I have two parameters, both strings, and I need to find the index of where the second string is found within the first string. I can do this when it's just one character but not when its more. Here is an example of the one I can do:
for i, j in enumerate(primary):
if final == j:
index = i
break
This is what I tried for the problem:
for i in range (0,len(initial),1):
for j in range (len(initial), i, -1):
if initial[i:j] == last:
index_str = i
break
However, it always returns "None" so can anyone help me with this?
I think that you need the index() method. Example usage (taken from this link)
str1 = "this is string example....wow!!!";
str2 = "exam";
print str1.index(str2);
The string.find method can do that.
"somestring".find("str")
http://docs.python.org/2/library/string.html#string.find
Related
I'm stuck on an exercise where I must obtain a string from a user and then print it in reverse with each letter on its own line. I've done this successfully with a for loop but I was wondering how to do so without it.
user_string = input("Please enter a string: ")
for letter in (user_string)[::-1]:
print(letter)
You can reverse and use str.join to rebuild the string.
print("\n".join(reversed(input("Please enter a string: "))))
If you know how many characters there are in the string or array (calculate using the length method len()) then you could do:
while i < arr_length:
with i incrementing at the end of every round.
The rest of the code would be the same but using i as an index.
One method would be to cast the string to a list and use the list.pop() method until there are no characters left in the list.
user_input = list(input())
while len(user_input) > 0:
print(user_input.pop())
list.pop() will remove the last item in the list and return it.
def reverse(s):
str = ""
for i in s:
str = i + str
return str
s = "Geeksforgeeks"
print("The original string is : ", end="")
print(s)
print("The reversed string(using loops) is : ", end="")
print(reverse(s))
Using index property we can easily reverse a given string
a = input()
print(a[::-1])
Given two string str1 and str2 which contain only 0 or 1, there
are some steps to change str1 to str2,
step1: find a substring of str1 of length 2 and reverse the substring, and str1 becomes str1' (str1' != str1)
step2: find a substring of str1' of length 3, and reverse the substring, and str1' becomes str1'' (str1'' != str1')
the following steps are similar.
the string length is in the range [2, 30]
Requirement: each step must be performed once and we can not skip
previous steps and perform the next step.
If it is possible to change str1 to str2, output the minimum steps required, otherwise, output -1
Example 1
str1 = "1010", str2 = "0011", the minimum step required is 2
first, choose substring in range [2, 3], "1010" --> "1001",
then choose substring in the range [0, 2], "1001" --> "0011"
Example 2
str1 = "1001", str2 = "0110", it is impossible to change str1 to str2,
because in step1, str1 can be changed to "0101" or "1010", but in step3, it is impossible to change a length3 substring to make it different. So the output is -1.
Example 3
str1 = "10101010", str2 = "00101011"， output is 7
I can not figure out example 3, because there are two many possibilities. Can anyone gives some hint on how to solve this problem? What is the type of this
problem? Is it dynamic programming?
This is in fact a dynamic programming problem. To solve it, we are going to try all possible permutations, but memoize the results along the way. It could seem that there are way too many options - there are 2^30 different binary strings of length 30, but keep in mind that reverting a string doesn't change number of zeroes and ones we have, so the upper bound is in fact 30 choose 15 = 155117520 when we have a string of 15 zeroes and ones. Around 150 million possible results is not too bad.
So starting with our start string, we are going to derive all possible string from each string we derived so far, until we generate end string. We are also going to track predecessors to reconstruct generation. Here's my code:
start = '10101010'
end = '00101011'
dp = [{} for _ in range(31)]
dp[1][start] = '' # Originally only start string is reachable
for i in range(2, len(start) + 1):
for s in dp[i - 1].keys():
# Try all possible reversals for each string in dp[i - 1]
for j in range(len(start) - i + 1):
newstr = s
newstr = newstr[:j] + newstr[j:j+i][::-1] + newstr[j+i:]
dp[i][newstr] = s
if end in dp[i]:
ans = []
cur = end
for j in range(i, 0, -1):
ans.append(cur)
cur = dp[j][cur]
print(ans[::-1])
exit(0)
print('Impossible!')
