Here are some example strings:
2222
333333
12345
111
123456789
12345678
The expected result is:
2@222
333@333
12@345
111
123@456@789
12@345@678
In other words, the character '@' should be inserted at the 4th, 8th, 12th, etc. positions counting from the end of the string.
I think this can be achieved using replace and other methods available in JavaScript.
To validate the output string, a regular expression has been created:
^(\d{1,3})(\.\d{3})*?$
If you're looking to extract and group specific digits in a string, consider using the regular expression below:
/(\d)(\d{3})$/
This regex will identify and group the first digit as \d and the last three digits as \d{3}, creating separate groups for each set of digits. By referencing these matched groups with $1 and $2 in your replacement string, you can manipulate the output effectively.
Take a look at the example code snippet provided:
const transform = str => str.replace(/(\d)(\d{3})$/, '$1@$2');
console.log(transform("2222")); // 2@222
console.log(transform("333333")); // 333@333
console.log(transform("12345")); // 12@345
console.log(transform("111")); // 111For longer strings with a size of N, an alternate approach would be to utilize methods like .match() and reverse the string before grouping the digits. Here's an example showcasing this technique:
const reverse = str => Array.from(str).reverse().join('');
const transform = str => {
return reverse(reverse(str).match(/(\d{1,3})/g).join('@'));
}
console.log(transform("2222")); // 2@222
console.log(transform("333333")); // 333@333
console.log(transform("12345")); // 12@345
console.log(transform("111")); // 111
console.log(transform("123456789")); // 123@456@789
console.log(transform("12345678")); // 12@345@678var numbers = [
'111',
'2222',
'333333',
'12345',
'123456789',
'1234567890123456'
];
console.log(numbers.map(function (number) {
return number.replace(/(?=(?:\B\d{3})+$)/g, '@');
}));To capture all the numbers, you can apply a pattern where an @ is added after every third digit from the right using a positive lookahead.
(?=(?:\B\d{3})+$)
(?= This denotes a Positive lookahead, checking what comes next
(?:\B\d{3})+ Repeats 1+ times and captures not a word boundary followed by 3 digits$ Asserts the end of the string) Closes the lookaheadconst regex = /^\d+$/;
["2222",
"333333",
"12345",
"111",
"123456789",
"12345678"
].forEach(s => console.log(
s.replace(/(?=(?:\B\d{3})+$)/g, "@")
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