Check if string contains a 3 character long substring of another string

Question

I need a regex pattern to check whether characters present in a string consecutively repeats 3 character in another string. Eg:

var string1 = "HelloWorld";    
var string2 = "Work";

Here the letters "Wor" in string1 repeats in string2, so it should return true.

Any help on this

maybe this can gives you and idea [`.*(Work?|W?ork).*`](https://regex101.com/r/XtA7M4/1) — Youcef LAIDANI, Dec 07 '17 at 12:21
what is the expected output for `HelloWorld` - `xWor`? true or false? — Salman A, Dec 07 '17 at 13:05
@paka I am asking if match from middle of string2 also counts as a match? `HelloWorld - xWor` does not match `x` but `Wor` matches, 3 characters. — Salman A, Dec 07 '17 at 13:22
@SalmanA yes it should be taken into consideration. It should just check for any 3 character match between two strings — Paka, Dec 07 '17 at 13:28

Jonny · Answer 1 · 2017-12-07T12:20:23.200

Code from https://www.garysieling.com/blog/javascript-function-find-overlap-two-strings:

function findOverlap(a, b) {
  if (b.length === 0) {
    return '';
  }

  if (a.endsWith(b)) {
    return b;
  }

  if (a.indexOf(b) >= 0) {
    return b;
  }

  return findOverlap(a, b.substring(0, b.length - 1));
}

Some test cases:

findOverlap("12345", "aaa") // ""
findOverlap("12345", "12") // "12"
findOverlap("12345", "345") // "345"
findOverlap("12345", "3456") // "345"
findOverlap("12345", "111") // "1"

To solve your particular problem you could:

const haveOverlap = (string1, string2) => findOverlap(string1, string2).length >= 3;
console.log(haveOverlap('HelloWorld', 'Work')); // true

It should have been `length >= 3` instead of `length > 3` of course. Fixed my code example. — Jonny, Dec 07 '17 at 12:20

gil.fernandes · Answer 2 · 2017-12-07T12:43:39.777

Another idea would be to concatenate the two strings, lowercase them and then apply this regular expression:

(\w{3}).+(\1)

function repetitions(s1, s2) {
  const s = s1.toLowerCase() + s2.toLowerCase()
  const r = /(\w{3}).+(\1)/
  const res = r.exec(s)
  return res !== null ? res[1] : "";
}

console.log(repetitions("HelloWorld", "Work"));
console.log(repetitions("HelloWo", "Work"));

Here is a more robust version which prevents finding the string repetition in either of the input strings:

function repetitions(s1, s2) {
  const replaceRegex = /(\w{3})(.*)(\1)/;
  const s = s1.toLowerCase().replace(replaceRegex, "$1") + " " + s2.toLowerCase().replace(replaceRegex, "$1");
  const r = /(\w{3}).+?(\1)/;
  const res = r.exec(s);
  return res !== null ? res[1] : "";
}

console.log(repetitions("HelloWorld", "Work"));
console.log(repetitions("HelloWo", "Work"));
console.log(repetitions("HelloWoHello", "Work"));
console.log(repetitions("HelloWoHello", "WorkWork"));
console.log(repetitions("HelloWo", "HelloWork"));

i think it will fail for the condition console.log(repetitions("HelloWoHello", "Work")); — Paka, Dec 07 '17 at 12:23
@paka In this case you need to eliminate the repetitions in the input strings. — gil.fernandes, Dec 07 '17 at 12:24
@paka check the updated answer on a suggestion to eliminate repetitions in input strings. — gil.fernandes, Dec 07 '17 at 12:44

score 2 · Accepted Answer · edited Dec 11 '17 at 12:17

Use split, substring and includes

var fn = function( string1, string2, matchChars ) {
   return !!string2.split("").find( function(item, index){
       if (index + matchChars <= string2.length ) 
       { 
          return string1.includes( string2.substring( index, index +  matchChars ) ); //check after each turn if the substring from index is included in string1 or not
       }
       return false;
   });
}

console.log( fn("HelloWorld", "Work", 3) );

var fn = function(string1, string2, matchChars) {
  return !!string2.split("").find(function(item, index) {
    if (index + matchChars <= string2.length) {
      return string1.includes(string2.substring(index, matchChars));
    }
    return false;
  });
}

console.log(fn("HelloWorld", "Work", 3));
console.log(fn("HelloWorld", "Wod", 3));

score 0 · Answer 4 · answered Dec 07 '17 at 13:17

Slightly longer than the other answers here, and nothing too clever about it. Will simply go through and find all matches, which returns an array, and will loop through that array and return true when the longer string contains one of the split strings.

function getAllConsecutives(string, number) {
  let matches = [],
    regex = "";

  for (let i = 0; i < (string.length - number); i++) {
    regex = new RegExp(`\\w{${i}}(\\w{${number}})`);
    matches.push(string.match(regex)[1])
  }

  return matches
}

function areThereConsecutives(string1, string2, number) {
  let short = string1.length < string2.length ? string1 : string2,
    long = string1.length < string2.length ? string2 : string1,
    consecutives = getAllConsecutives(short, number);

  for (let i = 0; i < consecutives.length; i++) {
    if (long.includes(consecutives[i])) {
      return true;
    }
  }

  return false
}

let string1 = "HelloWorld",
  string2 = "Work";

console.log(areThereConsecutives(string1, string2, 3))

Check if string contains a 3 character long substring of another string

4 Answers4