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President Donald Trump will hold court in court tonight as he delivers the speech of the Oval Office of

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c ++ – Largest chain lexicographically among authorized exchanges

Problem: Given a string str and a pair array that indicates which string
the indexes of the chain can be inverted, returns the lexicographer
the largest chain resulting from authorized exchanges. You can exchange
indices any number of times.


For str = "abdc" and peers = [[1, 4], [3, 4]], the output should be
swapLexOrder (str, peers) = "dbca".

By exchanging the given indices, you get the strings: "cbda", "cbad",
"dbac", "dbca". The lexicographically largest chain of this list is

My comments: My program has passed all the tests of the CodeSignal website, but since I am relatively new to algorithms, there is probably a more efficient way to solve it. In addition, I am looking for a general revision of the code: does my code indicate that I look like a beginner? Thank you.


std :: string swapLexOrder (std :: str string, std :: vector <std :: vector> pairs
if (pairs.size () == 0) returns str;
std :: vector <std :: set> pairpool; // pairpool: contains interchangeable index sets
std :: vector <std :: vector> stringpool; // stringpool: contains interchangeable character vectors

// create a pair structure
for (std :: size_t i = 0; i <couples.size (); i ++)
bool alrExists = false;
std :: set newset;
for (auto & p: pairpool)
for (auto ele: p)
if ((pairs[i][0]    == ele) || (pairs[i][1]    == ele))
if (! alrExists)
alrExists = true;
p.insert (pairs[i][0])
p.insert (pairs[i][1])
newSet = p;
if (p == newSet) pause;
p.insert (newSet.begin (), newSet.end ());
pairpool.erase (std :: remove (pairpool.begin (), pairpool.end (), newSet), pairpool.end ());
Pause; // really needed this breakout statement
if (! alrExists)
newSet.insert (pairs[i][0])
newSet.insert (pairs[i][1])
pairpool.push_back (newSet);

// create a sorted string pool structure
for (auto p: pairpool)
std :: vector newset;
for (auto ele: p)
newset.push_back (str.substr (ele - 1, 1));
std :: sort (newset.begin (), newset.end ());
stringpool.push_back (newset);

// use stringpool and pairpool to change the string only once
int counter = 0;
for (auto p: pairpool)
for (auto ele: p)
str.replace (ele - 1, 1, stringpool[counter].return());
stringpool[counter].pop_back ();
counter ++;

back str;

int main ()
std :: cost << swapLexOrder ("acxrabdz", {{1, 3}, {6, 8}, {3, 8}, {2, 7}}) << " n";
std :: string STR = "lvvyfrbhgiyexoirhunnuejzhesylojwbyatfkrv";
std :: vector <std :: vector> PAIR =
{13, 23},
{13, 28},
{15, 20},
{24, 29},
{6, 7},
{3, 4},
{21, 30},
{2, 13},
{12, 15},
{19, 23},
{10, 19},
{13, 14},
{6, 16},
{17, 25},
{6, 21},
{17, 26},
{5, 6},
{12, 24}
std :: cost << swapLexOrder (STR, PAIR) << " n";
returns 0;

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Among the javascript code snippet below, which one is considered the most appropriate?

Scenario 1:

elements.personWrapper.getElement (& # 39; span[name=personClass]& # 39;) .set (& # 39; innerHTML & # 39;; & # 39;);
} catch (TypeError) {
elements.personWrapper.getElement (& # 39; input[name=personClass]& # 39;) .set (& # 39; innerHTML & # 39;; & # 39;);

Scenario 2:

                                                                                var span_element = elements.personWrapper.getElement (& # 39; span[name=personClass]& # 39;);
var input_element = elements.personWrapper.getElement (& # 39; input[name=personClass]& # 39;);
if (span_element) {
span_element.set (& # 39; innerHTML & # 39;; & # 39;);
else if (input_element) {
input_element.set ('value', & # 39;); & # 39;);

The above code fragment serves the same purpose (please disregard legibility and naming conventions, they have been changed). So, I would like to know which one is the best. Anyone please provide your valuable suggestions.

Please note that one of the two is defined while the page is loading. The span and input elements are dynamically created. The goal is therefore to define innerHTML if it is an extent or value if it is an input type element.

I want to know which code snippet is the most efficient in terms of code execution and less prone to errors, thanks!

5 awards are spread among 20 students What is the probability that a particular student will receive 3 awards?

There are 5 prizes to be distributed to 20 students. What is the probability that a particular student will receive 3 awards?

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Just remember that online research is no longer limited to interactions in search engine websites. Cards are also essential, especially for stores.

Google takes the cake here too, but Here Maps are the most commonly used maps for in-dash GPS navigation systems. A good list can make all the difference.

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