Find same characters in 2 stringsFind ASCII codes and replace with charactersTypeahead Talent BuddyFind...
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Find same characters in 2 strings
Find ASCII codes and replace with charactersTypeahead Talent BuddyFind function for strings in cellsCleaning up and reformatting imported data in an Excel sheetNumber of sub strings with same first and last characterCodeEval solutions to find trailing stringsGenerate strings of random charactersBinary search for a missing or default value by a given formulaFind spanning tree with maximum edges with the same weightPHP class that behaves like enums
$begingroup$
I have a solution, which compares 2 given strings and returns the number of total matching characters.
I know I haven't initialized the variables and arrays and that this (correctly) yields warnings. I know the variables aren't self-explanatory as they should be.
function commonCharacterCount($s1, $s2) {
$a = (strlen($s1) > strlen($s2)) ? $s2 : $s1;
$b = (strlen($s1) > strlen($s2)) ? $s1 : $s2;
foreach(count_chars($a, 1) as $i => $v) {
$c[chr($i)] = $v;
}
foreach (count_chars($b, 1) as $i => $v) {
$d[chr($i)] = $v;
}
$t = 0;
foreach($c as $k => $v) {
if($c[$k] <= $d[$k]) {
$t += $c[$k];
} else {
$t += $d[$k];
}
}
return $t;
}
For instance, I have the two strings:
$s1 = "abacadeee"; and $s2 = "aabbccddee";, the expected output would be 7.
As required, this solution works so far and you can test it here:
sandbox
Which steps are unnecessary and how can I improve this algorithm?
performance php
edited Mar 14 at 15:11
Stephen Rauch
3,77061630
asked Mar 14 at 13:24
DasSaffeDasSaffe
1335
$endgroup$
add a comment |
$begingroup$
I have a solution, which compares 2 given strings and returns the number of total matching characters.
I know I haven't initialized the variables and arrays and that this (correctly) yields warnings. I know the variables aren't self-explanatory as they should be.
function commonCharacterCount($s1, $s2) {
$a = (strlen($s1) > strlen($s2)) ? $s2 : $s1;
$b = (strlen($s1) > strlen($s2)) ? $s1 : $s2;
foreach(count_chars($a, 1) as $i => $v) {
$c[chr($i)] = $v;
}
foreach (count_chars($b, 1) as $i => $v) {
$d[chr($i)] = $v;
}
$t = 0;
foreach($c as $k => $v) {
if($c[$k] <= $d[$k]) {
$t += $c[$k];
} else {
$t += $d[$k];
}
}
return $t;
}
For instance, I have the two strings:
$s1 = "abacadeee"; and $s2 = "aabbccddee";, the expected output would be 7.
As required, this solution works so far and you can test it here:
sandbox
Which steps are unnecessary and how can I improve this algorithm?
performance php
edited Mar 14 at 15:11
Stephen Rauch
3,77061630
asked Mar 14 at 13:24
DasSaffeDasSaffe
1335
$endgroup$
1
$begingroup$
what is the logic of common character count? from you test data characters that matches are a, b, c, d, e = 5. Again if I count occurrence the value is not 7. or even i count maximum occurrence the value is not 7
$endgroup$
– Muhammed Imran Hussain
Mar 14 at 15:22
add a comment |
$begingroup$
I have a solution, which compares 2 given strings and returns the number of total matching characters.
I know I haven't initialized the variables and arrays and that this (correctly) yields warnings. I know the variables aren't self-explanatory as they should be.
function commonCharacterCount($s1, $s2) {
$a = (strlen($s1) > strlen($s2)) ? $s2 : $s1;
$b = (strlen($s1) > strlen($s2)) ? $s1 : $s2;
foreach(count_chars($a, 1) as $i => $v) {
$c[chr($i)] = $v;
}
foreach (count_chars($b, 1) as $i => $v) {
$d[chr($i)] = $v;
}
$t = 0;
foreach($c as $k => $v) {
if($c[$k] <= $d[$k]) {
$t += $c[$k];
} else {
$t += $d[$k];
}
}
return $t;
}
For instance, I have the two strings:
$s1 = "abacadeee"; and $s2 = "aabbccddee";, the expected output would be 7.
