Is there a way to use regex to find out how many times a word can occured in a given string despite order of its characters Let's take "NEOTWONE" as our examples which should return a count of "4"
NEO --> 'ONE' --> +1
OTW --> 'TWO' --> +1
TWO --> 'TWO' --> +1
ONE --> 'ONE' --> +1
This is what I have so far and I couldn't get the regex to work properly.
const nums = ['ZERO','ONE','TWO','THREE','FOUR','FIVE','SIX','SEVEN','EIGHT','NINE'];
function amount(str,count=0) {
for (const n of nums) {
RegExp(`\\b[${str}]+\\b`,'g').test(n) && count++;
}
return count;
}
console.log(amount('ONE')); // 1
console.log(amount('ONEOTW')); // 2
console.log(amount('ONENO')); // 1
console.log(amount('NEOTWONE')); // 2
As you can see, the 2nd, 3rd & 4th example above did not render the correct outcome which should be:
console.log(amount('ONEOTW')); // 3
console.log(amount('ONENO')); // 2
console.log(amount('NEOTWONE')); // 4
I'm new to regex, any feedback will be greatly appreciated with explanation. Million thanks in advance :)
You'll have to create a huge regular expression for that:
First, create all the unique permutations of all input strings (nums).
Concatenate these into one regular expression, using | as separator, but use look-ahead, so that one character can be part of a match multiple times. So for instance, for "ONE", the regular expression would be:
(?=ONE|OEN|ENO|EON|NEO|NOE)
But then you would also include all permutations of "ZERO" and all the other words.
function* permutations(word) {
if (word.length <= 1) return yield word;
for (let i = 0; i < word.length; i++) {
for (let perm of permutations(word.slice(0, i) + word.slice(i + 1))) {
yield word[i] + perm;
}
}
}
function createRegex(words) {
const allPermutations = words.flatMap(word => [...new Set(permutations(word))]);
return RegExp("(?=" + allPermutations.join("|") + ")", "g");
}
function countMatches(regex, phrase) {
return phrase.match(regex)?.length ?? 0;
}
const nums = ['ZERO','ONE','TWO','THREE','FOUR','FIVE','SIX','SEVEN','EIGHT','NINE'];
const regex = createRegex(nums);
for (const test of ['ONE', 'ONEOTW', 'ONENO', 'NEOTWONE']) {
console.log(test, countMatches(regex, test));
}Note that for the second test the answer is 3, not 2, since "NEO" also counts.