Three Sum Problem All Unique Triplets in DSA Go - Deep Dive

Overview - Three Sum Problem All Unique Triplets

What is it?

The Three Sum Problem asks us to find all unique groups of three numbers in a list that add up to zero. Each group is called a triplet. We want to list all such triplets without repeating any. This problem helps us understand how to efficiently search and combine elements in arrays.

Why it matters

Without this problem-solving approach, checking all possible triplets would take too long, especially for big lists. This problem teaches us how to reduce unnecessary work and avoid duplicates, which is important in many real-world tasks like finding combinations or patterns in data. Without it, programs would be slower and less efficient.

Where it fits

Before learning this, you should know about arrays, sorting, and two-pointer techniques. After mastering this, you can explore more complex combination problems and optimization techniques in algorithms.

Mental Model

Core Idea

Find three numbers that sum to zero by sorting the list and using two pointers to efficiently find pairs that complement each chosen number.

Think of it like...

Imagine you have a list of people with different weights, and you want to find three people whose combined weight is exactly zero (like balancing a seesaw). You first arrange them by weight, then pick one person and look for two others who balance the total weight.

Sorted array: [-4, -1, -1, 0, 1, 2]
Choose first number (i): -4
Set two pointers: left = i+1, right = end
Move pointers inward to find pairs where sum + i == 0

  i
  ↓
[-4, -1, -1, 0, 1, 2]
     ↑           ↑
   left        right

If sum too small, move left rightwards
If sum too big, move right leftwards
If sum zero, record triplet and skip duplicates

Build-Up - 7 Steps

1

Foundation

Understanding the Problem Setup

Concept: Learn what the problem asks: find all unique triplets that sum to zero.

Given an array of integers, we want to find all groups of three numbers where the sum is zero. For example, in [-1, 0, 1, 2, -1, -4], the triplets are [-1, -1, 2] and [-1, 0, 1]. We must avoid duplicates.

Result

Clear understanding of the problem goal and example triplets.

Knowing exactly what to find helps focus on the right approach and avoid confusion about duplicates or sum conditions.

2

Foundation

Sorting the Array to Simplify Searching

Concept: Sort the array to make it easier to find pairs and skip duplicates.

Sorting arranges numbers from smallest to largest. For example, [-1, 0, 1, 2, -1, -4] becomes [-4, -1, -1, 0, 1, 2]. This order helps us use two pointers to find pairs efficiently and detect duplicates by comparing neighbors.

Result

Sorted array ready for efficient searching.

Sorting transforms the problem into a structured form, enabling faster and simpler pair searches.

3

Intermediate

Using Two Pointers to Find Pairs

🤔Before reading on: do you think moving both pointers inward always finds all pairs? Commit to your answer.

Concept: Use two pointers moving inward from both ends to find pairs that sum to a target.

After fixing one number, set two pointers: one just after it (left), one at the end (right). Calculate sum of these two plus fixed number. If sum is less than zero, move left pointer right to increase sum. If sum is more than zero, move right pointer left to decrease sum. If sum equals zero, record triplet and move both pointers skipping duplicates.

Result

Efficiently found pairs that complement the fixed number to sum zero.

Two pointers reduce the search from checking all pairs to a linear scan, greatly improving speed.

4

Intermediate

Skipping Duplicates to Ensure Unique Triplets

🤔Before reading on: do you think skipping duplicates is optional? Commit to your answer.

Concept: Avoid repeating the same triplet by skipping duplicate numbers during iteration and pointer moves.

When the fixed number is the same as the previous, skip it. Similarly, after finding a valid triplet, move left and right pointers past any duplicates. This prevents adding the same triplet multiple times.

Result

List of triplets with no duplicates.

Skipping duplicates is crucial to meet the problem's requirement and avoid redundant work.

5

Intermediate

Implementing the Complete Algorithm in Go

Concept: Combine sorting, iteration, two pointers, and duplicate skipping into a working Go function.

Write a Go function that sorts the array, loops through each number as fixed, uses two pointers to find pairs, and skips duplicates. Return all unique triplets found.

Result

A runnable Go function that outputs all unique triplets summing to zero.

Seeing the full code connects all concepts and shows how they work together in practice.

6

Advanced

Optimizing for Early Stopping and Edge Cases

🤔Before reading on: do you think checking all numbers is always necessary? Commit to your answer.

Concept: Improve efficiency by stopping early when fixed number is positive and handling small arrays.

Since array is sorted, if fixed number is greater than zero, no triplet can sum to zero (all numbers after are positive). Also handle arrays with fewer than three elements by returning empty result immediately.

Result

Algorithm runs faster by avoiding unnecessary checks.

Understanding problem constraints allows skipping impossible cases, saving time.

