<?php
function isAnagram($satu, $dua)
{
$satuAr = str_split($satu);
$duaAr = str_split($dua);
$jumlah = 0;
foreach ($satuAr as $k => $v) {
if (!in_array($v, $duaAr)) {
$jumlah++;
}
}
if ($jumlah != 0) {
echo "false";
} else {
echo "true";
}
}
#isAnagram("secure","resc ue");
isAnagram("conifers", "fir cones");
#isAnagram("google","facebook");
<?php
$wordPairs[] = array("cat", "act");
// happy path
$wordPairs[] = array("cat", "apt");
// not so happy path
$wordPairs[] = array("cat", "pact");
// not so happy path
$wordPairs[] = array("cat", "tact");
// not so happy path
$wordPairs[] = array("31121984", "12311984");
// happy path
foreach ($wordPairs as $words) {
if (isAnagram($words[0], $words[1])) {
echo "{$words['0']}, {$words['1']} are anagrams. \n";
} else {
echo "{$words['0']}, {$words['1']} are not anagrams. \n";
}
}
function isAnagram($word1, $word2)
{
if (empty($word1) || empty($word1) || strlen($word1) != strlen($word2)) {
// check for empty strings or if
// one of the words atleast has one or more chars more than the other
return false;
} else {
$charFreqMapWord1 = __buildFrequencyMap($word1);
print_r($charFreqMapWord1);
$charFreqMapWord2 = __buildFrequencyMap($word2);
print_r($charFreqMapWord2);
foreach ($charFreqMapWord1 as $char => $count) {