DEV Community: duccanhole

New lightbox package here!

duccanhole — Sat, 13 Jul 2024 13:20:29 +0000

Hello everyone,

I’m very excited to introduce you to my open-source project:
@duccanhole/lightbox
. This simple lightbox file viewer supports various types of files and was developed using vanilla JavaScript, with no dependencies.

As this is my first time publishing an npm package as an open-source developer, your feedback will be incredibly valuable to help me improve this project.

You can check out the project here.

Thank you for reading, and have a great day!

code everyday with me

duccanhole — Thu, 25 Nov 2021 05:01:00 +0000

--DAY 26--

Hi, I am going to make #100DaysOfCode Challenge. Everyday I will try solve 1 problem from leetcode or hackerrank. Hope you can go with me until end.
Now let's solve problem today:
Problem: Add Digits
Detail: here
My solution (javascript):

var addDigits = function(num) {
    if(num<=9) return num;
    let n=0;
    while(num>0){
        n+=num%10;
        num=Math.floor(num/10);
    }
    return addDigits(n);
};

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Tue, 23 Nov 2021 01:33:29 +0000

--DAY 26--

var reverseVowels = function(s) {
    s=s.split('');
    let vowel = ['a','e','i','o','u','A','E','I','O','U'];
    let i=0,j=s.length-1;
    while(i<j){
        if(!vowel.includes(s[i])) i++;
        else if(!vowel.includes(s[j])) j--;
        if(vowel.includes(s[i])&&vowel.includes(s[j])){
            [s[i],s[j]]=[s[j],s[i]];
            i++;
            j--;
        }
    }
    return s.join('');
};

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Mon, 22 Nov 2021 03:01:20 +0000

--DAY 25--

var isIsomorphic = function(s, t) {
    let map1 = new Map(), map2 = new Map();
    if (s.length != t.length) return false;
    for (let i = 0; i < s.length; i++) {
        if((map1.has(s[i])&&map1.get(s[i])!=t[i])||
           (map2.has(t[i])&&map2.get(t[i])!=s[i])
          ) return false;
        map1.set(s[i],t[i]);
        map2.set(t[i],s[i]);
    }
    return true;
};

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Sun, 21 Nov 2021 03:41:54 +0000

--DAY 25--

function utopianTree(n) {
    let height = 1;
    if(n==0) return 1;
    for(let i=1;i<=n;i++){
        if(i%2!=0) height*=2;
        else height++;
    }
    return height;
}

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Sat, 20 Nov 2021 04:06:17 +0000

--DAY 24--

Hi, I am going to make #100DaysOfCode Challenge. Everyday I will try solve 1 problem from leetcode or hackerrank. Hope you can go with me until end.
Now let's solve problem today:

Problem: Happy Number
Detail: here
Idea: After each step we split number, store new number to array, to check if number loops endlessly in a cycle, we need to check if array has any number appears twice or not.
- Example: 2
- step1: 2^2 = 4; array = [4];
- step2: 4^2 = 16; array = [4,16];
- step3: 1^2 + 6^2 = 37; array = [4,16,37];
- step4: 3^2 + 7^2 = 58; array = [4,16,37,58];
- step5: 5^2 + 8^2 = 89; array = [4,16,37,58,89];
- step6: 8^2 + 9^2 = 145; array = [4,16,37,58,89,145];
- step7: 1^2 + 4^2 + 5^2 = 42; array = [4,16,37,58,89,145,42];
- step8: 4^2 + 2^2 = 20; array = [4,16,37,58,89,145,42,20];
- step9: 2^2 + 0^2 = 4; array = [4,...,4];
- Because number 4 appears twice, we return false.
My solution(javascript):

var Sum=(n)=>{
    let tmp=0;
    while(n>0){
        tmp=tmp+(n%10)**2;
        n=parseInt(n/10);
    }
    return tmp;
}
var isHappy = function(n) {
    let ans=[],i=1;
    ans[0]=n;
    while(i){
        let nums = Sum(ans[i-1]);
        if(nums==1) return true;
        else if(ans.includes(nums)) return false;
        ans.push(nums);
        i++;
    }
};

