Four Sum Problem All Unique Quadruplets in DSA C - Execution Trace

Concept Flow - Four Sum Problem All Unique Quadruplets

Sort the array

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Fix first element i

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Fix second element j

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Set left = j+1, right = end

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While left < right

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Sum == target?

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Add quadruplet

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Skip duplicates

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Move pointers

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Repeat until all quadruplets found

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Return all unique quadruplets

The flow sorts the array, then uses two nested loops to fix the first two elements. Two pointers scan the rest to find quadruplets summing to target, skipping duplicates.

Execution Sample

DSA C

int* fourSum(int* nums, int numsSize, int target, int* returnSize) {
  // Sort nums
  // Loop i from 0 to numsSize-4
  // Loop j from i+1 to numsSize-3
  // Use left and right pointers
  // Find quadruplets summing to target
}

This code finds all unique quadruplets in nums that sum to target.

Execution Table

Step	Operation	i, j, left, right	Sum	Quadruplet Found	Visual State
1	Sort array	-	-	-	[1, 0, -1, 0, -2, 2] → [-2, -1, 0, 0, 1, 2]
2	Fix i=0	i=0	-	-	i points to -2
3	Fix j=1	i=0, j=1	-	-	j points to -1
4	Set left=2, right=5	i=0, j=1, left=2, right=5	-	-	left=2, right=5
5	Calculate sum	i=0, j=1, left=2, right=5	-2 + -1 + 0 + 2 = -1	No	No quadruplet
6	Sum < target(0), move left	left=3, right=5	-	-	left moves to 3 (index 3)
7	Calculate sum	i=0, j=1, left=3, right=5	-2 + -1 + 0 + 2 = -1	No	No quadruplet
8	Sum < target, move left	left=4, right=5	-	-	left moves to 4 (index 4)
9	Calculate sum	i=0, j=1, left=4, right=5	-2 + -1 + 1 + 2 = 0	Yes	Quadruplet [-2, -1, 1, 2] found
10	Skip duplicates, move left and right	left=5, right=4	-	-	left > right, inner loop ends
11	Increment j=2	i=0, j=2	-	-	j points to 0
12	Set left=3, right=5	i=0, j=2, left=3, right=5	-	-	left=3, right=5
13	Calculate sum	i=0, j=2, left=3, right=5	-2 + 0 + 0 + 2 = 0	Yes	Quadruplet [-2, 0, 0, 2] found
14	Skip duplicates, move left and right	left=4, right=4	-	-	left >= right, inner loop ends
15	Increment j=3	i=0, j=3	-	-	j points to 0 (duplicate)
16	Skip duplicate j, increment j=4	i=0, j=4	-	-	j points to 1
17	Set left=5, right=5	i=0, j=4, left=5, right=5	-	-	left == right, inner loop ends
18	Increment i=1	i=1	-	-	i points to -1
19	Fix j=2	i=1, j=2	-	-	j points to 0
20	Set left=3, right=5	i=1, j=2, left=3, right=5	-	-	left=3, right=5
21	Calculate sum	i=1, j=2, left=3, right=5	-1 + 0 + 0 + 2 = 1	No	No quadruplet
22	Sum > target, move right	left=3, right=4	-	-	right moves to 4 (index 4)
23	Calculate sum	i=1, j=2, left=3, right=4	-1 + 0 + 0 + 1 = 0	Yes	Quadruplet [-1, 0, 0, 1] found
24	Skip duplicates, move left and right	left=4, right=3	-	-	left > right, inner loop ends
25	Increment j=3	i=1, j=3	-	-	j points to 0 (duplicate)
26	Skip duplicate j, increment j=4	i=1, j=4	-	-	j points to 1
27	Set left=5, right=5	i=1, j=4, left=5, right=5	-	-	left == right, inner loop ends
28	Increment i=2	i=2	-	-	i points to 0
29	Fix j=3	i=2, j=3	-	-	j points to 0 (duplicate)
30	Skip duplicate j, increment j=4	i=2, j=4	-	-	j points to 1
31	Set left=5, right=5	i=2, j=4, left=5, right=5	-	-	left == right, inner loop ends
32	Increment i=3	i=3	-	-	i points to 0 (duplicate)
33	Skip duplicate i, increment i=4	i=4	-	-	i points to 1
34	Loop ends as i > numsSize-4	-	-	-	All quadruplets found
35	Return result	-	-	Quadruplets: [-2,-1,1,2], [-2,0,0,2], [-1,0,0,1]	Done

💡 i reaches numsSize-4, no more quadruplets possible

Variable Tracker

Variable	Start	After Step 2	After Step 11	After Step 18	After Step 28	After Step 33	Final
i	-	0	0	1	2	4	4
j	-	-	1	2	4	-	-
left	-	-	2	3	3	-	-
right	-	-	5	5	5	-	-
Quadruplets Found	[]	[[-2,-1,1,2]]	[[-2,-1,1,2],[-2,0,0,2]]	[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]	[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]	[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]	[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

Key Moments - 3 Insights

Why do we skip duplicates for i, j, left, and right pointers?

Skipping duplicates prevents adding the same quadruplet multiple times. For example, after finding a quadruplet at step 9, we move left and right pointers past duplicates (step 10) to avoid repeats.

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

Because the array is sorted, increasing left moves to a larger number to increase sum, and decreasing right moves to a smaller number to decrease sum. This is shown in steps 6-8 and 21-22.

Why does the inner while loop stop when left >= right?

When left meets or passes right, all pairs for current i and j are checked. For example, after step 14, left=4 and right=4, so loop ends.

Visual Quiz - 3 Questions

Look at the execution table at step 9. What quadruplet was found?

A [-2, -1, 1, 2]

B [-2, 0, 0, 2]

C [-1, 0, 0, 1]

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

💡 Hint

Check the 'Quadruplet Found' column at step 9 in the execution table.

At which step does the condition left >= right first occur, ending the inner loop?

A Step 10

B Step 14

C Step 24

D Step 27

💡 Hint

Look for when 'left' and 'right' pointers meet or cross in the execution table.

If the array was not sorted at step 1, how would the execution table change?

A Sum comparisons would be invalid, and duplicates might not be skipped correctly.

B The quadruplets found would be the same but in different order.

C The pointers left and right would not move.

D The loops for i and j would not run.

💡 Hint

Sorting is essential for correct sum checks and skipping duplicates, as shown in step 1.

Concept Snapshot

Four Sum Problem:
- Sort array first
- Use two nested loops to fix first two elements
- Use two pointers (left, right) to find pairs
- Move pointers based on sum vs target
- Skip duplicates for unique quadruplets
- Return all found quadruplets

Full Transcript

This visualization shows how to find all unique quadruplets in an array that sum to a target. First, the array is sorted. Then, two loops fix the first two elements. Two pointers scan the remaining elements to find pairs that complete the quadruplet. The sum is compared to the target, and pointers move accordingly. Duplicates are skipped to avoid repeated quadruplets. The process continues until all quadruplets are found. The execution table tracks each step, pointer positions, sums, and quadruplets found. Key moments clarify why duplicates are skipped, pointer movements, and loop termination conditions.