[LeetCode][python3]0015. 3Sum

Start the journey
N2I -2020.03.21

1. My first try

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        ans=[]
        nums.sort()
        #print(nums)
        for i in range(len(nums)-2):
            if nums[i]>0:
                break
            if i==0 or nums[i]>nums[i-1]:
                j=i+1
                k=len(nums)-1
                while j<k:
                    if nums[i]+nums[j]+nums[k] == 0:
                        ans.append([nums[i],nums[j],nums[k]])
                        j+=1
                        k-=1
                        while j<k and nums[j]==nums[j-1]:
                            j+=1
                        while j<k and nums[k]==nums[k+1]:
                            k-=1
                    elif -nums[i]>nums[j]+nums[k]:
                        j+=1
                    else:
                        k-=1
        #print(ans)
        return ans

N2I -2020.03.21

Explanation:

The classic way to solve this problem. The range between 500 ms to 1100 ms are probably the same method. First you sort the array, then search the array with three pointers.

2. A Recursive Solution

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        if not nums:
            return []
        nums = sorted(nums)
        res = []
        self.kSum(nums, 0, len(nums)-1, 3, 0, [], res)
        return res
        
    
    def kSum(self, nums, left, right, k, target, state, res):
        # if the selecting range is not enough
        if k < 2 or right - left + 1 < k:
            return
        if k * nums[left] > target or k * nums[right] < target:
            return 
        if k == 2:
            # Solving 2Sum problem
            while left < right:
                cur = nums[left] + nums[right]
                if cur == target:
                    res.append(state + [nums[left], nums[right]])
                    left += 1
                    while left < right and nums[left] == nums[left - 1]:
                        left += 1
                elif cur > target:
                    right -= 1
                else:
                    left += 1
        else:
            # k > 2, reduce the degree
            for i in range(left, right+1):
                if i == left or i > left and nums[i] != nums[i-1]:
                    self.kSum(nums, i+1, right, k-1, target-nums[i], state+[nums[i]], res)

N2I -2020.03.21

Explanation:

The solution is a classic recursive way to solve this kind of problems. It reduce its degree in every recursion. Puts any situation into a 2 Sum problem.

3. A Better Solution

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        ans = set()
        count = {}
        for i in nums:
            if i not in count:  count[i] = 1
            else:  count[i] += 1
        nums = sorted(count)
        for i, item_i in enumerate(nums):
            if item_i < 0:
                target = -item_i
                left = bisect.bisect_left(nums, target - nums[-1], i + 1)
                for item_j in nums[left:bisect.bisect_right(nums, target // 2, left)]:
                    item_k = target - item_j
                    if item_k in count and item_k != item_j:
                        ans.add((item_i, item_j, item_k))
            if count[item_i] > 1:
                if item_i == 0:
                    if count[0] > 2:
                        ans.add((0, 0, 0))
                else:
                    if -2 * item_i in count:
                        ans.add((item_i, item_i, -2 * item_i))
        return ans

N2I -2020.03.21

Note:

• bisect.bisect_left(a,x,lo=0,hi=len(a)) A function using binary search to find the target value x and return the index of it. If there are multiple x , return the first index. If there are none, return the index x should be insert in and remain the array sorted. a is the searching array. x is the target. lo and hi is the searching area of a.

print("example of bisect_left")
import bisect
a=[0,1,2,3,3,4,6,7]
x=4;print(bisect.bisect_left(a,x))
    #>>5
x=3;print(bisect.bisect_left(a,x))
    #>>3
x=5;print(bisect.bisect_left(a,x))
    #>>6
x=5;print(bisect.bisect_left(a,x,2,4))
    #>>4

Explanation:

The solution is probably the fastest solution among the solutions in Leetcode. It is a close methon to the first solution above, but it reduce the searching area to (target+nums[-1],target//2), that reduces time spent in every for loop.