3Sum

Problem Highlights

• 🔗 Leetcode Link: 3Sum
• 💡 Difficulty: Medium
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Array, Two Pointer
• 🗒️ Similar Questions: Number of Arithmetic Triplets

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Is this question similar to Two Sum and Two Sum II?
  • Yes, it's a good idea to take a first look at Two Sum and Two Sum II. Two Sum, Two Sum II and 3Sum share a similarity that the sum of elements must match the target exactly. A difference is that, instead of exactly one answer, we need to find all unique triplets that sum to zero.
• Can you modify the input array?
  • Yes, you can especially if you want to avoid copying it due to memory constraints.

HAPPY CASE
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Example 2:
Input: nums = [0,1,1]
Output: []

EDGE CASE
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Array problems, we want to consider the following approaches:

• Sort.
  • Sort the array so we can skip repeated values.
• Two pointer solutions (left and right pointer variables).
  • It requires the array to be first sorted. To make sure the result contains unique triplets, we need to skip duplicate values. It is easy to do because repeating values are next to each other in a sorted array. ), sorting the array would not change the overall time complexity. Repeat until we reach index 0 for nums2.
• Storing the elements of the array in a HashMap or a Set.
  • A HashMap or Set just complicates our code.
• Traversing the array with a sliding window. Similar to the two pointer solution.
  • A sliding window doesn't really help us here.

⚠️ Common Mistakes

• The requirements asks to return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. You may notice that there a solution where j == k is possible. i, j and k are indexes and not the elements of the array. You cannot use same element more than once in the triplet, but you can definitely use same number on different indexes. Remember you've to return triplet of elements and not triplet of indexes unlike the two sum problem where you've to return pair of indexes.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: use the two pointers pattern to make sure the result contains unique triplets

First, sort the array.
Have two pointers l and r where l starts from i + 1 and r in reverse. ie len(nums)-1. Here i is the index of n.
As the array is sorted, if n + nums[l] + nums[r] > 0 then decreare r or else decrease l.
If equal to zero, append to ans.


n == nums[i-1] explanation:
if 2 numbers are same, skip the number.
eg [-1,0,1,2,-1,-4]
After sorting, [-4,-1,-1,0,1,2]
Here -1 is repeated. So for i = 2, n = -1, nums[i-1] == n. Thus move to next step (or else it will create duplicates)

4: I-mplement

Implement the code to solve the algorithm.

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:

        nums.sort()  # will make spoting of duplicates easy

        triplets = []
        length = len(nums)

        for i in range(length-2):  # ignore last two

            # check if element is a duplicate. the first cannot be a duplicate
            if i > 0 and nums[i] == nums[i-1]:
                # skip handling an element if it's similar to the one before it
                # because it is sorted, we effectively skip duplicates
                continue

            # TWO SUM for a sorted array
            # 1. find elements that will add up to 0
            # 2. check inner elements
            left = i + 1
            right = length - 1
            while left < right:

                # will be used to check if the sum is equal to 0
                total = nums[i] + nums[left] + nums[right]

                # if total is less than 0 we try to increase it's value
                if total < 0:
                    left += 1  # moving left to a lerger value

                # if total is more than 0 we try to decrease it's value
                elif total > 0:
                    right -= 1  # moving right to a smaller value

                # 1. add list of elements to triplets
                # 2. check inner elements
                else:
                    # add elements to triplets
                    triplets.append([nums[i], nums[left], nums[right]])

                    # check inner elements
                    # 1. skip similar elements
                    # 2. move to inner elements

                    # skip:
                    # no need to continue with an element with the same value as l/r
                    # Skip all similar to the current left and right so that,
                    # when we are moving to the next element, we dont move to an element with the same value
                    while left < right and nums[left] == nums[left+1]:
                        left += 1
                    while left < right and nums[right] == nums[right-1]:
                        right -= 1

                    # move to inner elements
                    left += 1
                    right -= 1

        return triplets

class Solution {
    public List<List<Integer>> threeSum(int[] arr) {
        
        Arrays.sort(arr);
        
        List<List<Integer>> ls = new ArrayList<>();
        
        for (int i = 0; i<arr.length-2; i++) {
            if (i == 0 || (i>0 && arr[i-1]!=arr[i])) { //check for duplicates to avoid copy we've used arr[i-1]!=arr[i] instead of arr[i+1]!=arr[i] because we must take the duplicate value in condition i.e. in example 1 [-1,-1,2] is also and output so if we do arr[i+1]!=arr[i] then we'll skip to the next -1 and this output will not come
            int start = i+1;
            int end = arr.length-1;
			int sum = 0-arr[i];
            
            while (start<end) {
                if (arr[start] + arr[end] == sum) {
                    ls.add(Arrays.asList(arr[i], arr[start], arr[end]));
                    
                    while (start<end && arr[start] == arr[start+1]) start++;//avoid all the same values
                    while (start<end && arr[end] == arr[end-1]) end--;//avoid all the same values
                    
                    start++;
                    end--;
                } else if (arr[start] + arr[end] < sum) {
                    start++;
                } else end--;
            }
            }
        }
        
        return ls;
    }
    
    
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Given n elements,

• Time Complexity: O(n^2). Sorting the array takes O(nlog⁡n), so overall complexity is O(nlog⁡n+n2)+ n^2.
• Space Complexity: O(n)