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Nov 16, 2023LeifMessinger
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More JavaScript Posts

LeetCode Product Except Self

Sep 13, 2023C S

0 likes • 11 views

/**
* @param {number[]} nums
* @return {number[]}
*/
var productExceptSelf = function(nums) {
var output = [];
var leftMult = 1;
var rightMult = 1;
for (var i=nums.length - 1; i >= 0; i--) {
output[i] = rightMult;
rightMult *= nums[i];
console.log({output: JSON.stringify(output), i})
}
for (var j=0; j < nums.length; j++) {
output[j] *= leftMult;
leftMult *= nums[j];
console.log({output: JSON.stringify(output), j})
}
return output;
};
console.log(productExceptSelf([1, 2, 3, 4]))

Rings and Rods

Nov 19, 2022CodeCatch

0 likes • 1 view

// 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));

localStorage Cookie Consent

Oct 9, 2023Helper

0 likes • 259 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>
)
}

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]
}
}
}
};

Destructuring Assignment

Nov 19, 2022CodeCatch

0 likes • 0 views

//JavaScript program to swap two variables
//take input from the users
let a = prompt('Enter the first variable: ');
let b = prompt('Enter the second variable: ');
//using destructuring assignment
[a, b] = [b, a];
console.log(`The value of a after swapping: ${a}`);
console.log(`The value of b after swapping: ${b}`);

JS Test

Mar 9, 2021LeifMessinger

0 likes • 1 view

alert("bruh")