3630.two-sum-less-than-target.py

#  Tag: Binary Search, Two Pointers, Array
#  Time: O(NlogN)
#  Space: O(1)
#  Ref: Leetcode-1099
#  Note: -

#  Given an array of integers `nums` and an integer `target`, if there exists `i < j` such that `nums[i] + nums[j] < target`, then return **the maximum and  that satisfy this condition**.
#  If no `i, j` satisfying the above condition exists, then `-1` is returned.
#  
#  Example 1:
#  ```
#  Input:
#  nums = [2, 7, 11, 15], target = 24
#  Output:
#  22
#  Explanation:
#  7 + 15 = 22 < 24
#  11 + 15 = 26 > 24
#  22 is the maximum sum that satisfies the condition
#  ```
#  
#  Example 2:
#  ```
#  Input:
#  nums = [3, 5, 1, 9, 7], target = 3
#  Output:
#  -1
#  Explanation:
#  Cannot find two elements whose sum is less than 3
#  ```
#  
#  $1 \leq nums.length \leq 1000$
#  $-1000 \leq nums[i] \leq 1000$
#  $-2000 \leq target \leq 2000$

from typing import (
    List,
)

class Solution:
    """
    @param nums: An array of integer
    @param target: An integer
    @return: The sum of two numbers smaller than target
    """
    def two_sum_less_than_target(self, nums: List[int], target: int) -> int:
        # write your code here
        nums.sort()
        l = 0
        r = len(nums) - 1
        res = float('-inf')
        while l < r:
            if nums[l] + nums[r] < target:
                res = max(res, nums[l] + nums[r])
                l += 1
            else:
                r -= 1
        return -1 if res == float('-inf') else res