Continuous Subarray Sum

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

Note:

1. The length of the array won’t exceed 10,000.
2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

class Solution(object):
    def checkSubarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        sum, pre, s = 0, 0, set()

        for item in nums:
            sum += item
            if (k):
                sum %= k

            if (sum in s):
                return True

            s.add(pre)
            pre = sum

        return False

Convert a Number to Hexadecimal

Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, two’s complement method is used.

Note:

1. All letters in hexadecimal (a-f) must be in lowercase.
2. The hexadecimal string must not contain extra leading 0s. If the number is zero, it is represented by a single zero character '0'; otherwise, the first character in the hexadecimal string will not be the zero character.
3. The given number is guaranteed to fit within the range of a 32-bit signed integer.
4. You must not use any method provided by the library which converts/formats the number to hex directly.

Example 1:

Input:
26

Output:
"1a"

Example 2:

Input:
-1

Output:
"ffffffff"

/**
 * @param {number} num
 * @return {string}
 */
var toHex = function(num) {
  if (num > 0)
    return help(num);
  else if (num === 0)
    return '0';
  else {
    num = -num;
    return help(0xffffffff - num + 1);
  }

  function help(num) {
    let arr = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'];
    let ans = '';

    while (num) {
      let mod = num % 16;
      ans = arr[mod] + ans;
      num = ~~(num / 16);
    }

    return ans;
  }
};

Convert BST to Greater Tree

Given a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.

Example:

Input: The root of a Binary Search Tree like this:
              5
            /   \
           2     13

Output: The root of a Greater Tree like this:
             18
            /   \
          20     13

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def convertBST(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        def dfs(node):
            if node == None:
                return
            else:
                dfs(node.right)
                node.val += dfs.total
                dfs.total = node.val
                dfs(node.left)

        dfs.total = 0
        dfs(root)
        return root