Two pointers problems

n ^ 2 solution

class Solution(object):

def twoSum(self, nums, target):

for i in range(len(nums)):

for j in range(len(nums)):

if nums[i] + nums[j] == target and i != j:

return [i, j]

return [-1, -1]

hash map O(n)

class Solution(object):

def twoSum(self, nums, target):

m = {}

for i, n in enumerate(nums):

m[n] = i

for i in range(len(nums)):

diff = target - nums[i]

if diff in m and i != m[diff]:

return [m[diff], i]

return [-1, -1]

one pass

class Solution(object):

def twoSum(self, nums, target):

m = {}

for i in range(len(nums)):

diff = target - nums[i]

if diff in m:

return [m[diff], i]

m[nums[i]] = i

return [-1, -1]

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