7

Expert

Memory and Performance Considerations in Large Inputs

🤔Before reading on: do you think this algorithm uses extra memory proportional to input size? Commit to your answer.

Concept: Analyze memory use and performance trade-offs when input size grows very large.

The algorithm sorts in place (O(n log n)) and uses O(1) extra space besides output. However, output size can grow large if many triplets exist. Understanding this helps in systems with memory limits or when output size matters. Also, consider parallelizing or using approximate methods for huge data.

Result

Awareness of algorithm limits and practical considerations for big data.

Knowing resource use helps choose or adapt algorithms for real-world constraints.

Under the Hood

The algorithm first sorts the array to arrange numbers in order. Then it iterates through each number as a fixed point. For each fixed number, it uses two pointers starting just after the fixed number and at the end of the array. These pointers move inward based on the sum compared to zero. This two-pointer approach exploits the sorted order to find pairs in linear time. Duplicate numbers are skipped by comparing current values to previous ones, ensuring unique triplets. The sorting step dominates time complexity at O(n log n), and the two-pointer search runs in O(n^2) overall.

Why designed this way?

Originally, checking all triplets would take O(n^3) time, which is too slow for large inputs. Sorting and two pointers reduce this to O(n^2), a big improvement. Skipping duplicates avoids repeated work and redundant results. This design balances simplicity, efficiency, and correctness, making it practical for many applications.

Input array
  ↓
[unsorted]
  ↓ sort
[sorted array]
  ↓
For each index i:
  ├─ fixed number = arr[i]
  ├─ left = i+1
  └─ right = end

While left < right:
  ├─ sum = fixed + arr[left] + arr[right]
  ├─ if sum < 0: left++
  ├─ if sum > 0: right--
  └─ if sum == 0: record triplet, left++, right-- skipping duplicates

Myth Busters - 4 Common Misconceptions

Quick: Do you think sorting the array is optional for this problem? Commit to yes or no.

Common Belief: Some think we can find triplets without sorting by just checking all combinations.

Reality: Sorting is essential to efficiently find pairs with two pointers and to skip duplicates easily.

Why it matters: Without sorting, the algorithm would be much slower (O(n^3)) and harder to avoid duplicates, leading to poor performance and incorrect results.

Quick: Do you think duplicate triplets are automatically avoided without extra checks? Commit to yes or no.

Common Belief: People often believe that just finding triplets will naturally avoid duplicates.

Reality: Duplicates must be explicitly skipped by checking adjacent equal numbers during iteration and pointer moves.

Why it matters: Failing to skip duplicates causes repeated triplets in output, violating problem requirements and confusing users.

Quick: Do you think the two-pointer method works if the array is not sorted? Commit to yes or no.

Common Belief: Some assume two pointers can be used on any array regardless of order.

Reality: Two-pointer technique relies on sorted order to decide pointer moves based on sum comparisons.

Why it matters: Using two pointers on unsorted arrays leads to incorrect results and wasted effort.

Quick: Do you think the algorithm always needs to check every number as fixed? Commit to yes or no.

Common Belief: Many think every number must be fixed and checked for triplets.

Reality: Once the fixed number is positive, no triplet can sum to zero because the array is sorted.

Why it matters: Not stopping early wastes time checking impossible cases, reducing efficiency.

Expert Zone

1

The order of skipping duplicates is subtle: skip duplicates for the fixed number before the two-pointer search, and also skip duplicates for left and right pointers after finding a valid triplet.

2

The algorithm's time complexity is O(n^2), but the actual runtime depends heavily on input distribution and number of duplicates.

3

In some cases, using a hash set to store triplets can simplify duplicate checks but increases memory usage and may slow down performance.

When NOT to use

This approach is not suitable when the input is extremely large and memory is limited, or when approximate solutions are acceptable. Alternatives include hashing-based methods or probabilistic algorithms for approximate triplet sums.

Production Patterns

In production, this algorithm is used in recommendation systems to find balanced combinations, in financial software to detect zero-sum transactions, and in coding interviews as a classic problem to test problem-solving and optimization skills.

Connections

Two Sum Problem

Builds-on

Understanding the simpler Two Sum problem helps grasp the two-pointer technique used in Three Sum, as Three Sum fixes one number and reduces the problem to Two Sum.

Sorting Algorithms

Depends-on

Efficient sorting is critical for the Three Sum algorithm's performance, showing how sorting underpins many search and combination problems.

Chemical Reaction Balancing

Analogous pattern

Balancing chemical equations involves finding combinations of elements that sum to zero net change, similar to finding triplets summing to zero, illustrating cross-domain problem-solving patterns.

Common Pitfalls

#1 Not sorting the array before searching for triplets.

