Python Two Sum - Brute Force Approach

Question

I\'m new to Python and have just started to try out LeetCode to build my chops. On this classic question my code misses a test case.

The problem is as follows:

Answer 1

Bit different approach. We will build a dictionary of values as we need them, which is keyed by the values we are looking for.If we look for a value we track the index of that value when it first appears. As soon as you find the values that satisfy the problem you are done. The time on this is also O(N)

class Solution:
    def twoSum(self, nums, target):
        look_for = {}
        for n,x in enumerate(nums):
            try:
                return look_for[x], n
            except KeyError:
                look_for.setdefault(target - x,n)

test_case = Solution()
array = [1, 5, 7]
array2 = [3,2,4]
given_nums=[2,7,11,15]
print(test_case.twoSum(array, 6))
print(test_case.twoSum(array2, 6))
print(test_case.twoSum(given_nums,9))

output:

(0, 1)
(1, 2)
(0, 1)

Answer 2

import itertools

class Solution:
    def twoSum(self, nums, target):
        subsets = []
        for L in range(0, len(nums)+1):
            for subset in itertools.combinations(nums, L):
                if len(subset)!=0:
                    subsets.append(subset)
        print(subsets) #returns all the posible combinations as tuples, note not permutations!
        #sums all the tuples
        sums = [sum(tup) for tup in subsets]
        indexes = []
        #Checks sum of all the posible combinations
        if target in sums:
            i = sums.index(target)
            matching_combination = subsets[i] #gets the option
            for number in matching_combination:
                indexes.append(nums.index(number))
            return indexes
        else:
            return None


test_case = Solution()    
array = [1,2,3]
print(test_case.twoSum(array, 4))

I was trying your example for my own learning. I am happy with what I found. I used the itertools to make all the combination of the numbers for me. Then I used list manipulation to sum all the possible combination of numbers in your input array, then I just check in one shot if the target is inside the sum array or not. If not then return None, return the indexes otherwise. Please note that this approach will return all the three indexes as well, if they add up to the target. Sorry it took so long :)

Answer 3

This one is more comprehensive cohesive efficient one even so shorter lines of code.

nums = [6, 7, 11, 15, 3, 6, 5, 3,99,5,4,7,2]
target = 27
n = 0

for i in range(len(nums)):
    
    n+=1
    if n == len(nums):
      n == len(nums)
      
    else:
        if nums[i]+nums[n] == target:
          # to find the target position 
          print([nums.index(nums[i]),nums.index(nums[n])])
          # to get the actual numbers to add print([nums[i],nums[n]])

Answer 4

class Solution:
    def twoSum(self, nums, target):
            """
            :type nums: List[int]
            :type target: int
            :rtype: List[int]
            """
            ls=[]
            l2=[]
            for i in nums:
                ls.append(target-i)

            for i in range(len(ls)):
                if ls[i] in nums  :
                    if i!= nums.index(ls[i]):
                        l2.append([i,nums.index(ls[i])])            
            return l2[0]


x= Solution()
x.twoSum([-1,-2,-3,-4,-5],-8)

output

[2, 4]

Answer 5

A brute force solution is to double nest a loop over the list where the inner loop only looks at index greater than what the outer loop is currently on.

class Solution:
    def twoSum(self, nums, target):
        for i, a in enumerate(nums, start=0):
            for j, b in enumerate(nums[i+1:], start=0):
                if a+b==target:
                    return [i, j+i+1]

test_case = Solution()
array = [3, 2, 4]
print(test_case.twoSum(array, 6))

array = [1, 5, 7]
print(test_case.twoSum(array, 6))

array = [2, 7, 11, 15]
print(test_case.twoSum(array, 9))

Output:

[1, 2]
[0, 1]
[0, 1]