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Fri, 19 Nov 2021 02:27:59 +0000

--DAY 23--

var titleToNumber = function(str) {
    let colNum = '0ABCDEFGHIJKLMNOPQRSTUVWXYZ';
    str = str.split('');
    let result = 0, n = str.length-1;
    for(let i=0;i<str.length;i++){
        result+=((26**n)*colNum.indexOf(str[i]));
        n--;
    }
    return result;
};

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Wed, 17 Nov 2021 02:30:36 +0000

--DAY 22--

Hi, I am going to make #100DaysOfCode Challenge. Everyday I will try solve 1 problem from leetcode or hackerrank. Hope you can go with me until end.
Now let's solve problem today:

Problem: Excel Sheet Column Title
Detail: here
Idea: it's pretty similar to convert number to binary
My solution(javascript):

var convertToTitle = function(n) {
    let column = 'ZABCDEFGHIJKLMNOPQRSTUVWXYZ';
    let result = '';
    while(n>26){
        let tmp = n%26;
        result += column.charAt(tmp);
        if(n%26==0) n = Math.floor(n/27);
        else n = Math.floor(n/26);
    }
    result += column.charAt(n);
    return result.split('').reverse().join('');
};

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Tue, 16 Nov 2021 08:24:15 +0000

--DAY 21--

Hi, I am going to make #100DaysOfCode Challenge. Everyday I will try solve 1 problem from leetcode or hackerrank. Hope you can go with me until end.
Now let's solve problem today:

Problem: Climbing the Leaderboard
Detail: here
Idea:
- Remove the duplicate number of ranked
- Because player array was sorted, we will find the rank from bottom to top by traversing from tail to head of array
My solution(javascript):

function climbingLeaderboard(ranked, player) {
    ranked = [... new Set(ranked)];
    let ans=[],i=ranked.length-1;
    for(let score of player){
        while(i>=0){
            if(score<ranked[i]){
                ans.push(i+2);
                break;
            }
            i--;
        }
        if(i<0) ans.push(1);
    }
    return ans;
}

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Sun, 14 Nov 2021 07:21:28 +0000

--DAY 20--

var twoSum = function(numbers, target) {
    let i=0,j=numbers.length-1;
    while(i<j){
        if(numbers[i]+numbers[j]==target){
            return [i+1,j+1];
        }
        else if (numbers[i] + numbers[j] < target)
            i++;
        else
            j--;
    }
};

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Sat, 13 Nov 2021 03:05:16 +0000

--DAY 19--

Using sort:

function PickingNumbers(){
    let max=1,tmp=1,idx=0;
    for(let i=1;i<a.length;i++){
        if(a[i]-a[idx]<=1) tmp++;
        else{
            idx=i;
            tmp=1;
        }
        max=(tmp>max)?tmp:max;
    } 
    return max;
}

Using hash map:

function pickingNumbers(a) {
    let max=1;
    let map = new Map();
    for(let number of a){
        if(map.has(number)){
            map.set(number, map.get(number)+1);
        }
        else map.set(number, 1);
    }
    for(const [key,value] of map){
        let tmp=value;
        if(map.has(key+1)) tmp=map.get(key)+map.get(key+1);
        if(tmp>max) max=tmp;
    }
    return max;
}

-->If you have better solution or any question, please comment below. I will appreciate.

code every day with me

duccanhole — Fri, 12 Nov 2021 03:11:17 +0000

--DAY 18--

Hi, I am going to make #100DaysOfCode Challenge. Everyday I will try solve 1 problem from leetcode or hackerrank. Hope you can go with me until end.
Now let's solve problem today:

Problem: Best Time to Buy and Sell Stock
Detail: here
My solution(javascript):

var maxProfit = function(prices) {
    let profit=0,buy=100000;
    for(let x of prices){
        if(x<buy) buy=x;
        if(x-buy>profit) profit=(x-buy);
    }
    return profit;
};

-->If you have better solution or any question, please comment below. I will appreciate.