README_EN.md

259. 3Sum Smaller

中文文档

Description

Given an array of n integers nums and a 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.

Example:

Input: nums = [-2,0,1,3], and target = 2
Output: 2 
Explanation: Because there are two triplets which sums are less than 2:
             [-2,0,1]
             [-2,0,3]

Follow up: Could you solve it in O(n²) runtime?

Solutions

Python3

class Solution:
    def threeSumSmaller(self, nums: List[int], target: int) -> int:
        def threeSumSmaller(nums, start, end, target):
            count = 0
            while start < end:
                if nums[start] + nums[end] < target:
                    count += (end - start)
                    start += 1
                else:
                    end -= 1
            return count

        nums.sort()
        n, count = len(nums), 0
        for i in range(n - 2):
            count += threeSumSmaller(nums, i + 1, n - 1, target - nums[i])
        return count

Java

class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        Arrays.sort(nums);
        int n = nums.length;
        int count = 0;
        for (int i = 0; i < n - 2; ++i) {
            count += threeSumSmaller(nums, i + 1, n - 1, target - nums[i]);
        }
        return count;
    }

    private int threeSumSmaller(int[] nums, int start, int end, int target) {
        int count = 0;
        while (start < end) {
            if (nums[start] + nums[end] < target) {
                count += (end - start);
                ++start;
            } else {
                --end;
            }
        }
        return count;
    }
}

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