15. 3Sum

🟧 Medium

Given an integer array nums, 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.

Notice that the solution set must not contain duplicate triplets.

Example 1

Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter.

Example 2

Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0.

Example 3

Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.

Hint-1

So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand!

Hint-2

For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next number y, which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster?

Hint-3

The second train of thought for two-sum is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?

Constraints

• 3 <= nums.length <= 3000

• -10^5 <= nums[i] <= 10^5

Solution

My Solution

func threeSum(nums []int) [][]int {
    sort.Ints(nums)
    res := [][]int{}

    for i, num := range nums {
        if i > 0 && num == nums[i-1] {
            continue
        }

        l,r := i+1, len(nums) - 1
        for l < r {
            threeSum := num + nums[l] + nums[r]
            if threeSum > 0 {
                r -=1
            } else if threeSum < 0 {
                l+=1
            } else {
                res = append(res, []int{num, nums[l], nums[r]})
                l+=1

                for nums[l] == nums[l-1] && l < r {
                    l+=1
                }
            }
        }
    }


    return res
}

Optimal Solution

The optimal solution uses two key techniques:

1. Sorting for duplicate handling and efficient searching

2. Two-pointer technique for finding pairs that sum to target

Here's an optimized version with improved readability and performance:

func threeSum(nums []int) [][]int {
    // Sort the array first for efficient duplicate handling
    sort.Ints(nums)
    result := make([][]int, 0)
    n := len(nums)
    
    // Fix the first number and use two pointers for the remaining sum
    for i := 0; i < n-2; i++ {
        // Skip duplicates for first number
        if i > 0 && nums[i] == nums[i-1] {
            continue
        }
        
        // Early termination optimization
        if nums[i] > 0 {
            break // Since array is sorted, no three numbers can sum to 0
        }
        
        target := -nums[i]
        left := i + 1
        right := n - 1
        
        for left < right {
            currentSum := nums[left] + nums[right]
            
            if currentSum == target {
                result = append(result, []int{nums[i], nums[left], nums[right]})
                
                // Skip duplicates for second number
                for left < right && nums[left] == nums[left+1] {
                    left++
                }
                // Skip duplicates for third number
                for left < right && nums[right] == nums[right-1] {
                    right--
                }
                
                left++
                right--
            } else if currentSum < target {
                left++
            } else {
                right--
            }
        }
    }
    
    return result
}

Approach Analysis

The solution employs three main techniques:

1. Sorting:

   • Makes duplicate handling efficient

   • Enables two-pointer technique

   • Allows for early termination

2. Two-Pointer Technique:

   • Uses sorted array property

   • Efficiently finds pairs summing to target

   • Reduces time complexity from O(n²) to O(n)

3. Duplicate Handling:

   • Skip duplicates for first number

   • Skip duplicates for second and third numbers

   • Ensures unique triplets

Visualization of Both Approaches

Input array: [-1, 0, 1, 2, -1, -4]
After sorting: [-4, -1, -1, 0, 1, 2]

Step 1: First number = -4
[-4, -1, -1, 0, 1, 2]
 ^   L           R
Target sum = 4
Move L right as sum is too small

Step 2: First number = -1
[-4, -1, -1, 0, 1, 2]
     ^   L        R
Found triplet: [-1, -1, 2]

Step 3: Skip duplicate -1
[-4, -1, -1, 0, 1, 2]
         ^   L    R
Found triplet: [-1, 0, 1]

Final result: [[-1, -1, 2], [-1, 0, 1]]

Complexity Analysis

Time Complexity:

• Sorting: O(n log n)

• Two pointers: O(n²)

• Overall: O(n²)

Space Complexity:

• Result array: O(k) where k is number of valid triplets

• Sorting space: O(log n) to O(n) depending on sort implementation

• Overall: O(k)

Optimizations:

• Early termination when first number > 0

• Skip duplicates to avoid redundant calculations

• Two-pointer technique instead of nested loops

Why Solution Works

1. Sorting Enables Efficiency:

   • Makes duplicate detection simple

   • Allows two-pointer technique

   • Enables early termination

2. Two-Pointer Logic:

   • If sum too small, increase left pointer

   • If sum too large, decrease right pointer

   • Guarantees finding all valid pairs

3. Duplicate Handling:

   • Sorting puts duplicates adjacent

   • Skip duplicates for all three positions

   • Ensures unique triplets in result

When to Use

This approach is ideal when:

1. Need to find all combinations summing to target

2. Duplicates need to be handled

3. Order of triplets doesn't matter

4. Input array can be modified

Common variations:

• 4Sum problem

• K-sum problem

• Two-pointer problems

• Combination sum problems

Common Patterns & Applications

1. Two-Pointer Pattern:

   • Sorted array traversal

   • Meeting in middle

   • Opposite direction movement

2. Sorting + Two Pointers:

   • Finding combinations

   • Handling duplicates

   • Range-based problems

3. Sum Problems:

   • TwoSum variations

   • KSum problems

   • Target sum problems

Interview Tips

1. Key Discussion Points:

   • Why sorting first?

   • How to handle duplicates?

   • Time/space complexity tradeoffs

   • Optimization techniques

2. Common Follow-up Questions:

   • How to handle 4Sum?

   • Can we do it without sorting?

   • How to optimize for specific input patterns?

   • How to handle negative targets?

3. Edge Cases to Consider:

   • Empty array

   • Array with < 3 elements

   • All zeros

   • All same numbers

   • No valid triplets

4. Optimization Opportunities:

   • Early termination

   • Duplicate skipping

   • Memory management

   • Input validation

result

Leetcode: link