# 1099. Two Sum Less Than K

## Description

Given an array nums of integers and integer k, return the maximum sum such that there exists i < j with nums[i] + nums[j] = sum and sum < k. If no i, j exist satisfying this equation, return -1.

Example 1:

Input: nums = [34,23,1,24,75,33,54,8], k = 60
Output: 58
Explanation: We can use 34 and 24 to sum 58 which is less than 60.


Example 2:

Input: nums = [10,20,30], k = 15
Output: -1
Explanation: In this case it is not possible to get a pair sum less that 15.


Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 1000
• 1 <= k <= 2000

## Solutions

Solution 1: Sorting + Binary Search

We can first sort the array $nums$, and initialize the answer as $-1$.

Next, we enumerate each element $nums[i]$ in the array, and find the maximum $nums[j]$ in the array that satisfies $nums[j] + nums[i] < k$. Here, we can use binary search to speed up the search process. If we find such a $nums[j]$, then we can update the answer, i.e., $ans = \max(ans, nums[i] + nums[j])$.

After the enumeration ends, return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $nums$.

Solution 2: Sorting + Two Pointers

Similar to Solution 1, we can first sort the array $nums$, and initialize the answer as $-1$.

Next, we use two pointers $i$ and $j$ to point to the left and right ends of the array, respectively. Each time we judge whether $s = nums[i] + nums[j]$ is less than $k$. If it is less than $k$, then we can update the answer, i.e., $ans = \max(ans, s)$, and move $i$ one step to the right, otherwise move $j$ one step to the left.

After the enumeration ends, return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public int twoSumLessThanK(int[] nums, int k) {
Arrays.sort(nums);
int ans = -1;
int n = nums.length;
for (int i = 0; i < n; ++i) {
int j = search(nums, k - nums[i], i + 1, n) - 1;
if (i < j) {
ans = Math.max(ans, nums[i] + nums[j]);
}
}
return ans;
}

private int search(int[] nums, int x, int l, int r) {
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}

• class Solution {
public:
int twoSumLessThanK(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
int ans = -1, n = nums.size();
for (int i = 0; i < n; ++i) {
int j = lower_bound(nums.begin() + i + 1, nums.end(), k - nums[i]) - nums.begin() - 1;
if (i < j) {
ans = max(ans, nums[i] + nums[j]);
}
}
return ans;
}
};

• class Solution:
def twoSumLessThanK(self, nums: List[int], k: int) -> int:
nums.sort()
ans = -1
for i, x in enumerate(nums):
j = bisect_left(nums, k - x, lo=i + 1) - 1
if i < j:
ans = max(ans, x + nums[j])
return ans


• func twoSumLessThanK(nums []int, k int) int {
sort.Ints(nums)
ans := -1
for i, x := range nums {
j := sort.SearchInts(nums[i+1:], k-x) + i
if v := nums[i] + nums[j]; i < j && ans < v {
ans = v
}
}
return ans
}

• function twoSumLessThanK(nums: number[], k: number): number {
nums.sort((a, b) => a - b);
let ans = -1;
for (let i = 0, j = nums.length - 1; i < j; ) {
const s = nums[i] + nums[j];
if (s < k) {
ans = Math.max(ans, s);
++i;
} else {
--j;
}
}
return ans;
}