-07-20 19:20:25
Thank you very much!
A solution with array slice could be this
let parts = [];
let date = "2016,07,20,19,20,25";
let formatted = ((parts = date.split(",")).slice(0,3)).join("-") + ' ' + parts.slice(3).join(":")
You could also do it with String#replace and a function as the 2 argument;
let date = "2016,07,20,19,20,25";
date.replace(/,/g, (() => {
let count = 0;
return (match, position) => {
count += 1;
if(count == 3) return ' ';
else if(count < 3) return '-';
else return ':';
});
})())
Note: Both approaches assume that the format will always be the one provided
6 numbers seperated by commas
Related
I have a question about the following problem on Leetcode:
Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.
Example 1:
Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".
Example 2:
Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".
Example 3:
Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".
Example 4:
Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".
Note:
1 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and '#' characters.
Follow up:
Can you solve it in O(N) time and O(1) space?
My answer:
def backspace_compare(s, t)
if (s.match?(/[^#[a-z]]/) || t.match?(/[^#[a-z]]/)) || (s.length > 200 || t.length > 200)
return "fail"
else
rubular = /^[\#]+|([^\#](\g<1>)*[\#]+)/
if s.match?(/#/) && t.match?(/#/)
s.gsub(rubular, '') == t.gsub(rubular, '')
else
new_s = s.match?(/#/) ? s.gsub(rubular, '') : s
new_t = t.match?(/#/) ? t.gsub(rubular, '') : t
new_s == new_t
end
end
end
It works in the terminal and passes the given examples, but when I submit it on leetcode it tells me Time Limit Exceeded. I tried shortening it to:
rubular = /^[\#]+|([^\#](\g<1>)*[\#]+)/
new_s = s.match?(/#/) ? s.gsub(rubular, '') : s
new_t = t.match?(/#/) ? t.gsub(rubular, '') : t
new_s == new_t
But also the same error.
So far, I believe my code fulfills the O(n) time, because there are only two ternary operators, which overall is O(n). I'm making 3 assignments and one comparison, so I believe that fulfills the O(1) space complexity.
I have no clue how to proceed beyond this, been working on it for a good 2 hours..
Please point out if there are any mistakes in my code, and how I am able to fix it.
Thank you! :)
Keep in mind that with N <= 200, your problem is more likely to be linear coefficient, not algorithm complexity. O(N) space is immaterial for this; with only 400 chars total, space is not an issue. You have six regex matches, two of which are redundant. More important, regex is slow processing for such a specific application.
For speed, drop the regex stuff and do this one of the straightforward, brute-force ways: run through each string in order, applying the backspaces as appropriate. For instance, change both the backspace and the preceding letter to spaces. At the end of your checking, remove all the spaces in making a new string. Do this with both S and T; compare those for equality.
It may be easiest to start at the end of the string and work towards the beginning:
def process(str)
n = 0
str.reverse.each_char.with_object('') do |c,s|
if c == '#'
n += 1
else
n.zero? ? (s << c) : n -= 1
end
end.reverse
end
%w|ab#c ad#c ab## c#d# a##c #a#c a#c b|.each_slice(2) do |s1, s2|
puts "\"%s\" -> \"%s\", \"%s\" -> \"%s\" %s" %
[s1, process(s1), s2, process(s2), (process(s1) == process(s2)).to_s]
end
"ab#c" -> "ac", "ad#c" -> "ac" true
"ab##" -> "", "c#d#" -> "" true
"a##c" -> "c", "#a#c" -> "c" true
"a#c" -> "c", "b" -> "b" false
Let's look at a longer string.
require 'time'
alpha = ('a'..'z').to_a
#=> ["a", "b", "c",..., "z"]
s = (10**6).times.with_object('') { |_,s|
s << (rand < 0.4 ? '#' : alpha.sample) }
#=> "h####fn#fjn#hw###axm...#zv#f#bhqsgoem#glljo"
s.size
#=> 1000000
s.count('#')
#=> 398351
and see how long it takes to process.
