Question

1099. Two Sum Less Than K

中文文档

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 $nums$.

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

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;
  }
};

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;
}

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:
  def twoSumLessThanK(self, nums: List[int], k: int) -> int:
    nums.sort()
    i, j = 0, len(nums) - 1
    ans = -1
    while i < j:
      if (s := nums[i] + nums[j]) < k:
        ans = max(ans, s)
        i += 1
      else:
        j -= 1
    return ans

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

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

func twoSumLessThanK(nums []int, k int) int {
  sort.Ints(nums)
  ans := -1
  for i, j := 0, len(nums)-1; i < j; {
    if s := nums[i] + nums[j]; s < k {
      ans = max(ans, s)
      i++
    } else {
      j--
    }
  }
  return ans
}