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?
Companies:
Google
Related Topics:
Array, Two Pointers
Similar Questions:
// OJ: https://leetcode.com/problems/3sum-smaller/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
private:
int twoSumSmaller(vector<int>& nums, int start, int target) {
int i = start, j = nums.size() - 1, ans = 0;
while (i < j) {
if (nums[i] + nums[j] < target) {
ans += j - i;
++i;
} else --j;
}
return ans;
}
public:
int threeSumSmaller(vector<int>& nums, int target) {
int N = nums.size(), ans = 0;
sort(nums.begin(), nums.end());
for (int i = 0; i < N; ++i)
ans += twoSumSmaller(nums, i + 1, target - nums[i]);
return ans;
}
};