## Three sum algorithm solution

sum of three leetcode

3 sum zero interviewbit solution

2 sum

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triple sum hackerrank solution

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4 sum

Original Problem Statement:

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4], A solution set is: [[-1, 0, 1], [-1, -1, 2]]

Hello I solved the Two Sum problem on LeetCode some time back and I thought of using it to solve the three sum as well. My idea is that for every element find two elements in the remaining list which sum up to element * -1 to get 0. However, this code doesn't pass all the test, for example

Input: [-4,-2,-2,-2,0,1,2,2,2,3,3,4,4,6,6] Output: [[-4,-2,6],[-4,0,4],[-4,1,3],[-4,2,2]] Expected: [[-4,-2,6],[-4,0,4],[-4,1,3],[-4,2,2],[-2,-2,4],[-2,0,2]]

I don't really know what's wrong. Can someone be kind enough to explain the problem to me. Thank you

class Solution(object): def threeSum(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ def twoSum(self, nums, target): targ = target for index, i in enumerate(nums): targ -= i if targ in nums[index+1:]: return [nums[index], nums[nums[index+1:].index(targ)+index+1]] else: targ = target return None res = [] for index, i in enumerate(nums): target = i * -1 num = nums[:index] + nums [index+1:] ans = twoSum(self, num, target) if ans != None: temp = ans + [i] temp.sort() res.append(temp) print(res) import itertools res.sort() res = list(res for res,_ in itertools.groupby(res)) return res

Original Question: https://leetcode.com/problems/3sum/description/

The one I'm using to solve this: https://leetcode.com/problems/two-sum/description/

using `itertools`

.

import itertools stuff = [-1, 0, 1, 2, -1, -4] stuff.sort() ls = [] for subset in itertools.combinations(stuff, 3): if sum(list(subset))==0: # first I have sorted the list because of grouping # Ex: [-1, 0, 1] and [0, 1, -1] are build with the same element # so here is avoiding this. if list(subset) not in ls: ls.append(list(subset)) print(ls)

**input/output**

input : [-1, 0, 1, 2, -1, -4] output : [[-1, -1, 2], [-1, 0, 1]] input : [-4,-2,-2,-2,0,1,2,2,2,3,3,4,4,6,6] output: [[-4, -2, 6], [-4, 0, 4], [-4, 1, 3], [-4, 2, 2], [-2, -2, 4], [-2, 0, 2]]

**Solving the classic Two-Sum and Three-Sum Problem in JavaScript,** Now lets solve the Three-Sum Problem It's conjectured to be impossible to find an O(n^(2−ϵ)) algorithm for solving 3SUM, for any positive Then the algorithm will run until the last pointer points to the remaining term, giving the affirmative solution. Variants Non-zero sum. Instead of looking for numbers whose sum is 0, it is possible to look for numbers whose sum is any constant C in the following way: Subtract C/3 from all elements of the input array.

Here's another way of solving it which has **O(n^2)** time complexity and passes the LeetCode test. It counts the occurrences and then sorts `(number, count)`

tuples so `[-1, 0, 1, 2, -1, -4]`

becomes `[(-4, 1), (-1, 2), (0, 1), (1, 1), (2, 1)]`

. Then it iterates from beginning picking first trying to pick each number twice and third greater if possible and add this to result. Then it picks number once and tries to find two greater numbers which sum to 0.

from collections import Counter class Solution(object): def threeSum(self, nums): res = [] counts = Counter(nums) num_counts = sorted(counts.items()) # Handle the only case where we pick three same nums if counts[0] >= 3: res.append([0] * 3) for i, (first, first_count) in enumerate(num_counts): # Pick two of these and one greater if first_count >= 2 and first < 0 and -(first * 2) in counts: res.append([first, first, -(first * 2)]) # Pick one and two greater for j in range(i + 1, len(num_counts)): second, second_count = num_counts[j] # Pick two of these as second and third num if second_count >= 2 and -first == 2 * second: res.append([first, second, second]) # Pick this as second num and third which is greater third = -(first + second) if third > second and third in counts: res.append([first, second, third]) return res

**3SUM,** Three Number Sum Problem solution in Java. METHOD 1. Naive approach: Use three for loops. The naive approach is to just use three nested For example, given array S = [-1, 0, 1, 2, -1, -4], A solution set is: [[-1, 0, 1], [-1, -1, 2]] Hello I solved the Two Sum problem on LeetCode some time back and I thought of using it to solve the three sum as well. My idea is that for every element find two elements in the remaining list which sum up to element * -1 to get 0.