require 'time'
start_time = Time.now
(u = process(s)).size
#=> 203301
puts (Time.now - start_time).round(2)
#=> 0.28 (seconds)
u #=> "ffewuawhfa...qsgoeglljo"
As u will be missing the 398351 pound signs in s, plus an almost equal number of other characters removed by the pound signs, we would expect u.size to be about:
10**6 - 2 * s.count('#')
#=> 203298
In fact, u.size #=> 203301, meaning that, at the end, 203301 - 203298 #=> 3 pound signs were unable to remove a character from s.
In fact, process can be simplified. I leave that as an exercise for the reader.
class Solution {
public boolean backspaceCompare(String s, String t) {
try {
Stack<Character> st1 = new Stack<>();
Stack<Character> st2 = new Stack<>();
st1 = convertToStack(s);
st2 = convertToStack(t);
if (st1.size() != st2.size()) {
return false;
} else {
int length = st1.size();
for (int i = 0; i < length; i++) {
if (st1.peek() != st2.peek())
return false;
else {
st1.pop();
st2.pop();
}
if (st1.isEmpty() && st2.isEmpty())
return true;
}
}
} catch (Exception e) {
System.out.print(e);
}
return true;
}
public Stack<Character> convertToStack(String s){
Stack<Character> st1 = new Stack<>();
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != '#') {
st1.push(s.charAt(i));
} else if (st1.empty()) {
continue;
} else {
st1.pop();
}
}
return st1;
}
}
I try to come up with regular expression patter that fulfills such requirements.
it is US phone number format wit 3 groups
I have input strings like this
(999) 98__-9999 here there is extra _ at the end of second section which I want to delete
(999) 9_8_-9999 here there is extra _ at the end of second section I want to delete
(999) 9_-9999 here if second group length is < 3 and ends with _ there should be added _ to pad second group to 9__ (3 characters)
(999) 98-9999 here if second group length is equal to 3 or it ends with digit there shouldn't be any modifications
To sum up:
If secondGroup.length > 3 && secondGroup.lastCharacter == '_' I want to remove this last character
else if secondGroup.length < 3 && secondGroup.lastCharacter == '_' I wan to append "_" (or pad wit underscore to have 3 characters in total)
else leave second group as in the input string.
The same should be applied to first group. The difference are the different delimiters i.e. (xxx) in first group and \sxxx- in second group
Here is my Swift code I have used to achieve it in brute force way by manually manipulating the string: (length 4 instead of 3 takes into account first delimiter like ( or \s. )
var componentText = ""
let idx1 = newText.index(of: "(")
let idx2 = newText.index(of: ")")
if let idx1 = idx1, let idx2 = idx2 {
var component0 = newText[..<idx1]
var component1 = newText[idx1..<idx2]
if component1.count > 4 && component1.last == "_" {
component1.popLast()
} else if component1.count < 4 && component1.last == "_" {
component1.append("_")
}
componentText += "\(component0)\(component1))"
} else {
componentText = newText
}
let idx3 = newText.index(of: " ")
let idx4 = newText.index(of: "-")
if let idx2 = idx2, let idx3 = idx3, let idx4 = idx4 {
var component2 = newText[idx2..<idx3]
component2.popFirst()
var component3 = newText[idx3..<idx4]
var component4 = newText[idx4...]
if component3.count > 4 && component3.last == "_" {
component3.popLast()
} else if component3.count < 4 && component3.last == "_" {
component3.append("_")
}
componentText += "\(component2) \(component3)-\(component4)"
} else {
componentText = newText
}
newText = componentText != "" ? componentText : newText
I think that using regular expression this code could be more flexible and much shorter.
