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linear-list/array/3sum-smaller

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2023-12-01

3Sum Smaller

描述

Given an array of n integers nums and an integer target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

Follow up: Could you solve it in O($n^2$) runtime?

Example 1:

Input: nums = [-2,0,1,3], target = 2

Output: 2

Explanation: Because there are two triplets which sums are less than 2:

[-2,0,1]

[-2,0,3]

Example 2:

Input: nums = [], target = 0

Output: 0

Example 3:

Input: nums = [0], target = 0

Output: 0

Constraints:

  • n == nums.length
  • 0 <= n <= 300
  • -100 <= nums[i] <= 100
  • -100 <= target <= 100

分析

先排序,然后双指针左右夹逼,复杂度 $$O(n^2)$$。

代码

# 3Sum Smaller
# 先排序,然后双指针左右夹逼
# Time Complexity: O(n^2)
# Space Complexity: from O(logn) to O(n), depending on the
# implementation of the sorting algorithm
class Solution:
    def threeSumSmaller(self, nums: List[int], target: int) -> int:
        nums.sort()
        count = 0
        for i in range(len(nums)-2):
            count += self.twoSumSmaller(nums, i, target - nums[i])
        return count

    def twoSumSmaller(self, nums: List[int], i: int, target: int) -> int:
        count = 0
        left, right = i + 1, len(nums) - 1
        while left < right:
            if nums[left] + nums[right] < target:
                count += right - left
                left += 1
            else:
                right -= 1
        return count
// 3Sum Smaller
// 先排序,然后双指针左右夹逼
// Time Complexity: O(n^2)
// Space Complexity: from O(logn) to O(n), depending on the
// implementation of the sorting algorithm
class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        if(nums.length < 3) return 0;
        Arrays.sort(nums);
        int count = 0;
        for (int i = 0; i < nums.length - 2; i++) {
            count += twoSumSmaller(nums, i, target - nums[i]);
        }
        return count;
    }

    private int twoSumSmaller(int[] nums, int i, int target) {
        int count = 0;
        int left = i + 1, right = nums.length - 1;
        while (left < right) {
            if (nums[left] + nums[right] < target) {
                count += right - left;
                left++;
            } else {
                right--;
            }
        }
        return count;
    }
}
// 3Sum Smaller
// 先排序,然后双指针左右夹逼
// Time Complexity: O(n^2)
// Space Complexity: from O(logn) to O(n), depending on the
// implementation of the sorting algorithm
class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        if(nums.size() < 3) return 0;
        sort(nums.begin(), nums.end());
        int count = 0;
        for (int i = 0; i < nums.size() - 2; i++) {
            count += twoSumSmaller(nums, i, target - nums[i]);
        }
        return count;
    }
private:
    int twoSumSmaller(const vector<int>& nums, int i, int target) {
        int count = 0;
        int left = i + 1, right = nums.size() - 1;
        while (left < right) {
            if (nums[left] + nums[right] < target) {
                count += right - left;
                left++;
            } else {
                right--;
            }
        }
        return count;
    }
};

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