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

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 $\textit{nums}[i]$. Within the range $\textit{nums}[i+1:n-1]$, we use two pointers pointing to $\textit{nums}[j]$ and $\textit{nums}[k]$, where $j$ is the next element of $\textit{nums}[i]$ and $k$ is the last element of the array.

If $\textit{nums}[i] + \textit{nums}[j] + \textit{nums}[k] < \textit{target}$, then for any element $j \lt k' \leq k$, we have $\textit{nums}[i] + \textit{nums}[j] + \textit{nums}[k'] < \textit{target}$. 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 $j \geq k$.
If $\textit{nums}[i] + \textit{nums}[j] + \textit{nums}[k] \geq \textit{target}$, then for any element $j \leq j' \lt k$, it is impossible to make $\textit{nums}[i] + \textit{nums}[j'] + \textit{nums}[k] < \textit{target}$. Therefore, we move $k$ one position to the left and continue to find the next $k$ that meets the condition until $j \geq k$.

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 $\textit{nums}$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

JavaScript

/**
 * @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;
};