Lets take a word
qwerty
What I want is I need to insert periods (dots .) between the string. It can be any other character also.
For example,
q.werty
qw.erty
qwe.rty
qwer.ty
qwert.y
The above is for 1 period or dot. So 1 period combination for a 5 letter string will generate 5 outputs. (N-1)
Now for 2 periods (2 dots) (2 examples only):
q.w.erty
q.we.rty
q.wer.ty
q.wert.y
qw.e.rty
qw.er.ty
qw.ert.y
qwe.r.ty
qwe.rt.y
qwer.t.y
and so on..
NOTE: There must not be 2 consecutive dots between 2 letters in the string. Also, there must not be a period before starting character and/or after ending character.
Can anyone provide a Shell Script (sh, bash) for the above to list all the possible combinations and permutations. I have tried Googling and didn't find any worthwhile content to refer.
EDIT: Any help on how to start this on bash shell script would be great...
Your puzzle is fun so here's a code:
#!/bin/bash
t=qwerty
echo '---- one dot ----'
for (( i = 1; i < ${#t}; ++i )); do
echo "${t:0:i}.${t:i}"
done
echo '---- two dots ----'
for (( i = 1; i < (${#t} - 1); ++i )); do
for (( j = i + 1; j < ${#t}; ++j )); do
echo "${t:0:i}.${t:i:j - i}.${t:j}"
done
done
Output:
---- one dot ----
q.werty
qw.erty
qwe.rty
qwer.ty
qwert.y
---- two dots ----
q.w.erty
q.we.rty
q.wer.ty
q.wert.y
qw.e.rty
qw.er.ty
qw.ert.y
qwe.r.ty
qwe.rt.y
qwer.t.y
See the Bash Manual for everything.
I won't write the code, but I can guide you to the answer.
I assume you want to consider all possible number of dots, not just 1 or 2, but 3, 4, ... , up to the length of the string - 1.
For each character in the string up until the last, there are two possibilities: there is a dot or there is not a dot. So for an n character string, there are O(2^(n-1)) possibilities.
You could write a for loop that goes through all 2^(n-1) possibilities. Each one of these corresponds to a single output with dots after letters.
Let i be an iteration of the for loop. Then have an internal j loop that goes 1 to n-1. If the jth bit is 1, then put a dot after the jth letter.
Related
I have a list containing string patterns for digits 0-3. I am trying to print them onto the same line, so that print(digits1+col+digits[2]+col+digits[3]) prints '1 2 3' from the # pattern strings from the respective list index, but can only get the number patterns printed on their own.
# Create strings for each number 0-3 and store in digits list.
zero = '#'*3+'\n'+'#'+' '+'#'+'\n'+'#'+' '+'#'+'\n'+'#'+' '+'#'+'\n'+'#'*3
one = '#\n'.rjust(4)*6
two = '#'*3+'\n'+'#'.rjust(3)+'\n'+'#'*3+'\n'+'#'.ljust(3)+'\n'+'#'*3
three = '#'*3+'\n'+'#'.rjust(3)+'\n'+'#'*3+'\n'+'#'.rjust(3)+'\n'+'#'*3
digits = [zero, one, two, three]
col = '\n'.ljust(1)*6 # A divider column between each printed digit.
print(digits[1]+col+digits[2]+col+digits[3],end='')
The result of the above code.
One way to solve this is by reversing the digits matrix, right now each index in digits list has the complete digit values but if we keep horizontal values at each index it will print properly.
think it would be better represented in code...https://repl.it/#pavanskipo/DirectTriangularSlash
# Digits replaced horizntally
digits_rev = [digits[0].split("\n"),
digits[1].split("\n"),
digits[2].split("\n"),
digits[3].split("\n")]
for i in range(0, len(digits)+1):
print(digits_rev[0][i] + '\t' +
digits_rev[1][i] + '\t' +
digits_rev[2][i] + '\t' +
digits_rev[3][i])
click on the link and hit run, let me know if it works
I need to identify following patterns in string.