And for your third example, this gives us sequence ['10101010', '10101001', '10101100', '10100011', '00101011'] - from your str1 to str2. If you check differences between the strings, you'll see which transitions were made. So this transformation can be done in 4 steps rather than 7 like you suggested.
Lastly, this will be a bit slow for 30 in python, but if you rewrite it into C++, it's going to be a couple of seconds tops.
This Question can be solved using Backtracking. here is my C++ Code, Which runs smooth with my testcases. This Question Came in an OA of Persistent systems and i was a bit confused about the steps, but this is simple Backtracking. Wants your suggestions if Dp can Optimize my solution!.
//prabaljainn
#include <bits/stdc++.h>
using namespace std;
string s1,s2;
int ans=1e9; int n;
void rec(string s1,int level){
if(s1==s2){
ans = min(ans,level-2);
return;
}
for(int i=0; i<= n-level; i++){
reverse(s1.begin()+i, s1.begin()+i+level);
rec(s1,level+1);
reverse(s1.begin()+i, s1.begin()+i+level);
}
}
int main(){
cin>>s1>>s2;
n = s1.size();
rec(s1,2);
if(ans==1e9)
cout<<"-1"<<endl;
else
cout<<ans<<endl;
}
Happy coding
This problem can be solved using breadth-first search. The following solution uses a queue which stores a pair having the current string as the first member and current operation length(initially 2) as the second member. A set is used to store already visited strings to prevent entering redundant states. For current string, we reverse every substring of length k where k is current operation length and add it to the queue if it hasn't been seen before. If the current string equals the desired string then answer is 'current operation length-2'. If queue becomes empty, then the answer isn't possible.
string str1,str2;
cin>>str1>>str2;
queue<pair<string, int>> q;
set<string> s;
q.push({str1,2});
s.insert(str1);
while(!q.empty())
{
auto p=q.front();
q.pop();
if(p.first==str2)
{
cout<<p.second-2;
return 0;
}
if(p.second<=p.first.size())
{
for(int i=0;i<=p.first.size()-p.second;i++)
{
string x=p.first;
reverse(x.begin()+i,x.begin()+i+p.second);
if(s.find(x)==s.end())
{
q.push({x,p.second+1});
s.insert(x);
}
}
}
}
cout<<-1;
save str1 as start of BFS and at each step,reverse values of all substrings of length 2 and 3 and see if the new strings formed after reversing have been seen previously or not.....if not seen....push them in the queue and also maintain count of steps...if the string at the front of queue is str2 at any time...that step is the answer
My question is more or less similar to:
Is there a way to substring a string in Python?
but it's more specifically oriented.
How can I get a par of a string which is located between two known words in the initial string.
Example:
mySrting = "this is the initial string"
Substring = "initial"
knowing that "the" and "string" are the two known words in the string that can be used to get the substring.
Thank you!
You can start with simple string manipulation here. str.index is your best friend there, as it will tell you the position of a substring within a string; and you can also start searching somewhere later in the string:
>>> myString = "this is the initial string"
>>> myString.index('the')
8
>>> myString.index('string', 8)
20
Looking at the slice [8:20], we already get close to what we want:
>>> myString[8:20]
'the initial '
Of course, since we found the beginning position of 'the', we need to account for its length. And finally, we might want to strip whitespace:
>>> myString[8 + 3:20]
' initial '
>>> myString[8 + 3:20].strip()
'initial'
Combined, you would do this:
startIndex = myString.index('the')
substring = myString[startIndex + 3 : myString.index('string', startIndex)].strip()
If you want to look for matches multiple times, then you just need to repeat doing this while looking only at the rest of the string. Since str.index will only ever find the first match, you can use this to scan the string very efficiently:
searchString = 'this is the initial string but I added the relevant string pair a few more times into the search string.'