Using JavaScript & regex I want to split a string on every %20 that is not within quotes, example:
Here%20is%20"a%20statement%20"%20for%20Testing%20"%20The%20Values%20"
//easy to read version: Here is "a statement " for Testing " The Values "
______________ ______________
would return
{"Here","is","a statement ","for","Testing"," The Values "}
but it seems my regex are no longer strong enough to build the expression. Thanks for any help!
A way using the replace method, but without using the replacement result. The idea is to use a closure to fill the result variable at each occurence:
var txt = 'Here%20is%20"a%20statement%20"%20for%20Testing%20"%20The%20Values%20"';
var result = Array();
txt.replace(/%20/g, ' ').replace(/"([^"]+)"|\S+/g, function (m,g1) {
result.push( (g1==undefined)? m : g1); });
console.log(result);
Just try with:
var input = 'Here%20is%20"a%20statement%20"%20for%20Testing%20"%20The%20Values%20"',
tmp = input.replace(/%20/g, ' ').split('"'),
output = []
;
for (var i = 0; i < tmp.length; i++) {
var part = tmp[i].trim();
if (!part) continue;
if (i % 2 == 0) {
output = output.concat(part.split(' '));
} else {
output.push(part);
}
}
Output:
["Here", "is", "a statement", "for", "Testing", "The Values"]
I have a django project and I have a method that return the salary for the employee, the salary is being returned as a number like 5432 and i want to display it as five thousand four hundred and thirty two. I tried to use humanize but it works for only small digits any clue??
regards,
Here's some Javascript that you can use, courtesy of About.com's Banking page:
<script type="text/javascript">
// Convert numbers to words
// copyright 25th July 2006, by Stephen Chapman http://javascript.about.com
// permission to use this Javascript on your web page is granted
// provided that all of the code (including this copyright notice) is
// used exactly as shown (you can change the numbering system if you wish)
// American Numbering System
var th = ['','thousand','million', 'billion','trillion'];
// uncomment this line for English Number System
// var th = ['','thousand','million', 'milliard','billion'];
var dg = ['zero','one','two','three','four', 'five','six','seven','eight','nine']; var tn = ['ten','eleven','twelve','thirteen', 'fourteen','fifteen','sixteen', 'seventeen','eighteen','nineteen']; var tw = ['twenty','thirty','forty','fifty', 'sixty','seventy','eighty','ninety']; function toWords(s){s = s.toString(); s = s.replace(/[\, ]/g,''); if (s != String(parseFloat(s))) return 'not a number'; var x = s.indexOf('.'); if (x == -1) x = s.length; if (x > 15) return 'too big'; var n = s.split(''); var str = ''; var sk = 0; for (var i=0; i < x; i++) {if ((x-i)%3==2) {if (n[i] == '1') {str += tn[Number(n[i+1])] + ' '; i++; sk=1;} else if (n[i]!=0) {str += tw[n[i]-2] + ' ';sk=1;}} else if (n[i]!=0) {str += dg[n[i]] +' '; if ((x-i)%3==0) str += 'hundred ';sk=1;} if ((x-i)%3==1) {if (sk) str += th[(x-i-1)/3] + ' ';sk=0;}} if (x != s.length) {var y = s.length; str += 'point '; for (var i=x+1; i<y; i++) str += dg[n[i]] +' ';} return str.replace(/\s+/g,' ');}
I wondering how should be the regex string for the string containig '#'
e.g.
abc#def#ghj#ijk
I wanna get
#def
#ghj
#ijk
I tried #[\S]+ but it selects the whole #def#ghj#ijk Any ideas ?
Edit
The code below selects only #Me instead of #MessageBox. Why ?
var m = new RegExp('#[^\s#]+').exec('http://localhost/Lorem/10#MessageBox');
if (m != null) {
var s = '';
for (i = 0; i < m.length; i++) {
s = s + m[i] + "\n";
}
}
Edit 2
the double backslash solved that problem. '#[^\\s#]+'
Try #[^\s#]+ to match # followed by a sequence of one or mor characters which are neither # nor whitespace.
Match all characters that are not #:
#[^#]+