What if you could find all zero-sum triplets in seconds instead of hours of guesswork?
Why Three Sum Problem All Unique Triplets in DSA Typescript?
Imagine you have a list of numbers and you want to find all groups of three numbers that add up to zero. Doing this by checking every possible group by hand would be like trying to find three friends in a crowd who together have zero money--very tiring and confusing!
Manually checking every combination means trying many groups one by one. This takes a lot of time and you might miss some groups or count the same group twice. It's slow and easy to make mistakes, especially when the list is big.
The Three Sum problem uses a smart way to find all unique triplets that add up to zero. It sorts the list first, then uses two pointers to quickly find matching pairs for each number. This method is faster and avoids duplicates, making the search neat and reliable.
for (let i = 0; i < nums.length; i++) { for (let j = i + 1; j < nums.length; j++) { for (let k = j + 1; k < nums.length; k++) { if (nums[i] + nums[j] + nums[k] === 0) { console.log([nums[i], nums[j], nums[k]]); } } } }
nums.sort((a, b) => a - b); for (let i = 0; i < nums.length - 2; i++) { if (i > 0 && nums[i] === nums[i - 1]) continue; let left = i + 1, right = nums.length - 1; while (left < right) { const sum = nums[i] + nums[left] + nums[right]; if (sum === 0) { console.log([nums[i], nums[left], nums[right]]); while (nums[left] === nums[left + 1]) left++; while (nums[right] === nums[right - 1]) right--; left++; right--; } else if (sum < 0) left++; else right--; } }
This approach lets you quickly find all unique sets of three numbers that sum to zero, even in large lists, without missing or repeating any group.
Think of a budget team trying to find three expenses that cancel each other out to zero. Using this method, they can quickly spot all such triplets without checking every possible combination manually.
Manual checking is slow and error-prone for finding triplets.
Sorting and two-pointer technique speeds up the search and avoids duplicates.
This method reliably finds all unique triplets summing to zero.
Practice
Three Sum Problem All Unique Triplets in an array of integers?Solution
Step 1: Understand the problem requirement
The problem asks to find all unique sets of three numbers (triplets) in the array whose sum equals zero.Step 2: Differentiate from other sum problems
Unlike finding pairs or maximum sums, this problem specifically looks for triplets with sum zero and no duplicates.Final Answer:
Find all unique triplets that sum to zero -> Option BQuick Check:
Three Sum Problem = Find unique triplets summing to zero [OK]
- Confusing with pair sum problems
- Looking for maximum sum instead of zero sum
- Ignoring uniqueness of triplets
Solution
Step 1: Analyze the input parameter type
The input should be an array of numbers, sonums: number[]is correct.Step 2: Analyze the return type
The function returns an array of triplets, each triplet is an array of numbers, sonumber[][]is correct.Final Answer:
function threeSum(nums: number[]): number[][] -> Option AQuick Check:
Input array of numbers, output array of number triplets [OK]
- Using wrong input type like single number or string array
- Returning flat array instead of array of arrays
- Confusing input and output types
const nums = [-1, 0, 1, 2, -1, -4];
function threeSum(nums: number[]): number[][] {
nums.sort((a, b) => a - b);
const result: number[][] = [];
for (let i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
let left = i + 1, right = nums.length - 1;
while (left < right) {
const sum = nums[i] + nums[left] + nums[right];
if (sum === 0) {
result.push([nums[i], nums[left], nums[right]]);
while (left < right && nums[left] === nums[left + 1]) left++;
while (left < right && nums[right] === nums[right - 1]) right--;
left++;
right--;
} else if (sum < 0) {
left++;
} else {
right--;
}
}
}
return result;
}
console.log(threeSum(nums));Solution
Step 1: Sort the array
Sorted array is [-4, -1, -1, 0, 1, 2].Step 2: Find unique triplets summing to zero
Triplets found are [-1, -1, 2] and [-1, 0, 1].Final Answer:
[[-1, -1, 2], [-1, 0, 1]] -> Option CQuick Check:
Unique triplets summing zero = [[-1,-1,2],[-1,0,1]] [OK]
- Not sorting array before processing
- Including duplicate triplets
- Incorrect pointer movement causing wrong sums
function threeSum(nums: number[]): number[][] {
nums.sort((a, b) => a - b);
const result: number[][] = [];
for (let i = 0; i < nums.length - 2; i++) {
let left = i + 1, right = nums.length - 1;
while (left < right) {
const sum = nums[i] + nums[left] + nums[right];
if (sum === 0) {
result.push([nums[i], nums[left], nums[right]]);
left++;
right--;
} else if (sum < 0) {
left++;
} else {
right--;
}
}
}
return result;
}Solution
Step 1: Check for duplicate skipping
The code does not skip duplicate values for index i, which can cause duplicate triplets.Step 2: Confirm other parts are correct
Sorting is correct ascending, pointers initialized properly, and return is after loops.Final Answer:
Missing check to skip duplicate 'i' values -> Option AQuick Check:
Duplicate skipping for i missing = bug [OK]
- Not skipping duplicates for i index
- Skipping duplicates only for left or right pointers
- Incorrect pointer initialization
nums = [-2, 1, 0, -1, 2, -3], which of the following is the correct set of all unique triplets that sum to zero?Solution
Step 1: Sort the array
Sorted array is [-3, -2, -1, 0, 1, 2].Step 2: Find all unique triplets summing to zero
Triplets are [-3, 1, 2], [-2, 0, 2], and [-1, 0, 1].Final Answer:
[[-3, 1, 2], [-2, 0, 2], [-1, 0, 1]] -> Option DQuick Check:
All unique triplets summing zero = [[-3,1,2],[-2,0,2],[-1,0,1]] [OK]
- Including triplets with wrong sums
- Missing valid triplets
- Including duplicates or invalid triplets
Ready to prove it?
Pass the test and this pattern moves unknown