- "2N':'2N':'2N"
- "2N'-'2N'-'2N"
- "2N'/'2N'/'2N"
- "2N'/'2N'-'2N"
AND SO ON.....
basically i want this pattern if written in Simple language
2 NUMBERS [: / -] 2 NUMBERS [: / -] 2 NUMBERS
So is there anyway by which i could write one pattern which will cover all the possible scenarios ? or else i have to write total 9 patterns and had to match all 9 patterns to string.... and it is not the scenario in my code , i have to match 4, 2 number digits separated by [: / -] to string for which i have towrite total 27 patterns. So for understanding purpose i have taken 3 ,2 digit scenario...
Please help me...Thank you
Maybe you could try something like (Pick R83 style)
OK = X MATCH "2N1X2N1X2N" AND X[3,1]=X[6,1] AND INDEX(":/-",X[3,1],1) > 0
Where variable X is some input string like: 12-34-56
Should set variable OK to 1 if validation passes, else 0 for any invalid format.
This seems to get all your required validation into a single statement. I have assumed that the non-numeric characters have to be the same. If this is not true, the check could be changed to something like:
OK = X MATCH "2N1X2N1X2N" AND INDEX(":/-",X[3,1],1) > 0 AND INDEX(":/-",X[6,1],1) > 0
Ok, I guess the requirement of surrounding characters was not obvious to me. Still, it does not make it much harder. You just need to 'parse' the string looking for the first (I assume) such pattern (if any) in the input string. This can be done in a couple of lines of code. Here is a (rather untested ) R83 style test program:
PROMPT ":"
LOOP
LOOP
CRT 'Enter test string':
INPUT S
WHILE S # "" AND LEN(S) < 8 DO
CRT "Invalid input! Hit RETURN to exit, or enter a string with >= 8 chars!"
REPEAT
UNTIL S = "" DO
*
* Look for 1st occurrence of pattern in string..
CARDNUM = ""
FOR I = 1 TO LEN(S)-7 WHILE CARDNUM = ""
IF S[I,8] MATCH "2N1X2N1X2N" THEN
IF INDEX(":/-",S[I+2,1],1) > 0 AND INDEX(":/-",S[I+5,1],1) > 0 THEN
CARDNUM = S[I,8] ;* Found it!
END ELSE I = I + 8
END
NEXT I
*
CRT CARDNUM
REPEAT
There is only 7 or 8 lines here that actually look for the card number pattern in the source/test string.
Not quite perfect but how about 2N1X2N1X2N this gets you 2 number followed by 1 of any character followed by 2 numbers etc.
This might help:
BIG.STRING ="HELLO TILDE ~ CARD 12:34:56 IS IN THIS STRING"
TEMP.STRING = BIG.STRING
CONVERT "~:/-" TO "*~~~" IN TEMP.STRING
IF TEMP.STRING MATCHES '0X2N"~"2N"~"2N0X' THEN
FIRST.TILDE.POSN = INDEX(TEMP.STRING,"~",1)
CARD.STRING = BIG.STRING[FIRST.TILDE.POSN-2,8]
PRINT CARD.STRING
END
I have 2 question regarding searching for strings in MATLAB
If I have to find a string in a cell array of strings I can do the following to get the location of 'PO' in the cell array
find(strcmpi({'PO','FOO','PO1','FOO1','PO1','PO'},'PO'))
% 1 6
But, I really want to search for multiple strings ({'PO1', 'PO'}) at the same time (not using a for loop). What is the best way to do this?
Is there any function like histc() which can tell me how many times the string has occurred. Again for one string, I could do:
length(strfind({'PO','FOO','PO1','FOO1','PO1','PO'},'PO'))
But this obviously doesn't work for multiple strings at a time.
