Flatten an array
0 likes • Nov 19, 2022
JavaScript
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const celsiusToFahrenheit = (celsius) => celsius * 9/5 + 32;const fahrenheitToCelsius = (fahrenheit) => (fahrenheit - 32) * 5/9;// ExamplescelsiusToFahrenheit(15); // 59celsiusToFahrenheit(0); // 32celsiusToFahrenheit(-20); // -4fahrenheitToCelsius(59); // 15fahrenheitToCelsius(32); // 0
var invertTree = function(root) {const reverseNode = node => {if (node == null) {return null}reverseNode(node.left);reverseNode(node.right);let holdLeft = node.left;node.left = node.right;node.right = holdLeft;return node;}return reverseNode(root);};
const binomialCoefficient = (n, k) => {if (Number.isNaN(n) || Number.isNaN(k)) return NaN;if (k < 0 || k > n) return 0;if (k === 0 || k === n) return 1;if (k === 1 || k === n - 1) return n;if (n - k < k) k = n - k;let res = n;for (let j = 2; j <= k; j++) res *= (n - j + 1) / j;return Math.round(res);};binomialCoefficient(8, 2); // 28
// 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 Rodsfor (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));
new StringBuilder(hi).reverse().toString()
const powerset = arr =>arr.reduce((a, v) => a.concat(a.map(r => r.concat(v))), [[]]);powerset([1, 2]); // [[], [1], [2], [1, 2]]