I am trying to solve the twoSum LeetCode problem via a brute force algorithm. Essentially the question is to find the indexes of two numbers that equal to the target. So if I had a list [7,2,4,1] and my target was 9, I would return index of 7 & 9 (which in this ex is [0,1]) as the solution.

I have written this algorithm here:

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:

    if (len(nums) < 2):
        return False

    count = 0
    for i in range(len(nums)):
        for j in range (i 1, len(nums)):
            if nums[i]   nums[j] == target:
                ans =  str(nums.index(nums[j]))   str(nums.index(nums[i]))
                return ans
            count = count   1

It works for everything except the one scenario where two elements in a list are duplicates. So if the list contained [3,3] with target 6, my algorithm is returning index [0,0] instead of [0,1]. What is causing this issue and what would be the proper solution to fixing it?

CodePudding user response：

ans =  str(nums.index(nums[j]))   str(nums.index(nums[i]))

This happens because you try to obtain the index of 3 on this line twice, which will return index 0 twice. Why not return string of i and j itself since they are already indices? So:

ans =  str(j)   str(i)

CodePudding user response：

The problem is not from your algorithm but in the way you translate the answer in string (ans = str(nums.index(nums[j])) str(nums.index(nums[i])))

l.index(i) will return the first index of l containing the value i, so when the value is duplicated, nums.index(nums[j])==nums.index(nums[i]), I believe you actually want to have ans = str(j) str(i) instead