What you described is known as Levenshtein distance. There is library called Apache commons-text that you can use to do it for you. int ops = new LevenshteinDistance().apply(string1, string2) Here is source code 3 solved Calculate number of Updation, Insertion and Deletion on a One String to reach Another String [closed]
Modify the body of the main method like this: public static void main(String[] arg) { Address address = new Address(“The Game”, “Seventh street”, “Generic Town”, “State”, “1234”); System.out.println(address.toString()); } 1 solved Pls help me, how can I make the code printout the full details of the address as an output instead of empty output [closed]
If different identifiers can have same values, your third map will keep the last parsed one. E.g. : File 1 : localId1 => “aaaa” localId2 => “bbbb” localId3 => “cccc” localId4 => “aaaa” File 2: localId1 => “1111” localId2 => “2222” localId3 => “3333” localId4 => “4444” Your first and second maps will store this … Read more
Maybe: >>> myList = [“Bob1”, “Bob2”, “Bob3”, “Bob4”] >>> range(1, len(myList) + 1) [1, 2, 3, 4] 1 solved Converting string list into integer list beginning from 1
May be something like this will do Steps:- 1.Split the string by space delimiter. 2.Map through the Seq of strings created in the first step and split by . delimiter. val s = “abc.def ghi.jkl mno.pqr” val strS = s.split(” “).toSeq val firstStrS = strS.map { s=> val t = s.split(‘.’).toSeq t.head }.mkString(” “) Hope … Read more
print(eval(str(“value” + str(x)))) But must not use eval even if you know it’s completely secure. There is always a better way. 6 solved how to transform a string in a variable name ? python [duplicate]
str(tuple(str(fv[“v”]).split(‘,’))) will probably give you what you want but if you give more information there is definitely a better way to write this. What does fv[‘v’] consist of? 2 solved TypeError: cannot concatenate ‘str’ and ‘list’ objects. how to convert list to string [closed]
There are many uses of ! operator in Linux. See this article. In your case, it is probably to check if the a command from history is given as input. See the first point of the article. solved ‘!’ in input buffer in comparison in if statement
I think this is more of a subjective question. I would argue that its ok following the principle of duck typing: If it looks like a duck, swims like a duck, and quacks like a duck, then it probably is a duck. If you are looping through an array of strings and they all have … Read more
It strips the first two ‘[‘, it seems you have one long string, you have to split it first. datalist = data.split[‘,’] for x in datalist: # code here If you don’t want to split it and have it all in one string you need replace not strip (strip only works at the end and … Read more
I would probably consider doing this another way… Have an empty hashmap<String sub, int multiplier> Read in the name Generate the subname Fetch from or create subname in hashmap If new, set multiplier to 1 If exists, increment multiplier Return String.format(“%s%3d”, subname, multiplier x 5).toUpperCase() I would give you code but writing the above was … Read more
Different methods can be used for removing leading and trailing spaces, for converting multiple spaces to one and to remove spaces before exclamation mark, comma etc: mystr = ” Hello . To , the world ! ” print(mystr) mystr = mystr.strip() # remove leading and trailing spaces import re # regex module mystr = re.sub(r’\s+’,” … Read more
The difference is: names1 is declared simply as an array of string pointers without other defined characteristics. Using string literals here will put the string literals into a section in your executable file which is read-only, because the compiler can this way re-use them. For example, when you use char* a = “abc”; char* b … Read more
int n; cin >> n; cin.ignore(); // are you sure you tried this? for(int i = 0; i < n; i++) { string input; getline(cin, input); cout << “Line Entered: ” << input << endl; } 0 solved How to force getline() to input one line at a time [duplicate]
Hope the following function help you to replace first occurance void str_find_replace(char *str, char *find, char *replace) { size_t find_len,replace_len; char *str_end; find_len = strlen(find); replace_len = strlen(replace); str_end = strlen(str); while(*str){ if(!strncmp(str,find,find_len) { memmove(str+replace_len,str+find_len, str_end – (str + find_len) + 1); memcpy(str,replace,replace_len); return; } str++; } } If you want to replace all the … Read more