Three numbers in an array that sum up to a target sum in O(n)

Recently hashedin asked the triplet sum problem where three numbers are supposed to add up to a target sum. They said to do it in O(n).

I have tried doing it in O(n^2). First, i sorted the array and then for searching the combination i had to apply sliding window technique to all of the elements in the array. I'm not able to reduce it to O(n).

def threeNumberSum(array, targetSum):
    array.sort()
    l = len(array)
    trip = list()
    for i in range(l-2):
        left = i+1
        right = len(array)-1
        while left<right:
            currentSum = array[i] + array[left] + array[right]
            if currentSum == targetSum:
                trip.append([array[i], array[left], array[right]])
                left += 1
                right -= 1
            elif currentSum < targetSum:
                left += 1
            else:
                right -= 1
    return trip

This code actually returns all the combinations of sums possible. But according to the company, only one triplet is needed. All possible triplets are not needed

Thanks, everyone for trying to help. I have come up with an initial solution to do it in O(n). I am not yet sure if it's correct, but it's running on all the test case Here is a Github link to download the python file github.com/TripletInArrayWithTargetSum

def trip(arr, target):
    d = dict()
    n = len(arr)
    for i in arr:
        d[i] = 1

    i = 0
    j = n - 1
    while i<j:
        s = arr[i] + arr[j]     # Calculate the sum of the elements at i and j position
        diff = target - s       # Calculate the difference needed to complete the table
        if arr[i] != diff and arr[j] != diff and diff in d: # Check if the difference exists in the hashtable and as we cannot reuse elements
            return [arr[i], arr[j], diff]                   # diff should not be equal to elements at i or j and  then return the triplet if exists
        else:
            if s > target:                                  # If difference dosent exists then we adjust pointers
                j -= 1
            elif s == target:
                if arr[i+1] + arr[j] < target:
                    i+=1
                else:
                    j-=1
            else:
                i += 1

    return [None]

Download full file with test cases on Github