LeetCode 0259. 3Sum Smaller Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

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
• The input is generated such that the answer is less than or equal to 10⁹.

Solutions

Solution 1: Sorting + Two Pointers + Enumeration

Since the order of elements does not affect the result, we can sort the array first and then use the two-pointer method to solve this problem.

First, we sort the array and then enumerate the first element nums[i]. Within the range nums[i+1:n-1], we use two pointers pointing to nums[j] and nums[k], where j is the next element of nums[i] and k is the last element of the array.

• If nums[i] + nums[j] + nums[k] < target, then for any element j \lt k' ≤ k, we have nums[i] + nums[j] + nums[k'] < target. There are k - j such k', and we add k - j to the answer. Next, move j one position to the right and continue to find the next k that meets the condition until j ≥ k.
• If nums[i] + nums[j] + nums[k] ≥ target, then for any element j ≤ j' \lt k, it is impossible to make nums[i] + nums[j'] + nums[k] < target. Therefore, we move k one position to the left and continue to find the next k that meets the condition until j ≥ k.

After enumerating all i, we get the number of triplets that meet the condition.

The time complexity is O(n^2), and the space complexity is O(log n). Here, n is the length of the array nums.

PythonJavaC++GoTypeScriptJavaScript

class Solution: def threeSumSmaller(self, nums: List[int], target: int) -> int: nums.sort() ans, n = 0, len(nums) for i in range(n - 2): j, k = i + 1, n - 1 while j < k: x = nums[i] + nums[j] + nums[k] if x < target: ans += k - j j += 1 else: k -= 1 return ans(code-box)

class Solution { public int threeSumSmaller(int[] nums, int target) { Arrays.sort(nums); int ans = 0, n = nums.length; for (int i = 0; i + 2 < n; ++i) { int j = i + 1, k = n - 1; while (j < k) { int x = nums[i] + nums[j] + nums[k]; if (x < target) { ans += k - j; ++j; } else { --k; } } } return ans; } }(code-box)

class Solution { public: int threeSumSmaller(vector<int>& nums, int target) { ranges::sort(nums); int ans = 0, n = nums.size(); for (int i = 0; i + 2 < n; ++i) { int j = i + 1, k = n - 1; while (j < k) { int x = nums[i] + nums[j] + nums[k]; if (x < target) { ans += k - j; ++j; } else { --k; } } } return ans; } };(code-box)

func threeSumSmaller(nums []int, target int) (ans int) { sort.Ints(nums) n := len(nums) for i := 0; i < n-2; i++ { j, k := i+1, n-1 for j < k { x := nums[i] + nums[j] + nums[k] if x < target { ans += k - j j++ } else { k-- } } } return }(code-box)

function threeSumSmaller(nums: number[], target: number): number { nums.sort((a, b) => a - b); const n = nums.length; let ans = 0; for (let i = 0; i < n - 2; ++i) { let [j, k] = [i + 1, n - 1]; while (j < k) { const x = nums[i] + nums[j] + nums[k]; if (x < target) { ans += k - j; ++j; } else { --k; } } } return ans; }(code-box)

/** * @param {number[]} nums * @param {number} target * @return {number} */ var threeSumSmaller = function (nums, target) { nums.sort((a, b) => a - b); const n = nums.length; let ans = 0; for (let i = 0; i < n - 2; ++i) { let [j, k] = [i + 1, n - 1]; while (j < k) { const x = nums[i] + nums[j] + nums[k]; if (x < target) { ans += k - j; ++j; } else { --k; } } } return ans; };(code-box)