As required, this solution works so far and you can test it here:
sandbox
Which steps are unnecessary and how can I improve this algorithm?
performance php
edited Mar 14 at 15:11
Stephen Rauch
3,77061630
asked Mar 14 at 13:24
DasSaffeDasSaffe
1335
$endgroup$
I have a solution, which compares 2 given strings and returns the number of total matching characters.
I know I haven't initialized the variables and arrays and that this (correctly) yields warnings. I know the variables aren't self-explanatory as they should be.
function commonCharacterCount($s1, $s2) {
$a = (strlen($s1) > strlen($s2)) ? $s2 : $s1;
$b = (strlen($s1) > strlen($s2)) ? $s1 : $s2;
foreach(count_chars($a, 1) as $i => $v) {
$c[chr($i)] = $v;
}
foreach (count_chars($b, 1) as $i => $v) {
$d[chr($i)] = $v;
}
$t = 0;
foreach($c as $k => $v) {
if($c[$k] <= $d[$k]) {
$t += $c[$k];
} else {
$t += $d[$k];
}
}
return $t;
}
For instance, I have the two strings:
$s1 = "abacadeee"; and $s2 = "aabbccddee";, the expected output would be 7.
As required, this solution works so far and you can test it here:
sandbox
Which steps are unnecessary and how can I improve this algorithm?
performance php
performance php
edited Mar 14 at 15:11
Stephen Rauch
3,77061630
asked Mar 14 at 13:24
DasSaffeDasSaffe
1335
edited Mar 14 at 15:11
Stephen Rauch
3,77061630
asked Mar 14 at 13:24
DasSaffeDasSaffe
1335
edited Mar 14 at 15:11
Stephen Rauch
3,77061630
edited Mar 14 at 15:11
Stephen Rauch
3,77061630
edited Mar 14 at 15:11
Stephen Rauch
3,77061630
3,77061630
asked Mar 14 at 13:24
DasSaffeDasSaffe
1335
asked Mar 14 at 13:24
DasSaffeDasSaffe
1335
asked Mar 14 at 13:24
DasSaffeDasSaffe
1335
1335
1
$begingroup$
what is the logic of common character count? from you test data characters that matches are a, b, c, d, e = 5. Again if I count occurrence the value is not 7. or even i count maximum occurrence the value is not 7
$endgroup$
– Muhammed Imran Hussain
Mar 14 at 15:22
add a comment |
1
$begingroup$
what is the logic of common character count? from you test data characters that matches are a, b, c, d, e = 5. Again if I count occurrence the value is not 7. or even i count maximum occurrence the value is not 7
$endgroup$
– Muhammed Imran Hussain
Mar 14 at 15:22
1
1
$begingroup$
what is the logic of common character count? from you test data characters that matches are a, b, c, d, e = 5. Again if I count occurrence the value is not 7. or even i count maximum occurrence the value is not 7
$endgroup$
– Muhammed Imran Hussain
Mar 14 at 15:22
$begingroup$
what is the logic of common character count? from you test data characters that matches are a, b, c, d, e = 5. Again if I count occurrence the value is not 7. or even i count maximum occurrence the value is not 7
$endgroup$
– Muhammed Imran Hussain
Mar 14 at 15:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I would recommend a process like this:
Code: (Demo)
$string1 = "abacadeee";
$string2 = "aabbccddee";
$counts2 = count_chars($string2, 1);
$tally = 0;
foreach (array_intersect_key(count_chars($string1, 1), $counts2) as $charcode1 => $count1) {
$tally += min($counts2[$charcode1], $count1);
}
echo $tally;
count_chars() lends itself beautifully to this task, so using array functions onward is a sensible choice.
It is important to try to minimize iterations and not perform any useless iterations. By calling array_intersect_key() on the two count_chars() results, the foreach() loop is only going to iterate elements with keys which are shared between the two arrays. In doing this, you don't need to check which array is smaller (which is otherwise how you would choose which array to iterate).