If you want to find multiple strings, then just use the second output of ismember instead to tell you which string it is. If you really need case-insensitive matching, I've added the upper call to force all inputs to be upper-case. You can omit this if you think it's already uppercase.
data = {'PO','FOO','PO1','FOO1','PO1','PO', 'PO'};
[tf, inds] = ismember(upper(data), {'PO1', 'PO'});
% 2 0 1 0 1 2 2
You can then use the second output to determine which string was found where:
% PO1 Occurrences
find(inds == 1)
% 3 5
% PO Occurrences
find(inds == 2)
% 1 6 7
If you want the equivalent of histc, you can use accumarray to do that. We can pass it all of the values of inds that are non-zero (i.e. the ones that you were actually searching for).
accumarray(inds(tf).', ones(sum(tf), 1))
% 2 3
If instead you want to get the histogram of all strings (not just the ones you're searching for) you could do the following:
[strings, ~, inds] = unique(data, 'stable');
occurrences = accumarray(inds, ones(size(inds)));
% 'PO' [3]
% 'FOO' [1]
% 'PO1' [2]
% 'FOO1' [1]
We are given a string which consists of digits 0-9. We have to count number of sub-strings divisible by a number k. One way is to generate all the sub-strings and check if it is divisible by k but this will take O(n^2) time. I want to solve this problem in O(n*k) time.
1 <= n <= 100000 and 2 <= k <= 1000.
I saw a similar question here. But k was fixed as 4 in that question. So, I used the property of divisibility by 4 to solve the problem.
Here is my solution to that problem:
int main()
{
string s;
vector<int> v[5];
int i;
int x;
long long int cnt = 0;
cin>>s;
x = 0;
for(i = 0; i < s.size(); i++) {
if((s[i]-'0') % 4 == 0) {
cnt++;
}
}
for(i = 1; i < s.size(); i++) {
int f = s[i-1]-'0';
int s1 = s[i] - '0';
if((10*f+s1)%4 == 0) {
cnt = cnt + (long long)(i);
}
}
cout<<cnt;
}
But I wanted a general algorithm for any value of k.
This is a really interesting problem. Rather than jumping into the final overall algorithm, I thought I'd start with a reasonable algorithm that doesn't quite cut it, then make a series of modifications to it to end up with the final, O(nk)-time algorithm.
This approach combines together a number of different techniques. The major technique is the idea of computing a rolling remainder over the digits. For example, let's suppose we want to find all prefixes of the string that are multiples of k. We could do this by listing off all the prefixes and checking whether each one is a multiple of k, but that would take time at least Θ(n2) since there are Θ(n2) different prefixes. However, we can do this in time Θ(n) by being a bit more clever. Suppose we know that we've read the first h characters of the string and we know the remainder of the number formed that way. We can use this to say something about the remainder of the first h+1 characters of the string as well, since by appending that digit we're taking the existing number, multiplying it by ten, and then adding in the next digit. This means that if we had a remainder of r, then our new remainder is (10r + d) mod k, where d is the digit that we uncovered.
Here's quick pseudocode to count up the number of prefixes of a string that are multiples of k. It runs in time Θ(n):
remainder = 0
numMultiples = 0
for i = 1 to n: // n is the length of the string
remainder = (10 * remainder + str[i]) % k
if remainder == 0
numMultiples++
return numMultiples
We're going to use this initial approach as a building block for the overall algorithm.
So right now we have an algorithm that can find the number of prefixes of our string that are multiples of k. How might we convert this into an algorithm that finds the number of substrings that are multiples of k? Let's start with an approach that doesn't quite work. What if we count all the prefixes of the original string that are multiples of k, then drop off the first character of the string and count the prefixes of what's left, then drop off the second character and count the prefixes of what's left, etc? This will eventually find every substring, since each substring of the original string is a prefix of some suffix of the string.
Here's some rough pseudocode:
numMultiples = 0
for i = 1 to n:
remainder = 0
for j = i to n:
remainder = (10 * remainder + str[j]) % k
if remainder == 0
numMultiples++
return numMultiples
For example, running this approach on the string 14917 looking for multiples of 7 will turn up these strings:
String 14917: Finds 14, 1491, 14917
String 4917: Finds 49,
String 917: Finds 91, 917
String 17: Finds nothing
String 7: Finds 7
The good news about this approach is that it will find all the substrings that work. The bad news is that it runs in time Θ(n2).
But let's take a look at the strings we're seeing in this example. Look, for example, at the substrings found by searching for prefixes of the entire string. We found three of them: 14, 1491, and 14917. Now, look at the "differences" between those strings:
The difference between 14 and 14917 is 917.
