class Solution(object):

def findErrorNums(self, nums):

"""

:type nums: List[int]

:rtype: List[int]

"""

###### Method 0

"""

# Time limit exceeded, everything else passed

dup = miss = 0

for i in range(1,len(nums)+1):

if nums.count(i) == 0:

miss = i

if nums.count(i) == 2:

dup = i

return [dup,miss]

"""

###### Method 1

"""

# Pythonic using sums

# 100 pass - using total sum - the set sum which is unique => missing number

dup = sum(nums) - sum(set(nums)) # Total sum - unique sum == number left out or duplicate

miss = sum(range(1,len(nums)+1)) - sum(set(nums)) # total range sum - unique sum == missing number

return [dup,miss]

"""

###### Method 2

"""

# Extra array or dictionary 100 pass

counts = {}

dup = miss = 0

for n in nums:

if n not in counts:

counts[n] = 0

counts[n] += 1

#print counts

for i in range(1,len(nums)+1):

if i not in counts:

miss = i

if i in counts and counts[i] == 2:

dup = i

return [dup, miss]

"""

#### 100pass - Method 3 - Use sorting ==> you thought of sorting technique and sum technique but hesitated to implement them wondering if complexity will increase. First of all use brute force and bring about the solution in a naive manner then worry about everything else

nums = sorted(nums)

dup = 0

miss = 1

for i in range(len(nums)):

if nums[i] == nums[i-1]:

dup = nums[i]

else:

if nums[i] > nums[i-1] + 1:

miss = nums[i-1] + 1

if nums[len(nums)-1] != len(nums):

miss = len(nums)

return [dup,miss]

#### Lessons learnt:

# * Analysis and learn when to use if and if vs if and else

# * static values like miss = 1 is set deliberately by the logic when a programmer writes, analyze why??

# * FIRST == BRUTEFORCE

# * Always optimize using a logic you think of, dont avoid any thoughts like sum and sort ones you thought.