Wrong approach:

func threeSum(nums []int) [][]int {
var res [][]int
for i := 0; i < len(nums)-2; i++ {
for j := i+1; j < len(nums)-1; j++ {
for k := j+1; k < len(nums); k++ {
if nums[i]+nums[j]+nums[k] == 0 {
res = append(res, []int{nums[i], nums[j], nums[k]})
}
}
}
}
return res
}

Correct approach:

func threeSum(nums []int) [][]int {
sort.Ints(nums)
var res [][]int
for i := 0; i < len(nums)-2; i++ {
if i > 0 && nums[i] == nums[i-1] { continue }
left, right := i+1, len(nums)-1
for left < right {
sum := nums[i] + nums[left] + nums[right]
if sum == 0 {
res = append(res, []int{nums[i], nums[left], nums[right]})
for left < right && nums[left] == nums[left+1] { left++ }
for left < right && nums[right] == nums[right-1] { right-- }
left++
right--
} else if sum < 0 {
left++
} else {
right--
}
}
}
return res
}

Root cause:

Not understanding that sorting is necessary for the two-pointer technique and duplicate skipping.

#2 Failing to skip duplicates for the fixed number and pointers.

Wrong approach:

for i := 0; i < len(nums)-2; i++ {
left, right := i+1, len(nums)-1
for left < right {
sum := nums[i] + nums[left] + nums[right]
if sum == 0 {
res = append(res, []int{nums[i], nums[left], nums[right]})
left++
right--
} else if sum < 0 {
left++
} else {
right--
}
}
}

Correct approach:

for i := 0; i < len(nums)-2; i++ {
if i > 0 && nums[i] == nums[i-1] { continue }
left, right := i+1, len(nums)-1
for left < right {
sum := nums[i] + nums[left] + nums[right]
if sum == 0 {
res = append(res, []int{nums[i], nums[left], nums[right]})
for left < right && nums[left] == nums[left+1] { left++ }
for left < right && nums[right] == nums[right-1] { right-- }
left++
right--
} else if sum < 0 {
left++
} else {
right--
}
}
}

Root cause:

Not realizing duplicates cause repeated triplets and must be explicitly skipped.

#3 Not stopping early when fixed number is positive.

Wrong approach:

for i := 0; i < len(nums)-2; i++ {
// no check for nums[i] > 0
left, right := i+1, len(nums)-1
// two-pointer search
}

Correct approach:

for i := 0; i < len(nums)-2; i++ {
if nums[i] > 0 { break }
left, right := i+1, len(nums)-1
// two-pointer search
}

Root cause:

Not using the sorted property to optimize and stop impossible searches early.

Key Takeaways

The Three Sum problem finds all unique triplets in an array that sum to zero by combining sorting and two-pointer techniques.

Sorting the array is essential to efficiently find pairs and skip duplicates, reducing time complexity from cubic to quadratic.

Skipping duplicates for the fixed number and pointers ensures the output contains only unique triplets.

Early stopping when the fixed number is positive improves performance by avoiding unnecessary checks.

Understanding memory and performance trade-offs helps apply this algorithm effectively in real-world scenarios.

Practice

1. What is the main goal of the Three Sum Problem All Unique Triplets in data structures and algorithms?

easy

A. Count the number of pairs that sum to a target

B. Find all unique triplets in an array that sum to zero

C. Find the maximum element in the array

D. Sort the array in ascending order

Solution

1. Step 1: Understand the problem statement

   The problem asks to find all unique groups of three numbers in an array whose sum is zero.

2. Step 2: Identify the goal

   The goal is to find triplets (three numbers) that add up to zero without duplicates.

3. Final Answer:

   Find all unique triplets in an array that sum to zero -> Option B

4. Quick Check:

   Three Sum Problem = Find unique triplets summing to zero [OK]

Hint: Look for three numbers adding to zero without repeats [OK]

Common Mistakes:

• Confusing with pair sum problems
• Ignoring uniqueness of triplets
• Trying to find max or min instead of sum

2. Which of the following Go code snippets correctly initializes a slice of integers for solving the Three Sum problem?

easy

A. nums := []int{1, -1, 0, 2, -2}

B. nums := [1, -1, 0, 2, -2]

C. nums := (1, -1, 0, 2, -2)

D. nums := {1, -1, 0, 2, -2}

Solution

1. Step 1: Recall Go slice initialization syntax

   In Go, slices are initialized with square brackets and type, e.g., []int{...}.

2. Step 2: Check each option

   nums := []int{1, -1, 0, 2, -2} uses correct syntax: nums := []int{1, -1, 0, 2, -2}. Others use invalid syntax.

