What if you could find all groups of four numbers adding to your target without checking every single combination?
Why Four Sum Problem All Unique Quadruplets in DSA C#?
Imagine you have a big list of numbers and you want to find four numbers that add up to a special target number. Doing this by checking every possible group of four numbers by hand would take forever!
Manually checking all groups means trying many combinations, which is very slow and easy to mess up. It's like trying to find four friends in a huge crowd who together have a certain total height, but you have to check every group one by one.
The Four Sum problem uses smart ways to quickly find all unique groups of four numbers that add up to the target. It avoids repeating groups and skips unnecessary checks, making the search much faster and easier.
for (int i = 0; i < nums.Length; i++) { for (int j = i + 1; j < nums.Length; j++) { for (int k = j + 1; k < nums.Length; k++) { for (int l = k + 1; l < nums.Length; l++) { if (nums[i] + nums[j] + nums[k] + nums[l] == target) { // store quadruplet } } } } }
Array.Sort(nums); for (int i = 0; i < nums.Length - 3; i++) { if (i > 0 && nums[i] == nums[i - 1]) continue; for (int j = i + 1; j < nums.Length - 2; j++) { if (j > i + 1 && nums[j] == nums[j - 1]) continue; int left = j + 1, right = nums.Length - 1; while (left < right) { int sum = nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { // store quadruplet left++; right--; } else if (sum < target) left++; else right--; } } }
This method lets you quickly find all unique sets of four numbers that add up to a target, even in large lists, without wasting time on repeated or impossible groups.
Think of a shopping list where you want to pick exactly four items that together cost a certain amount of money. This method helps find all possible combinations without checking every single group manually.
Manual checking of all quadruplets is slow and error-prone.
Sorting and using two pointers speeds up finding unique quadruplets.
Efficiently finds all unique groups that sum to the target.
Practice
Four Sum Problem in data structures and algorithms?Solution
Step 1: Understand the problem statement
The Four Sum Problem asks to find all unique sets of four numbers in an array that add up to a specific target sum.Step 2: Differentiate from other problems
Unlike sorting or finding maximum, this problem focuses on quadruplets and their sums, not individual elements or pairs.Final Answer:
Find all unique quadruplets in an array that sum up to a target value -> Option DQuick Check:
Four Sum Problem = Find quadruplets with target sum [OK]
- Confusing with sorting or max element problems
- Looking for pairs instead of quadruplets
- Ignoring uniqueness of quadruplets
Solution
Step 1: Identify the data structure for storing quadruplets
Each quadruplet is a list of 4 integers, so a list of lists is needed to store multiple quadruplets.Step 2: Match with C# syntax
UsingList<List<int>>correctly represents multiple quadruplets, unlike a single list or dictionary.Final Answer:
var quadruplets = new List<List<int>>(); -> Option AQuick Check:
Use List of List for quadruplets [OK]
- Using List<int> for multiple quadruplets
- Using Dictionary instead of List
- Using fixed-size array for dynamic quadruplets
nums = [1, 0, -1, 0, -2, 2] and target 0, what does FourSum(nums, 0) return?Assuming
FourSum returns all unique quadruplets that sum to target.Solution
Step 1: Sort and find quadruplets summing to 0
Sorting nums gives [-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].Step 2: Verify uniqueness and completeness
All quadruplets are unique and sum exactly to 0, matching the expected output.Final Answer:
[[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]] -> Option CQuick Check:
Unique quadruplets sum to 0 = [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]] [OK]
- Not sorting array before processing
- Including duplicate quadruplets
- Missing some valid quadruplets
for (int i = 0; i < nums.Length - 3; i++) {
for (int j = i + 1; j < nums.Length - 2; j++) {
int left = j + 1, right = nums.Length - 1;
while (left < right) {
int sum = nums[i] + nums[j] + nums[left] + nums[right];
if (sum == target) {
result.Add(new List<int> { nums[i], nums[j], nums[left], nums[right] });
left++;
right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
}Solution
Step 1: Analyze duplicate handling
The code adds quadruplets without skipping duplicates, so repeated quadruplets can appear in the result.Step 2: Check other parts
Left and right pointers are correctly initialized; loop ranges are valid; sorting is assumed done before this snippet.Final Answer:
No check to skip duplicate quadruplets causing duplicates in result -> Option AQuick Check:
Missing duplicate skip = No check to skip duplicate quadruplets causing duplicates in result [OK]
- Not skipping duplicates for i, j, left, right
- Assuming sorting is done inside the loop
- Incorrect pointer updates causing infinite loops
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?Solution
Step 1: Understand the input and target
All elements are 2, and target is 8, so quadruplets of four 2's sum to 8.Step 2: Identify unique quadruplets
Only one unique quadruplet [2, 2, 2, 2] exists despite duplicates in input.Final Answer:
[[2, 2, 2, 2]] -> Option BQuick Check:
Only one unique quadruplet sums to 8 [OK]
- Counting duplicate quadruplets multiple times
- Returning empty list when quadruplet exists
- Returning triplets instead of quadruplets
Ready to prove it?
Pass the test and this pattern moves unknown
