Problem

Given an array of characters, compress it in-place.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.

After you are done modifying the input array in-place, return the new length of the array.

Follow up: Could you solve it using only O(1) extra space?

Example 1:

Input: [“a”,”a”,”b”,”b”,”c”,”c”,”c”]

Output: Return 6, and the first 6 characters of the input array should be: [“a”,”2”,”b”,”2”,”c”,”3”]

Explanation: “aa” is replaced by “a2”. “bb” is replaced by “b2”. “ccc” is replaced by “c3”.

Example 2:

Input: [“a”]

Output: Return 1, and the first 1 characters of the input array should be: [“a”]

Explanation: Nothing is replaced.

Example 3:

Input: [“a”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”]

Output: Return 4, and the first 4 characters of the input array should be: [“a”,”b”,”1”,”2”].

Explanation: Since the character “a” does not repeat, it is not compressed. “bbbbbbbbbbbb” is replaced by “b12”. Notice each digit has it’s own entry in the array.

Note:

All characters have an ASCII value in [35, 126]. 1 <= len(chars) <= 1000.

Pre analysis

Will keep a string with this logic and return array contents with string contents

Post analysis

This is one of dumbest solution proposed by me.

Another solution

    var compress = function(chars) {
        let endIndex = chars.length-1;
        while(endIndex>-1){
            let startIndex = endIndex-1;
            while(startIndex>-1 && chars[startIndex] === chars[endIndex]){
                startIndex--;
            }
            let count = endIndex-startIndex;
            if(count>1){
                chars.splice(startIndex+2,count-1,...(''+count));
            }

            endIndex = startIndex;
        }

        return chars.length;
    };