triplet-sum-close-to-target.py

# Problem Statement
# Given an array of unsorted numbers and a target number, find a triplet in the array whose sum is as close to the target number as possible, return the sum of the triplet. If there are more than one such triplet, return the sum of the triplet with the smallest sum.

# Example 1:

# Input: [-2, 0, 1, 2], target = 2
# Output: 1
# Explanation: The triplet[-2, 1, 2] has the closest sum to the target.
# Example 2:

# Input: [-3, -1, 1, 2], target = 1
# Output: 0
# Explanation: The triplet[-3, 1, 2] has the closest sum to the target.
# Example 3:

# Input: [1, 0, 1, 1], target = 100
# Output: 3
# Explanation: The triplet[1, 1, 1] has the closest sum to the target.



import math


def triplet_sum_close_to_target(arr, target_sum):
  arr.sort()
  smallest_difference = math.inf
  for i in range(len(arr)-2):
    left = i + 1
    right = len(arr) - 1
    while (left < right):
      target_diff = target_sum - arr[i] - arr[left] - arr[right]
      if target_diff == 0:  # we've found a triplet with an exact sum
        return target_sum  # return sum of all the numbers

      # the second part of the following 'if' is to handle the smallest sum when we have more than one solution
      if abs(target_diff) < abs(smallest_difference) or (abs(target_diff) == abs(smallest_difference) and target_diff > smallest_difference):
        smallest_difference = target_diff  # save the closest and the biggest difference

      if target_diff > 0:
        left += 1  # we need a triplet with a bigger sum
      else:
        right -= 1  # we need a triplet with a smaller sum

  return target_sum - smallest_difference


def main():
  print(triplet_sum_close_to_target([-2, 0, 1, 2], 2))
  print(triplet_sum_close_to_target([-3, -1, 1, 2], 1))
  print(triplet_sum_close_to_target([1, 0, 1, 1], 100))


main()