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Nov 19, 2022CodeCatch
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More JavaScript Posts

localStorage Cookie Consent

Oct 9, 2023Helper

0 likes • 261 views

import React, { useState, useEffect } from 'react';
import Link from 'next/link';
export default function CookieBanner() {
// Initialize with a default value (false if not previously set in localStorage)
const [cookieConsent, setCookieConsent] = useState(() =>
localStorage.getItem('cookieConsent') === 'true' ? true : false
);
useEffect(() => {
const newCookieConsent = cookieConsent ? 'granted' : 'denied';
window.gtag('consent', 'update', {
analytics_storage: newCookieConsent
});
localStorage.setItem('cookieConsent', String(cookieConsent));
}, [cookieConsent]);
return !cookieConsent && (
<div>
<div>
<p>
We use cookies to enhance the user experience.{' '}
<Link href='/privacy/'>View privacy policy</Link>
</p>
</div>
<div>
<button type="button" onClick={() => setCookieConsent(false)}>Decline</button>
<button type="button" onClick={() => setCookieConsent(true)}>Allow</button>
</div>
</div>
)
}

Longest Palindromic Substring

Nov 19, 2022CodeCatch

0 likes • 2 views

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"))

LeetCode Two Sum

Sep 13, 2023C S

0 likes • 2 views

/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function(nums, target) {
for(let i = 0; i < nums.length; i++) {
for(let j = 0; j < nums.length; j++) {
if(nums[i] + nums[j] === target && i !== j) {
return [i, j]
}
}
}
};

Composing components

Nov 19, 2022CodeCatch

0 likes • 3 views

function Welcome(props) {
return <h1>Hello, {props.name}</h1>;
}
function App() {
return (
<div>
<Welcome name="Sara" />
<Welcome name="Cahal" />
<Welcome name="Edite" />
</div>
);
}
ReactDOM.render(
<App />,
document.getElementById('root')
);

Tree Traversal

Oct 16, 2023C S

0 likes • 7 views

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);

geometric progression

Nov 19, 2022CodeCatch

0 likes • 0 views

const geometricProgression = (end, start = 1, step = 2) =>
Array.from({
length: Math.floor(Math.log(end / start) / Math.log(step)) + 1,
}).map((_, i) => start * step ** i);
geometricProgression(256); // [1, 2, 4, 8, 16, 32, 64, 128, 256]
geometricProgression(256, 3); // [3, 6, 12, 24, 48, 96, 192]
geometricProgression(256, 1, 4); // [1, 4, 16, 64, 256]