missing-number.py

"""

268. Missing Number
Easy

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

 

Example 1:

Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:

Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
Example 3:

Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
 

Constraints:

n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
All the numbers of nums are unique.
 

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

"""

# V0 
class Solution(object):
    def missingNumber(self, nums):
        return sum(range(len(nums)+1)) - sum(nums)

# V1 
# http://bookshadow.com/weblog/2015/08/24/leetcode-missing-number/
class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        return n * (n + 1) / 2 - sum(nums)

# V1' 
# http://bookshadow.com/weblog/2015/08/24/leetcode-missing-number/
# IDEA : BIT MANIPULATION 
# IDEA : XOR 
# https://stackoverflow.com/questions/14526584/what-does-the-xor-operator-do/14526640
# XOR is a binary operation, it stands for "exclusive or", that is to say the resulting bit evaluates to one if only exactly one of the bits is set.
# "XOR" is same as "^"
# a | b | a ^ b
# --|---|------
# 0 | 0 | 0
# 0 | 1 | 1
# 1 | 0 | 1
# 1 | 1 | 0
class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        a = reduce(operator.xor, nums)
        b = reduce(operator.xor, range(len(nums) + 1))
        return a ^ b

# V2 
class Solution(object):
    def missingNumber(self, nums):
        return sum(range(len(nums)+1)) - sum(nums)

# V3 
import operator
from functools import reduce
class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return reduce(operator.xor, nums, \
                      reduce(operator.xor, range(len(nums) + 1)))