Increasing Decreasing Strings

Increasing Decreasing String - LeetCode

INTRODUCTION

You are given a string s. Reorder the string using the following algorithm:

1. Pick the smallest character from s and append it to the result.
2. Pick the smallest character from s which is greater than the last appended character to the result and append it.
3. Repeat step 2 until you cannot pick more characters.
4. Pick the largest character from s and append it to the result.
5. Pick the largest character from s which is smaller than the last appended character to the result and append it.
6. Repeat step 5 until you cannot pick more characters.
7. Repeat the steps from 1 to 6 until you pick all characters from s.

In each step, If the smallest or the largest character appears more than once you can choose any occurrence and append it to the result.

Return the result string after sorting s with this algorithm.

Example 1:

Input: s = "aaaabbbbcccc"
Output: "abccbaabccba"
Explanation: After steps 1, 2 and 3 of the first iteration, result = "abc"
After steps 4, 5 and 6 of the first iteration, result = "abccba"
First iteration is done. Now s = "aabbcc" and we go back to step 1
After steps 1, 2 and 3 of the second iteration, result = "abccbaabc"
After steps 4, 5 and 6 of the second iteration, result = "abccbaabccba"

Example 2:

Input: s = "rat"
Output: "art"
Explanation: The word "rat" becomes "art" after re-ordering it with the mentioned algorithm.

Constraints:

• 1 <= s.length <= 500
• s consists of only lowercase English letters.

APPROACH

1. First, we need to split the given string into an array and sort it. Because, for the first iteration, we are going from “smallest” and appending it to the output.

let sArray = s.split('');
    sArray.sort();
    let answer = "";

1. Use the while loop to iterate until there is no element left in the original array.

while(sArray.length > 0 ){
}

1. Inside, filter the original array and append it to the output if each element doesn’t match the last index of the answer string, if it does, append it to the original array.

while(sArray.length > 0 ){
        sArray = sArray.filter((element,index)=>{
            if(index === 0 || element !== answer[answer.length-1]){
                answer += element;
                return false;
            }
            return true;
        })
}

1. reverse the filtered array after the first iteration and repeat until there is no element left in the Array.

while(sArray.length > 0 ){
        sArray = sArray.filter((element,index)=>{
            if(index === 0 || element !== answer[answer.length-1]){
                answer += element;
                return false;
            }
            return true;
        })
         sArray.reverse();
    }

1. return answer string

var sortString = function(s) {
    let sArray = s.split('');
    sArray.sort();
    let answer = "";
    while(sArray.length > 0 ){
        sArray = sArray.filter((element,index)=>{
            if(index === 0 || element !== answer[answer.length-1]){
                answer += element;
                return false;
            }
            return true;
        })
         sArray.reverse();
    }
   return answer;
};

SOLUTION

/**
 * @param {string} s
 * @return {string}
 */
var sortString = function(s) {
    let sArray = s.split('');
    sArray.sort();
    let answer = "";
    while(sArray.length > 0 ){
        sArray = sArray.filter((element,index)=>{
            if(index === 0 || element !== answer[answer.length-1]){
                answer += element;
                return false;
            }
            return true;
        })
         sArray.reverse();
    }
   return answer;
};