Today is an example of The non-Designer's Design Book on bottom of Page 72-73 and Distribute Candies to People on leetcode and leetcode's October LeetCoding Challenge with cpp.

The non-Designer's Design Book

my answer:

• [F] change font
• [F] add background for the shop name
• [F] use circle for shop name decoration
• [T] use left align and right align
• [T] use conspicuous way to display SOMEBODY NEEDS A HUG
• [F] use · to separate contact information
• [T] remove abundant TELEPHONE
• [T] use sentence case for URL
• [F] use part of pie image

Distribute Candies to People

Description

We distribute some number of candies, to a row of n = num_people people in the following way:

We then give 1 candy to the first person, 2 candies to the second person, and so on until we give n candies to the last person.

Then, we go back to the start of the row, giving n + 1 candies to the first person, n + 2 candies to the second person, and so on until we give 2 * n candies to the last person.

This process repeats (with us giving one more candy each time, and moving to the start of the row after we reach the end) until we run out of candies. The last person will receive all of our remaining candies (not necessarily one more than the previous gift).

Return an array (of length num_people and sum candies) that represents the final distribution of candies.

Example 1:

Input: candies = 7, num_people = 4
Output: [1,2,3,1]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0,0].
On the third turn, ans[2] += 3, and the array is [1,2,3,0].
On the fourth turn, ans[3] += 1 (because there is only one candy left), and the final array is [1,2,3,1].

Example 2:

Input: candies = 10, num_people = 3
Output: [5,2,3]
Explanation: 
On the first turn, ans[0] += 1, and the array is [1,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0].
On the third turn, ans[2] += 3, and the array is [1,2,3].
On the fourth turn, ans[0] += 4, and the final array is [5,2,3].

Constraints:

• 1 <= candies <= 10^9
• 1 <= num_people <= 1000

Solution

using binary search to find after which turn candies is over, then distribute these turns to people.

class Solution {
public:
  vector<int> distributeCandies(int candies, int num_people) {
    vector<int> answer(num_people);
    int begin = 0, end=sqrt(candies)*1.6;
    while (begin < end) {
      int mid = (begin + end)/2;
      if(1LL*mid*(mid+1)/2 > candies) {
        end = mid;
      } else {
        begin = mid + 1;
      }
    }
    int turns = end - 1;
    int rounds = turns / num_people, remainder = turns % num_people, last = candies - 1LL*turns*(turns+1)/2;
    for(int i = 0; i < num_people; ++i) {
      answer[i] = (i+1) * rounds + 1LL*(rounds-1)*rounds*num_people/2;
      if(i < remainder) {
        answer[i] += rounds*num_people + i+1;
      }
    }
    answer[remainder] += last;
    return answer;
  }
};

October LeetCoding Challenge 6

Description

Insert into a Binary Search Tree

You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

Example 1:

Input: root = [4,2,7,1,3], val = 5
Output: [4,2,7,1,3,5]
Explanation: Another accepted tree is:

Example 2:

Input: root = [40,20,60,10,30,50,70], val = 25
Output: [40,20,60,10,30,50,70,null,null,25]

Example 3:

Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
Output: [4,2,7,1,3,5]

Constraints:

• The number of nodes in the tree will be in the range [0, 104].
• -108 <= Node.val <= 108
• All the values Node.val are unique.
• -108 <= val <= 108
• It's guaranteed that val does not exist in the original BST.

Solution

my code is somewhat ugly, rewrite it later.

class Solution {
public:
  TreeNode* insertIntoBST(TreeNode* root, int val) {
    if(!root) {
      root = new TreeNode(val);
    }
    if(val < root->val) {
      if(root->left) {
        insertIntoBST(root->left, val);
      } else {
        root->left = new TreeNode(val);
      }
    } else if(val > root->val) {
      if(root->right) {
        insertIntoBST(root->right, val);
      } else {
        root->right = new TreeNode(val);
      }
    }
    return root;
  }
};