3216. Lexicographically Smallest String After a Swap using JavaScript

Problem

Given a string s containing only digits, return the 

lexicographically smallest string that can be obtained after swapping adjacent digits in s with the same parity at most once.

Digits have the same parity if both are odd or both are even. For example, 5 and 9, as well as 2 and 4, have the same parity, while 6 and 9 do not.

Example 1:

Input: s = “45320”

Output: “43520”

Explanation:

s[1] == '5' and s[2] == '3' both have the same parity, and swapping them results in the lexicographically smallest string.

Example 2:

Input: s = “001”

Output: “001”

Explanation:

There is no need to perform a swap because s is already the lexicographically smallest.

Constraints:

• 2 <= s.length <= 100
• s consists only of digits.

Solution

First attempt : Time complexity O(n)

/**
 * @param {string} s
 * @return {string}
 */
var getSmallestString = function(s) {
    let smallestnumber = s;
    let duplicate = s.split("");
    if(s.length == 2){
        //smallestnumber = s
        if(s[0] % 2 == 0 && s[1] % 2 == 0){
            let number = duplicate[0];
            duplicate[0] = duplicate[1];
            duplicate[1] = number;
            if(duplicate.join("") < smallestnumber){
                smallestnumber = duplicate.join("");
            }
            duplicate = s.split("");
        }
        else if(s[0] % 2 != 0 && s[1] % 2 != 0){
            let number = duplicate[0];
            duplicate[0] = duplicate[1];
            duplicate[1] = number;
            if(duplicate.join("") < smallestnumber){
                smallestnumber = duplicate.join("");
            }
            duplicate = s.split("");
        }
        else{
            return smallestnumber;
        }
    }
    else{
    for(let i = 0; i < s.length - 1; i++){
        if(s[i] % 2 == 0 && s[i+1] % 2 == 0){
            let number = duplicate[i];
            duplicate[i] = duplicate[i+1];
            duplicate[i+1] = number;
            if(duplicate.join("") < smallestnumber){
                smallestnumber = duplicate.join("");
            }
            duplicate = s.split("");
        }
        else if(s[i] % 2 != 0 && s[i+1] % 2 != 0){
            let number = duplicate[i];
            duplicate[i] = duplicate[i+1];
            duplicate[i+1] = number;
            if(duplicate.join("") < smallestnumber){
                smallestnumber = duplicate.join("");
            }
            duplicate = s.split("");
        }
    }

    }
    return smallestnumber;
};