I'd like to add the answer you claimed(in the comment) to post:

class Solution(object): def threeSum(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ def twoSum(self, nums, target): targ = target for index, i in enumerate(nums): targ -= i # if targ in nums[index+1:]: # return [nums[index], nums[nums[index+1:].index(targ)+index+1]] _num = nums[:index] + nums[index+1:] if targ in _num: return [i, targ] else: targ = target return None res = [] for index, i in enumerate(nums): target = i * -1 num = nums[:index] + nums [index+1:] ans = twoSum(self, num, target) if ans != None: temp = ans + [i] temp.sort() res.append(temp) print(res) import itertools res.sort() res = list(res for res,_ in itertools.groupby(res)) return res

I have only run it my brain and hope it is right.

**Three Number Sum Problem (Find all triplets with the given sum ,** The solution for 3sum, a popular tech interview question. vector<vector<int>> threeSum(const vector<int>& nums) (the values that k skipped over) j0+ki>0 (otherwise the algorithm wouldn't have skipped over those k). 1 Answer1. active oldest votes. 4. A couple of improvements you can make to your algorithm: 1) Use sets instead of a list for your solution. Using a set will insure that you don't have any duplicate and you don't have to do a if new_solution not in solutions: check. 2) Add an edge case check for an all zero list.

One of the approach is using the HashSet, What I have try here:

public List<List<Integer>> threeSum(int[] nums) { Set<List<Integer>> set = new HashSet<>(); Arrays.sort(nums); for (int i = 0; i < nums.length - 1; i++) { int j = i + 1; int k = nums.length - 1; while (j < k) { int sum = nums[i] + nums[j] + nums[k]; if (sum == 0) { set.add(Arrays.asList(nums[i], nums[j++], nums[k--])); } else if (sum > 0) { k--; } else if (sum < 0) { j++; } } } return new ArrayList<>(set); } }

**3sum Solution - Top Interview Questions and Tricks #1,** Find all unique triplets in the array which gives the sum of zero. Given array nums = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ] So, we essentially need to find three numbers x, y, and z such that they add up to the given value. Solution. Submissions. Discuss (112) 923. 3Sum With Multiplicity. Medium. Add to List. Given an integer array A, and an integer target, return the number of tuples i, j, k such that i < j < k and A [i] + A [j] + A [k] == target. As the answer can be very large, return it modulo 10^9 + 7.

**3Sum,** using itertools . import itertools stuff = [-1, 0, 1, 2, -1, -4] stuff.sort() ls = [] for subset in itertools.combinations(stuff, 3): if sum(list(subset))==0: # first Find all unique triplets in the array which gives the sum of zero. Note: The solution set must not contain duplicate triplets. Example: Given array nums = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]

**Three sum algorithm solution,** The solution set must not contain duplicate triplets. For example public List<List<Integer>> threeSum(int[] nums) { Arrays.sort(nums); Category >> Algorithms. Find subarray with given sum | Set 1 (Non-negative Numbers) | GeeksforGeeks - Duration: 2:50. GeeksforGeeks 28,762 views

**LeetCode – 3Sum,** The following code implements this simple method using three nested loops. Algorithm: Given an array of length n and a sum s; Create three nested loop first loop runs from start to end (loop counter i), second loop runs from i+1 to end (loop counter j) and third loop runs from j+1 to end (loop counter k)

##### Comments

- Please include the description of the problem in your question, not as a link to another site. If that site goes down, or changes URL patterns, your question will not be understood in the future.
- @Phrogz Noted and changes made, thanks for the input
- I figured out the problem is that the iteration misses out on any combination that might occur with the same element twice. I'll try to figure a way around it and post the answer here.
- Perfect, I figured a couple of ways to do it. However, I wanted to do it by using the two sum method. Your method works for the test case provided although I get a time limit exceeded when I submit it on LeetCode for this input [-7,-10,-1,3,0,-7,-9,-1,10,8,-6,4,14,-8,9,-15,0,-4,-5,9,11,3,-5,-8,2,-6,-14,7,-14,10,5,-6,7,11,4,-7,11,11,7,7,-4,-14,-12,-13,-14,4,-13,1,-15,-2,-12,11,-14,-2,10,3,-1,11,-5,1,-2,7,2,-10,-5,-8,-10,14,10,13,-2,-9,6,-7,-7,7,12,-5,-14,4,0,-11,-8,2,-6,-13,12,0,5,-15,8,-12,-1,-4,-15,2,-5,-9,-7,12,11,6,10,-6,14,-12,9,3,-10,10,-8,-2,6,-9,7,7,-7,4,-8,5,-4,8,0,3,11,0,-10,-9]
- Ran out of space, this is not a problem with the code but with LeetCode so I'll accept your answer. Thank you for the solution
- @someRandomGuy, Thanks for appreciate and happy codding +1.
- As pointed out by Alex Hall in stackoverflow.com/questions/50900899/…,
`itertools.combinations(stuff, 3)`

contains O(n^3) elements, so this solution has cubic time complexity. (Was the reason LeetCode didn't accept this solution by any chance "Time Limit Exceeded"?)