Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get. For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4). Note: You may assume that n is not less than 2 and not larger than 58.

There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it’s off or turning off if it’s on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds. Example: Given n = 3.

Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0. Example 1: Given [“abcw”, “baz”, “foo”, “bar”, “xtfn”, “abcdef”] Return 16 The two words can be “abcw”, “xtfn”. Example 2: Given [“a”, “ab”, “abc”, “d”, “cd”, “bcd”, “abcd”] Return 4 The two words can be “ab”, “cd”.

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity. Example: Given 1->2->3->4->5->NULL, return 1->3->5->2->4->NULL. Note: The relative order inside both the even and odd groups should remain as it was in the input.

Follow up for “Remove Duplicates”: What if duplicates are allowed at most twice? For example, Given sorted array nums = [1,1,1,2,2,3], Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn’t matter what you leave beyond the new length. Solution: public class Solution { public int removeDuplicates(int[] nums) { int i=0, j=0; while(j<nums.length){ while(j+2<nums.length && nums[j]==nums[j+2]) j++;

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. For example, Given input array nums = [1,1,2], Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn’t matter what you leave beyond the new length.

Given an encoded string, return it’s decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k.

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node’s value. If it is a null node, we record using a sentinel value such as #. _9_ / \ 3 2 / \ / \ 4 1 # 6 / \ / \ / \ # # # # # # For example, the above binary tree can be serialized to the string “9,3,4,#,#,1,#,#,2,#,6,#,#”, where # represents a null node.

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it. Note: You may assume k is always valid, 1 ≤ k ≤ BST’s total elements. Follow up: What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine? Hint: Try to utilize the property of a BST. What if you could modify the BST node's structure?

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public void flatten(TreeNode root) { TreeNode cur = root; while(cur!=null){ if(cur.left!=null){ TreeNode r = cur.left; while(r.right!=null) r = r.right; r.right = cur.right; cur.right = cur.left; cur.left = null; } cur = cur.right; }