Good morning! Here's our prompt for today.

Let's define the "compacting" of a string as taking an input like the following: 'sstttrrrr'.

We want to take the number of sequential appearances (number of times it shows up in a row) of each letter:

1s: 2
2t: 3
3r: 4

And denote them next to the original character in the string, getting rid of the duplicates. In our above example, our output is: 's2t3r4'.

How long is this compacted string? Could you write a method that takes in a string and returns the length of the compacted one? Could you solve it using only O(1) extra space?

Here are few other examples:

1compactLength('aabbbbbbbbbbbbb')
2// 5 because we end up with `a2b13`

Another to consider:

1compactLength('s')
2// 1 because we still end up with just `s`

Constraints

• Length of the given string <= 100000
• The string comprises of ASCII characters
• Expected time complexity : O(n)
• Expected space complexity : O(1)

Try to solve this here or in Interactive Mode.

How do I practice this challenge?

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  console.log('PASSED: ' + "compactLength('aabbbbbbbbbbbbb') should return 5");

var assert = require('assert');

​

/**

 * @param {character[]} str

 * @return {number}

 */

function compactLength(str) {

  // fill this in

  return str.length;

}

​

try {

  assertIsFunction(compactLength, 'compactLength is a function');

​

  console.log('PASSED: ' + '`compactLength` is a function');

} catch (err) {

  console.log(err);

}

​

try {

  assert.equal(compactLength('a'), 1);

​

  console.log('PASSED: ' + "compactLength('a') should return 1");

} catch (err) {

  console.log(err);

}

​

try {

  assert.equal(compactLength('abb'), 3);

We'll now take you through what you need to know.

How do I use this guide?