$tally is incremented by the lesser of the two counts for each char.
p.s. calling chr() is irrelevant to your objective.
answered Mar 18 at 12:50
mickmackusamickmackusa
1,739218
$endgroup$
add a comment |
Your Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
I would recommend a process like this:
Code: (Demo)
$string1 = "abacadeee";
$string2 = "aabbccddee";
$counts2 = count_chars($string2, 1);
$tally = 0;
foreach (array_intersect_key(count_chars($string1, 1), $counts2) as $charcode1 => $count1) {
$tally += min($counts2[$charcode1], $count1);
}
echo $tally;
count_chars() lends itself beautifully to this task, so using array functions onward is a sensible choice.
It is important to try to minimize iterations and not perform any useless iterations. By calling array_intersect_key() on the two count_chars() results, the foreach() loop is only going to iterate elements with keys which are shared between the two arrays. In doing this, you don't need to check which array is smaller (which is otherwise how you would choose which array to iterate).
$tally is incremented by the lesser of the two counts for each char.
p.s. calling chr() is irrelevant to your objective.
answered Mar 18 at 12:50
mickmackusamickmackusa
1,739218
$endgroup$
add a comment |
$begingroup$
I would recommend a process like this:
Code: (Demo)
$string1 = "abacadeee";
$string2 = "aabbccddee";
$counts2 = count_chars($string2, 1);
$tally = 0;
foreach (array_intersect_key(count_chars($string1, 1), $counts2) as $charcode1 => $count1) {
$tally += min($counts2[$charcode1], $count1);
}
echo $tally;
count_chars() lends itself beautifully to this task, so using array functions onward is a sensible choice.
It is important to try to minimize iterations and not perform any useless iterations. By calling array_intersect_key() on the two count_chars() results, the foreach() loop is only going to iterate elements with keys which are shared between the two arrays. In doing this, you don't need to check which array is smaller (which is otherwise how you would choose which array to iterate).
$tally is incremented by the lesser of the two counts for each char.
p.s. calling chr() is irrelevant to your objective.
answered Mar 18 at 12:50
mickmackusamickmackusa
1,739218
$endgroup$
add a comment |
$begingroup$
I would recommend a process like this:
Code: (Demo)
$string1 = "abacadeee";
$string2 = "aabbccddee";
$counts2 = count_chars($string2, 1);
$tally = 0;
foreach (array_intersect_key(count_chars($string1, 1), $counts2) as $charcode1 => $count1) {
$tally += min($counts2[$charcode1], $count1);
}
echo $tally;
count_chars() lends itself beautifully to this task, so using array functions onward is a sensible choice.
It is important to try to minimize iterations and not perform any useless iterations. By calling array_intersect_key() on the two count_chars() results, the foreach() loop is only going to iterate elements with keys which are shared between the two arrays. In doing this, you don't need to check which array is smaller (which is otherwise how you would choose which array to iterate).
$tally is incremented by the lesser of the two counts for each char.
p.s. calling chr() is irrelevant to your objective.
answered Mar 18 at 12:50
mickmackusamickmackusa
1,739218
$endgroup$
I would recommend a process like this:
Code: (Demo)
$string1 = "abacadeee";
$string2 = "aabbccddee";
$counts2 = count_chars($string2, 1);
$tally = 0;
foreach (array_intersect_key(count_chars($string1, 1), $counts2) as $charcode1 => $count1) {
$tally += min($counts2[$charcode1], $count1);
}
echo $tally;
count_chars() lends itself beautifully to this task, so using array functions onward is a sensible choice.
It is important to try to minimize iterations and not perform any useless iterations. By calling array_intersect_key() on the two count_chars() results, the foreach() loop is only going to iterate elements with keys which are shared between the two arrays. In doing this, you don't need to check which array is smaller (which is otherwise how you would choose which array to iterate).
$tally is incremented by the lesser of the two counts for each char.
p.s. calling chr() is irrelevant to your objective.
answered Mar 18 at 12:50
mickmackusamickmackusa
1,739218
answered Mar 18 at 12:50
mickmackusamickmackusa
1,739218
answered Mar 18 at 12:50
mickmackusamickmackusa
1,739218
answered Mar 18 at 12:50
mickmackusamickmackusa
1,739218
1,739218
add a comment |
add a comment |
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$begingroup$
what is the logic of common character count? from you test data characters that matches are a, b, c, d, e = 5. Again if I count occurrence the value is not 7. or even i count maximum occurrence the value is not 7
$endgroup$
– Muhammed Imran Hussain
Mar 14 at 15:22