3. Final Answer:

   nums := []int{1, -1, 0, 2, -2} -> Option A

4. Quick Check:

   Go slice syntax = []int{...} [OK]

Hint: Use []int{...} to declare integer slices in Go [OK]

Common Mistakes:

• Using parentheses or braces instead of curly braces
• Missing type declaration []int
• Using array literal syntax without type

3. What will be the output of the following Go code snippet that finds unique triplets summing to zero?

func threeSum(nums []int) [][]int {
    sort.Ints(nums)
    var res [][]int
    for i := 0; i < len(nums)-2; i++ {
        if i > 0 && nums[i] == nums[i-1] {
            continue
        }
        left, right := i+1, len(nums)-1
        for left < right {
            sum := nums[i] + nums[left] + nums[right]
            if sum == 0 {
                res = append(res, []int{nums[i], nums[left], nums[right]})
                for left < right && nums[left] == nums[left+1] {
                    left++
                }
                for left < right && nums[right] == nums[right-1] {
                    right--
                }
                left++
                right--
            } else if sum < 0 {
                left++
            } else {
                right--
            }
        }
    }
    return res
}

fmt.Println(threeSum([]int{-1, 0, 1, 2, -1, -4}))

medium

A. [[-4, 2, 2], [-1, 0, 1]]

B. [[0, 1, 2], [-1, -1, 2]]

C. [[-1, -1, 2], [-1, 0, 1]]

D. [[-1, 1, 1], [0, 0, 0]]

Solution

1. Step 1: Sort and iterate through the array

   The sorted array is [-4, -1, -1, 0, 1, 2]. The code skips duplicates and uses two pointers to find triplets summing to zero.

2. Step 2: Identify unique triplets summing to zero

   Triplets found are [-1, -1, 2] and [-1, 0, 1]. These are appended to the result.

3. Final Answer:

   [[-1, -1, 2], [-1, 0, 1]] -> Option C

4. Quick Check:

   Unique triplets summing zero = [[-1,-1,2],[-1,0,1]] [OK]

Hint: Sort array, use two pointers to find zero sum triplets [OK]

Common Mistakes:

• Not sorting array before processing
• Not skipping duplicates causing repeated triplets
• Incorrect pointer movement logic

4. Identify the bug in the following Go code snippet for the Three Sum problem:

func threeSum(nums []int) [][]int {
    sort.Ints(nums)
    var res [][]int
    for i := 0; i < len(nums)-2; i++ {
        left, right := i+1, len(nums)-1
        for left < right {
            sum := nums[i] + nums[left] + nums[right]
            if sum == 0 {
                res = append(res, []int{nums[i], nums[left], nums[right]})
                left++
                right--
            } else if sum < 0 {
                left++
            } else {
                right--
            }
        }
    }
    return res
}

medium

A. Returning res before the loop ends

B. Incorrect initialization of left and right pointers

C. Using sort.Ints instead of sort.Sort

D. Missing check to skip duplicate elements for i, left, and right

Solution

1. Step 1: Analyze duplicate handling

   The code does not skip duplicate values for i, left, or right, which can cause duplicate triplets in the result.

2. Step 2: Confirm other parts are correct

   Initialization of pointers and sorting are correct. The return statement is after loops, so no premature return.

3. Final Answer:

   Missing check to skip duplicate elements for i, left, and right -> Option D

4. Quick Check:

   Skipping duplicates is essential to avoid repeated triplets [OK]

Hint: Always skip duplicates to ensure unique triplets [OK]

Common Mistakes:

• Not skipping duplicates causing repeated triplets
• Incorrect pointer updates after finding a triplet
• Sorting with wrong function

5. Given the input array nums := []int{-2, 0, 1, 2, -1, -1, 3}, which of the following is the correct set of all unique triplets that sum to zero?

hard

A. [[-2, -1, 3], [-2, 0, 2], [-1, -1, 2], [-1, 0, 1]]

B. [[-2, 0, 2], [-1, 1, 1], [0, 0, 0]]

C. [[-2, -1, 3], [-1, 0, 1], [1, 1, -2]]

D. [[-1, -1, 2], [-2, 1, 1], [0, 1, -1]]

Solution

1. Step 1: Sort the array and find triplets

   Sorted array: [-2, -1, -1, 0, 1, 2, 3]. Find unique triplets summing to zero by two-pointer method.

2. Step 2: Identify all unique triplets

   Triplets are [-2, -1, 3], [-2, 0, 2], [-1, -1, 2], and [-1, 0, 1]. These cover all unique zero sums.

3. Final Answer:

   [[-2, -1, 3], [-2, 0, 2], [-1, -1, 2], [-1, 0, 1]] -> Option A

4. Quick Check:

   All unique triplets summing zero = [[-2, -1, 3], [-2, 0, 2], [-1, -1, 2], [-1, 0, 1]] [OK]

Hint: Sort array, then find triplets with two pointers skipping duplicates [OK]

Common Mistakes:

• Missing some triplets due to not skipping duplicates
• Including triplets with repeated elements incorrectly
• Not sorting array before processing