Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead. Example: Input: s = 7, nums = [2,3,1,2,4,3] Output: 2 Explanation: the subarray [4,3] has the minimal length under the problem constraint. Follow up: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1. Example 1 Input: coins = [1, 2, 5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1 Example 2 Input: coins = [2], amount = 3 Output: -1 Note You may assume that you have an infinite number of each kind of coin.
Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product. Example 1 Input: [2,3,-2,4] Output: 6 Explanation: [2,3] has the largest product 6. Example 2 Input: [-2,0,-1] Output: 0 Explanation: The result cannot be 2, because [-2,-1] is not a subarray. Solution # time: `O(n)` class Solution: def maxProduct(self, nums: List[int]) -> int: n = len(nums) if n == 0: return 0 res = currMax = currMin = nums[0] for i in range(1, n): newCurrMax = max( max(currMax * nums[i], currMin * nums[i]), nums[i] ) newCurrMin = min( min(currMax * nums[i], currMin * nums[i]), nums[i] ) currMax, currMin = newCurrMax, newCurrMin res = max(currMax, res) return res Solution (need to return the subarray) class Solution: def maxProduct(self, nums: List[int]) -> int: n = len(nums) if n == 0: return 0 res = prevMin = prevMax = nums[0] maxPos = 0 minLens = [1 for _ in range(n)] maxLens = [1 for _ in range(n)] print(prevMin, prevMax) for i in range(1, n): num = nums[i] currMin, currMax = prevMin, prevMax if num > 0: if prevMax * num > num: currMax *= num maxLens[i] = maxLens[i-1] + 1 else: currMax = num if prevMin * num < num: currMin *= num minLens[i] = minLens[i-1] + 1 else: currMin = num else: if prevMin * num > num: currMax = prevMin * num maxLens[i] = minLens[i-1] + 1 else: currMax = num if prevMax * num < num: currMin = prevMax * num minLens[i] = maxLens[i-1] + 1 else: currMin = num prevMin, prevMax = currMin, currMax if prevMax > res: maxPos = i res = prevMax print(prevMin, prevMax) print(minLens) print(maxLens) maxLen = maxLens[maxPos] maxSubarr = nums[maxPos+1-maxLen:maxPos+1] print(maxSubarr) return res
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle [ [2], [3,4], [6,5,7], [4,1,8,3] ] The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). Note Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess. The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note You can only move either down or right at any point in time. Example Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation Because the path 1→3→1→1→1 minimizes the sum. Solution 1 (DP with n extra space) Python class Solution: def minPathSum(self, grid: List[List[int]]) -> int: m = len(grid) n = len(grid[0]) if m == 0 or n == 0: return 0 sums = [[0 for _ in range(n)] for _ in range(m)] sums[0][0] = grid[0][0] for i in range(1, m): sums[i][0] = sums[i-1][0] + grid[i][0] for i in range(1, n): sums[0][i] = sums[0][i-1] + grid[0][i] for i in range(1, m): for j in range(1, n): sums[i][j] = min(sums[i-1][j], sums[i][j-1]) + grid[i][j] return sums[-1][-1] Java
Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0. If there are multiple solutions, return any subset is fine. Example 1 Input: [1,2,3] Output: [1,2] (of course, [1,3] will also be ok) Example 2 Input: [1,2,4,8] Output: [1,2,4,8] Solution class Solution: def largestDivisibleSubset(self, nums: List[int]) -> List[int]: if not nums or len(nums) == 0: return [] nums.
You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note Given n will be a positive integer. Example 1 Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps Example 2 Input: 3 Output: 3 Explanation: There are three ways to climb to the top.
Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle. Example 1 Input: [2,2,3,4] Output: 3 Explanation: Valid combinations are: 2,3,4 (using the first 2) 2,3,4 (using the second 2) 2,2,3 Note The length of the given array won’t exceed 1000. The integers in the given array are in the range of [0, 1000].
Given inorder and postorder traversal of a tree, construct the binary tree. Note You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 20 / \ 15 7 Hint Solution: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ // the idea is to start from the rightmst part of the postorder because that is always the root; then divide the inorder as per the value of the root call the right child first because after accesing the parent in postorder the right child is encountered first and then the left child class Solution { int postOrderIndex; public TreeNode buildTree(int[] inorder, int[] postorder) { postOrderIndex = postorder.