Four Sum Problem All Unique Quadruplets in DSA Javascript - Time & Space Complexity

Time Complexity: Four Sum Problem All Unique Quadruplets

O(n³)

Understanding Time Complexity

We want to understand how the time needed to find all sets of four numbers that add up to a target grows as the list gets bigger.

How does the number of steps change when the input list grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function fourSum(nums, target) {
  nums.sort((a, b) => a - b);
  const quadruplets = [];
  const n = nums.length;
  for (let i = 0; i < n - 3; i++) {
    for (let j = i + 1; j < n - 2; j++) {
      let left = j + 1, right = n - 1;
      while (left < right) {
        const sum = nums[i] + nums[j] + nums[left] + nums[right];
        if (sum === target) {
          quadruplets.push([nums[i], nums[j], nums[left], nums[right]]);
          left++;
          right--;
        } else if (sum < target) {
          left++;
        } else {
          right--;
        }
      }
    }
  }
  return quadruplets;
}
    

This code finds all groups of four numbers in the array that add up to the target value.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops and a two-pointer scan inside the innermost loop.
• How many times: The outer loop runs about n times, the next loop about n times, and the two-pointer scan runs up to n times for each pair, making the main work repeat roughly n x n x n times.

How Execution Grows With Input

As the input size grows, the number of steps grows roughly with the cube of the input size.

Input Size (n)	Approx. Operations
10	About 1,000
100	About 1,000,000
1000	About 1,000,000,000

Pattern observation: When input size increases 10 times, operations increase about 1,000 times, showing cubic growth.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows roughly with the cube of the input size, so doubling the input makes the work about eight times bigger.

Common Mistake

[X] Wrong: "Since there are four numbers, the time complexity must be O(n⁴)."

[OK] Correct: The two-pointer method reduces the innermost search from a loop to a linear scan, so the complexity is O(n³), not O(n⁴).

Interview Connect

Understanding this problem's time complexity helps you explain how nested loops and clever scanning affect performance, a skill useful in many coding challenges.

Self-Check

"What if we replaced the two-pointer scan with a nested loop for the last two numbers? How would the time complexity change?"

Practice

1. What is the main goal of the Four Sum Problem in data structures and algorithms?

easy

A. Find all unique quadruplets in an array that sum up to a target value

B. Sort the array in ascending order

C. Find the maximum element in the array

D. Count the number of pairs with a given sum

Solution

1. Step 1: Understand the problem statement

   The Four Sum Problem asks us to find all unique groups of four numbers in an array that add up to a specific target sum.

2. Step 2: Differentiate from other problems

   This is different from sorting, finding max, or counting pairs because it specifically looks for quadruplets with a sum equal to the target.

3. Final Answer:

   Find all unique quadruplets in an array that sum up to a target value -> Option A

4. Quick Check:

   Four Sum Problem = Find quadruplets summing to target [OK]

Hint: Look for four numbers adding to target, not sorting or max [OK]

Common Mistakes:

• Confusing with sorting or max element problems
• Looking for pairs instead of quadruplets
• Ignoring uniqueness of quadruplets

2. Which of the following is the correct way to start sorting an array nums in JavaScript before solving the Four Sum Problem?

easy

A. nums.sort()

B. nums.sort((a, b) => b - a)

C. nums.sorted()

D. nums.sort((a, b) => a - b)

Solution

1. Step 1: Understand JavaScript sort behavior

   By default, nums.sort() sorts elements as strings, which can give wrong order for numbers.

2. Step 2: Use numeric comparison function

   Using nums.sort((a, b) => a - b) sorts numbers in ascending order correctly, which is needed for the Four Sum algorithm.

3. Final Answer:

   nums.sort((a, b) => a - b) -> Option D

4. Quick Check:

   Numeric ascending sort = nums.sort((a, b) => a - b) [OK]

Hint: Use numeric compare function for sorting numbers [OK]

Common Mistakes:

• Using default sort without compare function
• Using descending order when ascending is needed
• Using non-existent method like sorted()

3. What will be the output of the following code snippet?

const nums = [1, 0, -1, 0, -2, 2];
const target = 0;
// Assume findFourSum returns all unique quadruplets summing to target
const result = findFourSum(nums, target);
console.log(result);

Given findFourSum correctly implements the Four Sum algorithm, what is the expected output?

medium

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

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

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

D. [[0, 0, 0, 0]]

Solution

1. Step 1: Sort the array and find quadruplets

   Sorted array is [-2, -1, 0, 0, 1, 2]. The unique quadruplets summing to 0 are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1].

2. Step 2: Verify quadruplets sum to target

   Each quadruplet sums to 0, matching the target, and all are unique sets.

3. Final Answer:

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

4. Quick Check:

   Unique quadruplets summing to 0 = [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]] [OK]

Hint: Sort array first, then find quadruplets summing to target [OK]

Common Mistakes:

• Not sorting array before processing
• Including duplicate quadruplets
• Incorrect quadruplet sums

4. Identify the error in the following snippet for solving the Four Sum Problem:

function fourSum(nums, target) {
  nums.sort();
  const quadruplets = [];
  for (let i = 0; i < nums.length - 3; i++) {
    for (let j = i + 1; j < nums.length - 2; j++) {
      let left = j + 1;
      let right = nums.length - 1;
      while (left < right) {
        const sum = nums[i] + nums[j] + nums[left] + nums[right];
        if (sum === target) {
          quadruplets.push([nums[i], nums[j], nums[left], nums[right]]);
          left++;
          right--;
        } else if (sum < target) {
          left++;
        } else {
          right--;
        }
      }
    }
  }
  return quadruplets;
}

medium

A. The loops do not cover all indices correctly

B. The array is not sorted numerically before processing

C. The quadruplets array is not returned

D. The sum calculation is incorrect

Solution

1. Step 1: Check sorting method

   The code uses nums.sort() without a compare function, which sorts elements as strings, causing incorrect order for numbers.

2. Step 2: Impact on algorithm correctness

   Incorrect sorting leads to wrong quadruplets or missed solutions because the two-pointer approach depends on sorted numeric order.

3. Final Answer:

   The array is not sorted numerically before processing -> Option B

4. Quick Check:

   Missing numeric sort compare function = The array is not sorted numerically before processing [OK]

Hint: Always use numeric compare in sort for number arrays [OK]

Common Mistakes:

• Using default sort without compare function
• Not returning the result array
• Incorrect loop ranges (not here though)
• Wrong sum calculation (correct here)

5. Given an array nums = [2, 2, 2, 2, 2] and target = 8, what is the correct output of the Four Sum Problem algorithm that finds all unique quadruplets?

hard

A. [[2, 2, 2]]

B. [[2, 2, 2, 2], [2, 2, 2, 2]]

C. [[2, 2, 2, 2]]

D. []

Solution

1. Step 1: Understand the input and target

   All elements are 2, and target is 8. Four 2's sum to 8 exactly.

2. Step 2: Identify unique quadruplets

   Even though there are multiple 2's, only one unique quadruplet [2, 2, 2, 2] exists that sums to 8.

3. Final Answer:

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

4. Quick Check:

   Unique quadruplet summing to 8 = [[2, 2, 2, 2]] [OK]

Hint: Duplicates form one unique quadruplet if sum matches target [OK]

Common Mistakes:

• Listing duplicate quadruplets multiple times
• Returning empty when quadruplet exists
• Confusing triplets with quadruplets