String Manipulation Recursive Function

String Manipulation Recursive Function

By : Robert Orzechowski
Date : October 25 2020, 09:10 AM
help you fix your problem You need to transform until input == result. When input == result that means it can no longer transformed. See below for changes.
code :
def transformations(J):
    if len(J) <= 1: # I made it less than or equal
        return J
    front_part = ""
    back_part = ""
    if J[0] == J[1]:
        back_part = J[2:]
        return transformations(back_part)
        front_part = J[0]
        back_part = J[1:]
        # See below
        result =  front_part + transformations(back_part)
        # If it's same we have done all transformations.
        if result == J:
            return result
        else: # try to perform more transformations
            return transformations(result)

tests = [
    ["abba", ""],
    ["ab", "ab"],
    ["aba", "aba"],
    ["aabbbccaaba", "a"]

for inp, expected in tests:
    actual = transformations(inp)
    print("trans(%s) == %s" % (inp, actual), "Test Passed =", actual == expected)
trans(abba) ==  Test Passed = True
trans(ab) == ab Test Passed = True
trans(aba) == aba Test Passed = True
trans(aabbbccaaba) == a Test Passed = True

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SQL Function for String manipulation

SQL Function for String manipulation

By : mbcraft
Date : March 29 2020, 07:55 AM
wish of those help Personally I like T-SQL's STUFF() function for this. One of the main reasons is that I don't have to care about how long the string can be, swap out 8000 for 4000 if it is NVARCHAR, etc.
code :
SELECT STUFF(@source, 1, 1, '');
SELECT SUBSTRING(@source, PATINDEX('%[^,]%', @source), 8000); 
SELECT STUFF(@source, 1, PATINDEX('%[^,]%', @source)-1, '');
char * array manipulation on recursive function

char * array manipulation on recursive function

By : Khine Lae
Date : March 29 2020, 07:55 AM
hop of those help? I have a program that calculates the edit distance of two strings. it also outputs all the edit operations to obtain the complete transformation. i wrote a recursive function that explores the matrix filled by the edit distance calculation function and reconstructs the path , You can add few line in match_out
code :
void match_out(char *s1, char *s2, char **edit ,int i, int j)
    if (s1[i] == s2[j]) 
        printf("M no edit needed \n" );
        printf("S subst %c with %c ",s1[i] , s2[j]);
        //**from here**
        printf("%s -> ",s1);
        //upto here
char str[]= {'p','a','r','e','n','t'};
char * str = "parent";
Call by reference and ostream and string manipulation recursive

Call by reference and ostream and string manipulation recursive

By : DrunkenPhD
Date : March 29 2020, 07:55 AM
it fixes the issue It's recursive because textBox calls itself. Your program will run if you:
A) Remove both calls to textBox within textBox
code :
void textBox(ostream & out, string text)
    int const textSize = text.size() + 1;

    out << '+' << setfill('-') << setw(textSize) << right << '+' << endl;
    out << '|' << text << '|' << endl;
    out << '+' << setfill('-') << setw(textSize) << right << '+' << endl;

    out << setfill(' ') ;
void textBoxWithInfo(ostream & out, string text)
    textBox(cout, "Message to screen");

    { // Braces to deallocate + close myFile faster
        ofstream myFile("myFile.txt");
        textBox(myFile, "Message to file");

    textBox(out, text);
segmentaion fault on large recursive string manipulation

segmentaion fault on large recursive string manipulation

By : Sandra Scheurwater
Date : March 29 2020, 07:55 AM
may help you . You have stack overflow because of too deep recursion (several thousands calls deep). You could easily implement the algorithm using a loop instead.
Optimize recursive string manipulation function with JavaScript

Optimize recursive string manipulation function with JavaScript

By : Phước
Date : March 29 2020, 07:55 AM
it helps some times The major problem with the naive recursive algorithm that you have presented is that it is called very often on the same input s - exponentially often even, and exactly that is what causes the noticeable slowdown on larger strings.
What you can do against this is to use memoisation - remember the result for a specific input in a lookup table.
code :
function maxSubstring(s, t, prevResults = new Map()) {
    function result(x) { prevResults.set(s, x); return x; }
    if (prevResults.has(s))
        return prevResults.get(s); // memoisation

    const first = s.indexOf(t);
    if (first == -1)
        return result(0);
    const withoutFirst = s.slice(0, first) + s.slice(first + t.length);

    const last = s.lastIndexOf(t);
    if (last == first) // only one match
        return result(1 + maxSubstring(withoutFirst, t, prevResults));

    if (t.lastIndexOf(t.charAt(t.length-1), t.length-1) == -1 // last character of t is found nowhere else in t
        || !t.includes(s.charAt(first+t.length))) // character after the match can never be part of a match
        // so this match is always removed in the optimal sequence and it doesn't matter whether as first or last
        return result(1 + maxSubstring(withoutFirst, t, prevResults));

    const withoutLast = s.slice(0, last) + s.slice(last + t.length);
    if (t.indexOf(t.charAt(0), 1) == -1 // first character of t is found nowhere else in t
        || !t.includes(s.charAt(last - 1))) // character before the match can never be part of a match
        // so this match is always removed and it doesn't matter when
        return result(1 + maxSubstring(withoutLast, t, prevResults));

    return result(1 + Math.max(maxSubstring(withoutFirst, t, prevResults),
                               maxSubstring(withoutLast, t, prevResults)));
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