# 259. 3Sum Smaller

## Description

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.

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 <= 3500
• -100 <= nums[i] <= 100
• -100 <= target <= 100

## Solutions

• class Solution {
public int threeSumSmaller(int[] nums, int target) {
Arrays.sort(nums);
int ans = 0;
for (int i = 0, n = nums.length; i < n; ++i) {
int j = i + 1;
int k = n - 1;
while (j < k) {
int s = nums[i] + nums[j] + nums[k];
if (s >= target) {
--k;
} else {
ans += k - j;
++j;
}
}
}
return ans;
}
}

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

• class Solution:
def threeSumSmaller(self, nums: List[int], target: int) -> int:
nums.sort()
ans, n = 0, len(nums)
for i in range(n):
j, k = i + 1, n - 1
while j < k:
s = nums[i] + nums[j] + nums[k]
if s >= target:
k -= 1
else:
ans += k - j
j += 1
return ans


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

• /**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var threeSumSmaller = function (nums, target) {
nums.sort((a, b) => a - b);
let ans = 0;
for (let i = 0, n = nums.length; i < n; ++i) {
let j = i + 1;
let k = n - 1;
while (j < k) {
s = nums[i] + nums[j] + nums[k];
if (s >= target) {
--k;
} else {
ans += k - j;
++j;
}
}
}
return ans;
};