正则表达式：如何计算给定字符串中一个或多个单词的排列？

Question

Is there a way to use regex to count the permutations of a word or words in a given string? 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. Million thanks in advance :)

UPDATE:

Inspired by Trincot, this is a "non-generator" version of the solution:

function permutations(arr) {
  return arr.length === 1
    ? arr
    : arr.flatMap((v,i,a) => permutations([...a.slice(0,i),...a.slice(i+1)]).map((d) => `${v}${d}`));
}

function createRegex(arr) {
  const p = arr.flatMap((v) => permutations([...v]));
  return `(?=${p.join('|')})`;
}

const nums = ['ZERO','ONE','TWO','THREE','FOUR','FIVE','SIX','SEVEN','EIGHT','NINE'];
const regex = createRegex(nums);
const amount = (str) => str.match(RegExp(regex,'g'))?.length || 0;

console.log(amount('TEN')); // 0
console.log(amount('ONE')); // 1
console.log(amount('ONEOTW')); // 3
console.log(amount('ONENO')); // 2
console.log(amount('NEOTWONE')); // 4
console.log(amount('NEOTWONEINEIGHTOWSVEEN')); // 8

Answer 1

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.