startWord = 'the'
endWord = 'string'
results = []
index = 0
while True:
try:
startIndex = searchString.index(startWord, index)
endIndex = searchString.index(endWord, startIndex)
results.append(searchString[startIndex + len(startWord):endIndex].strip())
# move the index to the end
index = endIndex + len(endWord)
except ValueError:
# str.index raises a ValueError if there is no match; in that
# case we know that we’re done looking at the string, so we can
# break out of the loop
break
print(results)
# ['initial', 'relevant', 'search']
You can also try something like this:
mystring = "this is the initial string"
mystring = mystring.strip().split(" ")
for i in range(1,len(mystring)-1):
if(mystring[i-1] == "the" and mystring[i+1] == "string"):
print(mystring[i])
I suggest using a combination of list, split and join methods.
This should help if you are looking for more than 1 word in the substring.
Turn the string into array:
words = list(string.split())
Get the index of your opening and closing markers then return the substring:
open = words.index('the')
close = words.index('string')
substring = ''.join(words[open+1:close])
You may want to improve a bit with the checking for the validity before proceeding.
If your problem gets more complex, i.e multiple occurrences of the pair values, I suggest using regular expression.
import re
substring = ''.join(re.findall(r'the (.+?) string', string))
The re should store substrings separately if you view them in list.
I am using the spaces between the description to rule out the spaces between words, you can modify to your needs as well.
I have two string arrays and I want to find where each string from the first array is in the second array, so i tried this:
for i = 1:length(array1);
cmp(i) = strfind(array2,array1(i,:));
end
This doesn't seem to work and I get an error: "must be one row".
Just for the sake of completeness, an array of strings is nothing but a char matrix. This can be quite restrictive because all of your strings must have the same number of elements. And that's what #neerad29 solution is all about.
However, instead of an array of strings you might want to consider a cell array of strings, in which every string can be arbitrarily long. I will report the very same #neerad29 solution, but with cell arrays. The code will also look a little bit smarter:
a = {'abcd'; 'efgh'; 'ijkl'};
b = {'efgh'; 'abcd'; 'ijkl'};
pos=[];
for i=1:size(a,1)
AreStringFound=cellfun(#(x) strcmp(x,a(i,:)),b);
pos=[pos find(AreStringFound)];
end
But some additional words might be needed:
pos will contain the indices, 2 1 3 in our case, just like #neerad29 's solution
cellfun() is a function which applies a given function, the strcmp() in our case, to every cell of a given cell array. x will be the generic cell from array b which will be compared with a(i,:)
the cellfun() returns a boolean array (AreStringFound) with true in position j if a(i,:) is found in the j-th cell of b and the find() will indeed return the value of j, our proper index. This code is more robust and works also if a given string is found in more than one position in b.
strfind won't work, because it is used to find a string within another string, not within an array of strings. So, how about this:
a = ['abcd'; 'efgh'; 'ijkl'];
b = ['efgh'; 'abcd'; 'ijkl'];
cmp = zeros(1, size(a, 1));
for i = 1:size(a, 1)
for j = 1:size(b, 1)
if strcmp(a(i, :), b(j, :))
cmp(i) = j;
break;
end
end
end
cmp =
2 1 3
I understand that str = str.replace('x', '') will eliminate all the x's.
But let's say I have a string jxjrxxtzxz and I only want to delete the first and last x making the string jjrxxtzz. This is not string specific. I want to be able to handle all strings, and not just that specific example.
edit: assume that x is the only letter I want to remove. Thank you!
One fairly straight forward way is to just use find and rfind to locate the characters to remove;
s = 'jxjrxxtzxz'
# Remove first occurrence of 'x'
ix = s.find('x')
if ix > -1:
s = s[:ix]+s[ix+1:]
# Remove last occurrence of 'x'
ix = s.rfind('x')
if ix > -1:
s = s[:ix]+s[ix+1:]
Not pretty but this will work:
def remove_first_last(c, s):
return s.replace(c,'', 1)[::-1].replace(c,'',1)[::-1]
Usage:
In [1]: remove_first_last('x', 'jxjrxxtzxz')
Out[1]: 'jjrxxtzz'