259. 3Sum Smaller π
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 $\textit{nums}$.
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