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getURLParameters

0 likes • Nov 19, 2022 • 0 views
JavaScript
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More JavaScript Posts

Redux Slice

CHS
0 likes • Mar 11, 2021 • 0 views
JavaScript
import { createSlice } from "@reduxjs/toolkit";
const alert = createSlice({
name: "alert",
initialState: {
msg: "",
status: "",
},
reducers: {
set_alert: (state, action) => {
return {
...state,
msg: action.payload.msg,
status: action.payload.status,
};
},
clear_alert: (state, action) => {
return {
...state,
msg: "",
status: "",
};
},
},
});
export default alert.reducer;
const { set_alert, clear_alert } = alert.actions;
export const setAlert = (msg, status) => dispatch => {
dispatch(set_alert({ msg, status }));
setTimeout(() => dispatch(clear_alert()), 4000);
};
export const clearAlert = () => dispatch => {
dispatch(clear_alert());
};

Longest Palindromic Substring

0 likes • Nov 19, 2022 • 0 views
JavaScript
const longestPalindrome = (s) => {
if(s.length === 1) return s
const odd = longestOddPalindrome(s)
const even = longestEvenPalindrome(s)
if(odd.length >= even.length) return odd
if(even.length > odd.length) return even
};
const longestOddPalindrome = (s) => {
let longest = 0
let longestSubStr = ""
for(let i = 1; i < s.length; i++) {
let currentLongest = 1
let currentLongestSubStr = ""
let left = i - 1
let right = i + 1
while(left >= 0 && right < s.length) {
const leftLetter = s[left]
const rightLetter = s[right]
left--
right++
if(leftLetter === rightLetter) {
currentLongest += 2
currentLongestSubStr = s.slice(left+1, right)
} else break
}
if(currentLongest > longest) {
if(currentLongestSubStr) longestSubStr = currentLongestSubStr
else longestSubStr = s[i]
longest = currentLongest
}
}
return longestSubStr
}
const longestEvenPalindrome = (s) => {
let longest = 0
let longestSubStr = ""
for(let i = 0; i < s.length - 1; i++) {
if(s[i] !== s[i+1]) continue;
let currentLongest = 2
let currentLongestSubStr = ""
let left = i - 1
let right = i + 2
while(left >= 0 && right < s.length) {
const leftLetter = s[left]
const rightLetter = s[right]
left--
right++
if(leftLetter === rightLetter) {
currentLongest += 2
currentLongestSubStr = s.slice(left+1, right)
} else break
}
if(currentLongest > longest) {
if(currentLongestSubStr) longestSubStr = currentLongestSubStr
else longestSubStr = s[i] + s[i+1]
longest = currentLongest
}
}
return longestSubStr
}
console.log(longestPalindrome("babad"))

Tree Traversal

CHS
0 likes • Oct 16, 2023 • 7 views
JavaScript
class TreeNode {
constructor(data, depth) {
this.data = data;
this.children = [];
this.depth = depth;
}
}
function buildNaryTree(hosts) {
const nodeMap = {};
// Step 1: Create a map of nodes using fqdn as the key
hosts.forEach((host) => {
nodeMap[host.fqdn] = new TreeNode(host, 0); // Initialize depth to 0
});
// Step 2: Iterate through the array and add each node to its parent's list of children
hosts.forEach((host) => {
if (host.displayParent) {
if (nodeMap[host.displayParent]) {
const parent = nodeMap[host.displayParent];
const node = nodeMap[host.fqdn];
node.depth = parent.depth + 1; // Update the depth
parent.children.push(node);
} else {
console.error(`Parent with fqdn ${host.displayParent} not found for ${host.fqdn}`);
}
}
});
// Find the root nodes (nodes with no parent)
const rootNodes = hosts.filter((host) => !host.displayParent).map((host) => nodeMap[host.fqdn]);
return rootNodes;
}
function treeToObjects(node) {
const result = [];
function inOrder(node) {
if (!node) {
return;
}
// Visit the current node and add it to the result
result.push(node.data);
// Visit children nodes first
node.children.forEach((child) => {
inOrder(child);
});
}
inOrder(node);
return result;
}
// Usage example
const hosts = [
{
fqdn: 'fqdn_1',
display_name: 'Host 1',
},
{
fqdn: 'fqdn_2',
display_name: 'Host 2',
displayParent: 'fqdn_1',
},
{
fqdn: 'fqdn_3',
display_name: 'Host 3'
},
{
fqdn: 'fqdn_4',
display_name: 'Host 4',
displayParent: 'fqdn_3',
},
{
fqdn: 'fqdn_5',
display_name: 'Host 5',
displayParent: 'fqdn_1',
},
];
const tree = buildNaryTree(hosts);
// Function to convert the tree to an array of objects
function convertTreeToArray(nodes) {
const result = [];
nodes.forEach((node) => {
result.push(...treeToObjects(node));
});
return result;
}
// Convert the tree back to an array of objects
const arrayFromTree = convertTreeToArray(tree);
console.log(arrayFromTree);

Rings and Rods

0 likes • Nov 19, 2022 • 1 view
JavaScript
// There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9.
// You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where:
// The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B').
// The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9').
// For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1.
// Return the number of rods that have all three colors of rings on them.
let rings = "B0B6G0R6R0R6G9";
var countPoints = function(rings) {
let sum = 0;
// Always 10 Rods
for (let i = 0; i < 10; i++) {
if (rings.includes(`B${i}`) && rings.includes(`G${i}`) && rings.includes(`R${i}`)) {
sum+=1;
}
}
return sum;
};
console.log(countPoints(rings));

copyString.js

0 likes • Nov 8, 2023 • 5 views
JavaScript
function copyString(text){
async function navigatorClipboardCopy(text){
if(document.location.protocol != "https:"){
return false;
}
return new Promise((resolve, reject)=>{
navigator.clipboard.writeText(text).then(()=>{resolve(true);}, ()=>{resolve(false);});
});
}
function domCopy(text){
if(!(document.execCommand)){
console.warn("They finally deprecated document.execCommand!");
}
const input = document.createElement('textarea');
input.value = text;
document.body.appendChild(input);
input.select();
const success = document.execCommand('copy');
document.body.removeChild(input);
return success;
}
function promptCopy(){
prompt("Copy failed. Might have ... somewhere. Check console.", text);
console.log(text);
}
function done(){
alert("Copied to clipboard");
}
navigatorClipboardCopy(text).catch(()=>{return false;}).then((success)=>{
if(success){
done();
}else{
if(domCopy(text)){
done();
}else{
promptCopy();
}
}
});
}

levenshtein distance

0 likes • Nov 19, 2022 • 1 view
JavaScript
const levenshteinDistance = (s, t) => {
if (!s.length) return t.length;
if (!t.length) return s.length;
const arr = [];
for (let i = 0; i <= t.length; i++) {
arr[i] = [i];
for (let j = 1; j <= s.length; j++) {
arr[i][j] =
i === 0
? j
: Math.min(
arr[i - 1][j] + 1,
arr[i][j - 1] + 1,
arr[i - 1][j - 1] + (s[j - 1] === t[i - 1] ? 0 : 1)
);
}
}
return arr[t.length][s.length];
};
levenshteinDistance('duck', 'dark'); // 2