https://leetcode.com/problems/3sum-smaller/
3Sum Smaller
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.
For example, given nums = [-2, 0, 1, 3], and target = 2.
Return 2. 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?
Code:
class Solution(object):
def threeSumSmaller(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if nums is None or len(nums) <=2:
return 0
nums.sort()
cnt = 0
for k in xrange(2, len(nums)):
i, j = 0, k-1
while i < j:
if nums[i] + nums[j] + nums[k] < target:
cnt += j - i
i += 1
j = k-1
else:
j -= 1
return cnt
Idea:
This is O(N^2) solution. You first sort nums in O(nlogn). Then, for every `k` in xrange(2, len(nums)), you start to pivot `i` at 0, and j on the left of `k`. Since nums is already a sorted array, if nums[i] + nums[j] + nums[k] < target, then fixing `i` and `k`, moving the index `j` between`i+1` and `j` can always satisfy the inequality. On the other hand, if nums[i] + nums[j] + nums[k] >= target, you need to move j one position left and retry. The algorithm is O(N^2) because for each `k`, i and j will be traversed until they meet together in the middle of nums, i.e. `i` and `j` will be traversed together in O(N).