Reverse a string in Java
0 likes • Nov 20, 2022
JavaScript
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const getTimezone = () => Intl.DateTimeFormat().resolvedOptions().timeZone;// ExamplegetTimezone();
//JavaScript program to swap two variables//take input from the userslet a = prompt('Enter the first variable: ');let b = prompt('Enter the second variable: ');// XOR operatora = a ^ bb = a ^ ba = a ^ bconsole.log(`The value of a after swapping: ${a}`);console.log(`The value of b after swapping: ${b}`);
//JavaScript program to swap two variables//take input from the userslet 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: ${b}`);
// There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9.// You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where:// The first character of the ith pair denotes the ith ring's color ('R', 'G', 'B').// The second character of the ith pair denotes the rod that the ith ring is placed on ('0' to '9').// For example, "R3G2B1" describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1.// Return the number of rods that have all three colors of rings on them.let rings = "B0B6G0R6R0R6G9";var countPoints = function(rings) {let sum = 0;// Always 10 Rodsfor (let i = 0; i < 10; i++) {if (rings.includes(`B${i}`) && rings.includes(`G${i}`) && rings.includes(`R${i}`)) {sum+=1;}}return sum;};console.log(countPoints(rings));
class SequentialQueue{ //if you want it to go backwards, too badnext(){return this.i++;}constructor(start = 0){this.i = start;}}const que = new SequentialQueue(0);for(let i = 0; i < 10; i++){console.log(que.next());}
// Time Complexity : O(N)// Space Complexity : O(1)var isMonotonic = function(nums) {let isMono = null;for(let i = 1; i < nums.length; i++) {if(isMono === null) {if(nums[i - 1] < nums[i]) isMono = 0;else if(nums[i - 1] > nums[i]) isMono = 1;continue;}if(nums[i - 1] < nums[i] && isMono !== 0) {return false;}else if(nums[i - 1] > nums[i] && isMono !== 1) {return false;}}return true;};let nums1 = [1,2,2,3]let nums2 = [6,5,4,4]let nums3 = [1,3,2]console.log(isMonotonic(nums1));console.log(isMonotonic(nums2));console.log(isMonotonic(nums3));