The difference between 14 and 1491 is 91
The difference between 1491 and 14917 is 7.
Notice that the difference of each of these strings is itself a substring of 14917 that's a multiple of 7, and indeed if you look at the other strings that we've matched later on in the run of the algorithm we'll find these other strings as well.
This isn't a coincidence. If you have two numbers with a common prefix that are multiples of the same number k, then the "difference" between them will also be a multiple of k. (It's a good exercise to check the math on this.)
So this suggests another route we can take. Suppose that we find all prefixes of the original string that are multiples of k. If we can find all of them, we can then figure out how many pairwise differences there are among those prefixes and potentially avoid rescanning things multiple times. This won't find everything, necessarily, but it will find all substrings that can be formed by computing the difference of two prefixes. Repeating this over all suffixes - and being careful not to double-count things - could really speed things up.
First, let's imagine that we find r different prefixes of the string that are multiples of k. How many total substrings did we just find if we include differences? Well, we've found k strings, plus one extra string for each (unordered) pair of elements, which works out to k + k(k-1)/2 = k(k+1)/2 total substrings discovered. We still need to make sure we don't double-count things, though.
To see whether we're double-counting something, we can use the following technique. As we compute the rolling remainders along the string, we'll store the remainders we find after each entry. If in the course of computing a rolling remainder we rediscover a remainder we've already computed at some point, we know that the work we're doing is redundant; some previous scan over the string will have already computed this remainder and anything we've discovered from this point forward will have already been found.
Putting these ideas together gives us this pseudocode:
numMultiples = 0
seenRemainders = array of n sets, all initially empty
for i = 1 to n:
remainder = 0
prefixesFound = 0
for j = i to n:
remainder = (10 * remainder + str[j]) % k
if seenRemainders[j] contains remainder:
break
add remainder to seenRemainders[j]
if remainder == 0
prefixesFound++
numMultiples += prefixesFound * (prefixesFound + 1) / 2
return numMultiples
So how efficient is this? At first glance, this looks like it runs in time O(n2) because of the outer loops, but that's not a tight bound. Notice that each element can only be passed over in the inner loop at most k times, since after that there aren't any remainders that are still free. Therefore, since each element is visited at most O(k) times and there are n total elements, the runtime is O(nk), which meets your runtime requirements.
I have been given a binary string of length n and i need to find the minimum numbers of operations to perform such that string does not contain more than k consecutive equal characters.
Only kind of operation I am allowed to perform is to flip any ith character of the string. flipping a character means changing a '1' to '0' or a '0' to '1'.
for example:
if n = 4 , k = 1 and string = 1001
then Answer:
string = 1010 and minimum operations = 2
I need to also find the new string.
can anyone tell me an efficient algorithm for solving problem considering n <=10^5
There's one way:
if k>1:
if k+1 matching characters are found:
if a[k+1]==a[k+2]:
flip a[k+1]
else if a[k+1]!=a[k+2]:
flip a[k]
for k=1 you can do it!
Here flipping means from 1 to 0 and vice-versa
For k=1 there are only two possible output strings - the one beginning with 0 and the one beginning with 1. You can check which of them is closer to the input string.
For larger k, you can just look at every sequence of k+1 identical characters, and fix it internally - without changing the characters at either end. For a sequence of k' > k you would need floor(k'/(k+1)) flips. It should not be hard to show that this is optimal.
Running time is linear and extra space is constant.
There are 2 cases:
1)For k>1
We have 2 possibilities.
a)one that is starting with 0:
eg:0101010101
b)one that is starrting with 1
eg:10101010.....
We should now calculate the distance(the number of different elements between the 2 strings)for each possiblity.Then the ans will be the one that has minimum changes.
2)for k>1
res2=0;res1=1;
c1=A[i];//it represents the last elemnet
i=1;
while(A[i]!='\0'){
if(A[i]==c1){
res1++;//the no of consecutive elements
if(res1>k){
if(A[i]==A[i+1])
flip(i);//it flips the ith element
else
flip(i-1);
res2++;//it counts the no of changes
res1=1;
}
}
else
res1=1;
c1=A[i];
i++;
}