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# Kirk Chukkchu791

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Last active May 16, 2019
View raindrops.rb
 # Raindrops Convert a number to a string, the contents of which depend on the number's factors. - If the number has 3 as a factor, output 'Pling'. - If the number has 5 as a factor, output 'Plang'. - If the number has 7 as a factor, output 'Plong'. - If the number does not have 3, 5, or 7 as a factor, just pass the number's digits straight through.
Created May 15, 2019
View questions.md

Backend App - Auction

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``````ruby '2.6.3'
``````
Created Apr 23, 2019
View tree.rb
 ass TreeNode attr_accessor :value, :left, :right def initialize(value) @value = value end end tree = TreeNode.new(5) tree.left = TreeNode.new(3)
Created Mar 19, 2019
View minCostClimbingStairs.js
 var minCostClimbingStairs = function(cost) { for (let i = 2; i < cost.length; i++) { cost[i] += Math.min(cost[i-1], cost[i-2]); } return Math.min(cost[cost.length-1], cost[cost.length-2]); };
Created Mar 19, 2019
recursive ways
View house_robber3.js
 var rob = function(nums) { }
Last active Mar 19, 2019
another solution in linear time
View house_robber2.js
 var rob = function(nums) { var prevEvenMax = 0; var prevOddMax = 0; for (var i = 0; i < nums.length; i++) { if (i % 2 === 0) { prevEvenMax = Math.max(prevEvenMax + nums[i], prevOddMax); } else { prevOddMax = Math.max(prevEvenMax, prevOddMax + nums[i]);
Created Mar 19, 2019
linear time solution
View house_robber.js
 var rob = function(profits) { let prevMaxProfit = 0; let currMaxProfit = 0; for (let i = 0; i < profits.length; i++) { const prevPlusProfit = prevMaxProfit + profits[i]; prevMaxProfit = currMaxProfit; currMaxProfit = Math.max(currMaxProfit, prevPlusProfit); }
Last active Mar 18, 2019
linear solving of maximum subarray problem (dynamic programming)
View maximum_subarray.js
 // Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. // Example: // Input: [-2,1,-3,4,-1,2,1,-5,4], // Output: 6 // Explanation: [4,-1,2,1] has the largest sum = 6. // Follow up: // If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
Created Feb 24, 2019
Using insertion sort
View squares_of_sorted_array.js
 // Given an array of integers A sorted in non-decreasing order, // return an array of the squares of each number, also in sorted non-decreasing order. // array = [-4,-1,0,3,10] const sortedSquares = (array) => { // square the elements in the array const squaredArray = array.map(el => el * el); // use insertion sort to sort the array from min to max
Created Feb 6, 2019
View two_sums
 # nums = [2, 3, 11, 22], target = 9, # Because nums[0] + nums[1] = 2 + 7 = 9, #return [0, 1] def two_sum(nums, target) nums.each do |i| nums.each do |x| if i + x == target && nums.index(i) != nums.index(x) return [nums.index(i), nums.index(x)] end
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