The Four Sum problem asks to find all unique groups of four numbers in an array that add up to a given target sum.
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beginner
Why do we sort the array before solving the Four Sum problem?
Sorting helps to efficiently skip duplicates and use two-pointer technique to find pairs that sum to a target.
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intermediate
What technique is commonly used to find quadruplets in the Four Sum problem?
We use nested loops for the first two numbers and then two pointers to find the remaining two numbers that complete the quadruplet.
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intermediate
How do we ensure quadruplets are unique in the Four Sum problem?
By skipping duplicate values during iteration and after finding a valid quadruplet, we move pointers past duplicates.
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intermediate
What is the time complexity of the Four Sum solution using sorting and two pointers?
The time complexity is O(n^3) because we have two nested loops and a two-pointer scan inside.
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What is the first step before applying the two-pointer technique in the Four Sum problem?
ARemove duplicates
BReverse the array
CShuffle the array
DSort the array
✗ Incorrect
Sorting the array helps to efficiently find pairs and skip duplicates.
How many nested loops are used before applying the two-pointer approach in Four Sum?
ATwo
BFour
CThree
DOne
✗ Incorrect
Two nested loops fix the first two numbers; then two pointers find the other two.
What is the main reason to skip duplicates in Four Sum?
ATo reduce time complexity
BTo avoid repeated quadruplets
CTo sort faster
DTo use less memory
✗ Incorrect
Skipping duplicates ensures each quadruplet is unique in the result.
What data structure is typically used to store the quadruplets found?
AVector of vectors (list of lists)
BHashMap
CStack
DQueue
✗ Incorrect
A vector of vectors stores multiple quadruplets as lists of four numbers.
What is the overall time complexity of the Four Sum solution using sorting and two pointers?
AO(n^4)
BO(n^2)
CO(n^3)
DO(n log n)
✗ Incorrect
Two nested loops plus two pointers give O(n^3) time complexity.
Explain the step-by-step approach to solve the Four Sum problem with unique quadruplets.
Think about how sorting and two pointers help find quadruplets efficiently.
You got /5 concepts.
Describe how to handle duplicates when finding quadruplets in the Four Sum problem.
Duplicates appear when consecutive numbers are the same after sorting.
You got /4 concepts.
Practice
(1/5)
1. What is the main goal of the Four Sum Problem in data structures and algorithms?
easy
A. Count the number of pairs with a given sum
B. Find all unique quadruplets in an array that sum up to a target value
C. Find the maximum element in the array
D. Sort the array in ascending order
Solution
Step 1: Understand the problem statement
The Four Sum Problem asks to find all unique groups of four numbers (quadruplets) in an array that add up to a specific target sum.
Step 2: Identify the goal
The goal is not sorting or counting pairs, but specifically finding quadruplets whose sum equals the target.
Final Answer:
Find all unique quadruplets in an array that sum up to a target value -> Option B
Quick Check:
Four Sum Problem = Find quadruplets with target sum [OK]
Hint: Look for four numbers adding to target, not sorting or counting [OK]
Common Mistakes:
Confusing Four Sum with sorting problems
Looking for pairs instead of quadruplets
Ignoring uniqueness of quadruplets
2. Which Rust syntax correctly declares a vector of integers named nums?
easy
A. let nums = {1, 2, 3, 4};
B. let nums = [1, 2, 3, 4];
C. let nums = (1, 2, 3, 4);
D. let nums = vec![1, 2, 3, 4];
Solution
Step 1: Identify Rust vector syntax
In Rust, vectors are created using vec![elements] macro, which creates a growable array.
Step 2: Check each option
let nums = vec![1, 2, 3, 4]; uses vec![1, 2, 3, 4] which is correct. let nums = [1, 2, 3, 4]; is an array, not a vector. let nums = (1, 2, 3, 4); is a tuple. let nums = {1, 2, 3, 4}; is invalid syntax.
Final Answer:
let nums = vec![1, 2, 3, 4]; -> Option D
Quick Check:
Vector syntax = vec![...] [OK]
Hint: Use vec! macro for vectors, arrays use brackets only [OK]
Common Mistakes:
Using array syntax when vector is needed
Confusing tuples with vectors
Using curly braces which are invalid here
3. What is the output of the following Rust code snippet that finds quadruplets summing to 0?
let nums = vec![1, 0, -1, 0, -2, 2];
let target = 0;
// Assume function four_sum returns Vec of quadruplets
let result = four_sum(nums, target);
println!("{:?}", result);
Assuming four_sum is correctly implemented, what is the printed output?
medium
A. [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]
B. [[1, 0, -1, 0]]
C. [[0, 0, 0, 0]]
D. []
Solution
Step 1: Understand the input and target
The input array is [1, 0, -1, 0, -2, 2] and target sum is 0. We want all unique quadruplets that sum to 0.
Step 2: Identify all quadruplets summing to 0
Possible quadruplets are [-2, -1, 1, 2], [-2, 0, 0, 2], and [-1, 0, 0, 1]. These are unique and sum to 0.
Quadruplets sum to 0 = [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]] [OK]
Hint: Sort array and find quadruplets summing to target [OK]
Common Mistakes:
Returning incomplete quadruplets
Ignoring duplicates
Returning empty list incorrectly
4. Identify the bug in this Rust code snippet for the Four Sum problem:
fn four_sum(nums: Vec<i32>, target: i32) -> Vec {
let mut nums = nums;
nums.sort();
let mut result = Vec::new();
for i in 0..nums.len() {
for j in i+1..nums.len() {
let mut left = j + 1;
let mut right = nums.len() - 1;
while left < right {
let sum = nums[i] + nums[j] + nums[left] + nums[right];
if sum == target {
result.push(vec![nums[i], nums[j], nums[left], nums[right]]);
left += 1;
right -= 1;
} else if sum < target {
left += 1;
} else {
right -= 1;
}
}
}
}
result
}
What is the main issue causing duplicate quadruplets in the result?
medium
A. The vector nums is not sorted before processing
B. The while loop condition left < right is incorrect and should be left <= right
C. The code does not skip duplicate values for i, j, left, and right indices after finding a quadruplet
D. The function does not return the result vector
Solution
Step 1: Analyze duplicate handling in loops
The code finds quadruplets but does not skip over duplicate values for indices i, j, left, and right after adding a quadruplet, causing duplicates in the result.
Step 2: Check other options
The vector is sorted correctly, the while loop condition is correct, and the function returns the result vector properly.
Final Answer:
The code does not skip duplicate values for i, j, left, and right indices after finding a quadruplet -> Option C
Quick Check:
Skipping duplicates needed to avoid repeated quadruplets [OK]
Hint: Skip duplicates after adding quadruplet to avoid repeats [OK]
Common Mistakes:
Not skipping duplicates for all pointers
Changing loop conditions incorrectly
Forgetting to sort the array
5. Given the array nums = [2, 2, 2, 2, 2] and target 8, what is the correct output of a Four Sum function that returns all unique quadruplets summing to the target?
hard
A. [[2, 2, 2, 2]]
B. [[2, 2, 2, 2], [2, 2, 2, 2]]
C. []
D. [[2, 2, 2]]
Solution
Step 1: Understand the input and target
The array contains five 2's and the target sum is 8. We need quadruplets of four numbers summing to 8.
Step 2: Identify unique quadruplets
Only one unique quadruplet exists: [2, 2, 2, 2]. Although multiple combinations exist, duplicates must be avoided.