Three Sum Problem All Unique Triplets in DSA C - Execution Trace

Concept Flow - Three Sum Problem All Unique Triplets

Sort the array

↓

For each element i

↓

Set left = i+1, right = end

↓

While left < right

↓

Sum == 0?

↓

Add triplet

↓

Skip duplicates

↓

Repeat until all i processed

↓

Return all unique triplets

Sort the array, then for each element use two pointers to find pairs that sum to zero with it, skipping duplicates.

Execution Sample

DSA C

int threeSum(int* nums, int numsSize) {
  // Sort nums
  // For i in 0 to numsSize-3
  //   left = i+1, right = numsSize-1
  //   while left < right
  //     sum = nums[i]+nums[left]+nums[right]
  //     if sum == 0 add triplet and skip duplicates
  //     else move pointers
}

Find all unique triplets in the array that sum to zero.

Execution Table

Step	Operation	i	left	right	Sum	Triplets Found	Pointer Changes	Array State
1	Sort array	-	-	-	-	[]	-	[-4, -1, -1, 0, 1, 2]
2	Set i=0	0	-	-	-	[]	-	[-4, -1, -1, 0, 1, 2]
3	Set left=1, right=5	0	1	5	-	[]	-	[-4, -1, -1, 0, 1, 2]
4	Calculate sum	0	1	5	-4 + (-1) + 2 = -3	[]	-	[-4, -1, -1, 0, 1, 2]
5	Sum < 0, move left	0	2	5	-	[]	left++	[-4, -1, -1, 0, 1, 2]
6	Calculate sum	0	2	5	-4 + (-1) + 2 = -3	[]	-	[-4, -1, -1, 0, 1, 2]
7	Sum < 0, move left	0	3	5	-	[]	left++	[-4, -1, -1, 0, 1, 2]
8	Calculate sum	0	3	5	-4 + 0 + 2 = -2	[]	-	[-4, -1, -1, 0, 1, 2]
9	Sum < 0, move left	0	4	5	-	[]	left++	[-4, -1, -1, 0, 1, 2]
10	Calculate sum	0	4	5	-4 + 1 + 2 = -1	[]	-	[-4, -1, -1, 0, 1, 2]
11	Sum < 0, move left	0	5	5	-	[]	left++	[-4, -1, -1, 0, 1, 2]
12	left >= right, move i	1	-	-	-	[]	-	[-4, -1, -1, 0, 1, 2]
13	Set left=2, right=5	1	2	5	-	[]	-	[-4, -1, -1, 0, 1, 2]
14	Calculate sum	1	2	5	-1 + (-1) + 2 = 0	[]	-	[-4, -1, -1, 0, 1, 2]
15	Sum == 0, add triplet	1	2	5	0	[[-1,-1,2]]	-	[-4, -1, -1, 0, 1, 2]
16	Skip duplicates left	1	3	5	-	[[-1,-1,2]]	left++	[-4, -1, -1, 0, 1, 2]
17	Skip duplicates right	1	3	4	-	[[-1,-1,2]]	right--	[-4, -1, -1, 0, 1, 2]
18	Calculate sum	1	3	4	-1 + 0 + 1 = 0	[[-1,-1,2]]	-	[-4, -1, -1, 0, 1, 2]
19	Sum == 0, add triplet	1	3	4	0	[[-1,-1,2],[-1,0,1]]	-	[-4, -1, -1, 0, 1, 2]
20	Skip duplicates left	1	4	4	-	[[-1,-1,2],[-1,0,1]]	left++	[-4, -1, -1, 0, 1, 2]
21	left >= right, move i	2	-	-	-	[[-1,-1,2],[-1,0,1]]	-	[-4, -1, -1, 0, 1, 2]
22	i=2 is duplicate of i=1, skip i	3	-	-	-	[[-1,-1,2],[-1,0,1]]	-	[-4, -1, -1, 0, 1, 2]
23	Set left=4, right=5	3	4	5	-	[[-1,-1,2],[-1,0,1]]	-	[-4, -1, -1, 0, 1, 2]
24	Calculate sum	3	4	5	0 + 1 + 2 = 3	[[-1,-1,2],[-1,0,1]]	-	[-4, -1, -1, 0, 1, 2]
25	Sum > 0, move right	3	4	4	-	[[-1,-1,2],[-1,0,1]]	right--	[-4, -1, -1, 0, 1, 2]
26	left >= right, move i	4	-	-	-	[[-1,-1,2],[-1,0,1]]	-	[-4, -1, -1, 0, 1, 2]
27	i=4, no enough elements for triplet, end	-	-	-	-	[[-1,-1,2],[-1,0,1]]	-	[-4, -1, -1, 0, 1, 2]

💡 i reached end where no triplets possible, all unique triplets found

Variable Tracker

Variable	Start	After Step 2	After Step 15	After Step 23	Final
i	-	0	1	3	4
left	-	-	2	4	-
right	-	-	5	5	-
Triplets Found	[]	[]	[[-1,-1,2]]	[[-1,-1,2],[-1,0,1]]	[[-1,-1,2],[-1,0,1]]
Array State	[-4,-1,-1,0,1,2]	[-4,-1,-1,0,1,2]	[-4,-1,-1,0,1,2]	[-4,-1,-1,0,1,2]	[-4,-1,-1,0,1,2]

Key Moments - 3 Insights

Why do we skip duplicate values for i and left pointer?

Skipping duplicates avoids repeated triplets. See steps 22 and 16 where duplicates are skipped to keep triplets unique.

Why do we move left pointer when sum < 0 and right pointer when sum > 0?

Because the array is sorted, moving left pointer increases sum, moving right pointer decreases sum. See steps 4-11 and 24-25.

Why stop when left >= right?

Because left and right pointers cross means all pairs for current i are checked. See step 11 and 20.

Visual Quiz - 3 Questions

Look at the execution table at step 15, what triplet is added?

A [-1, -1, 2]

B [-4, 0, 4]

C [0, 1, -1]

D [-1, 0, 1]

💡 Hint

Check the 'Triplets Found' column at step 15 in the execution table.

At which step does the left pointer move after finding the first triplet?

A Step 16

B Step 14

C Step 18

D Step 20

💡 Hint

Look at the 'Pointer Changes' column after step 15 in the execution table.

If the array was not sorted, which part of the execution would fail?

A Skipping duplicates

B Moving pointers based on sum comparison

C Adding triplets

D Setting initial pointers

💡 Hint

Refer to steps 4-11 and 24-25 where pointer moves depend on sum comparison assuming sorted array.

Concept Snapshot

Three Sum Problem:
- Sort array first
- For each element i, use two pointers left=i+1 and right=end
- Calculate sum = nums[i]+nums[left]+nums[right]
- If sum == 0, add triplet and skip duplicates
- If sum < 0, move left pointer right
- If sum > 0, move right pointer left
- Repeat until all i processed
- Return all unique triplets

Full Transcript

The Three Sum Problem finds all unique triplets in an array that sum to zero. First, the array is sorted to allow two-pointer technique. For each element i, two pointers left and right are set to the next element and the end of the array. The sum of nums[i], nums[left], and nums[right] is calculated. If the sum is zero, the triplet is added to the result and pointers skip duplicates. If the sum is less than zero, the left pointer moves right to increase sum. If the sum is greater than zero, the right pointer moves left to decrease sum. This continues until left meets right. Then i moves forward, skipping duplicates. The process repeats until all elements are processed. The final result contains all unique triplets that sum to zero.