Collection of array manipulation algorithms including searching, sorting, and sum-based problems.
Methods
# static twoSum(nums, target) → {Array.<number>}
Finds two distinct indices in an array whose corresponding values sum to the specified target.
This implementation uses a hash map to achieve linear time complexity by storing previously seen numbers and their indices. When the complement of the current number exists in the map, the solution is found.
Algorithm
- Iterate through the array once.
- For each number, compute its complement (
target - num). - If the complement is already stored in the lookup map, return both indices.
- Otherwise, store the current number and continue.
Constraints
- The input array may contain positive, negative, and zero values.
- Array elements may be duplicated, but returned indices must be distinct.
- The input array is not modified.
Edge Cases
- Returns an empty array if no valid pair is found.
- Works with negative numbers and large values.
- Handles repeated numbers correctly (e.g.,
[3,3], target6).
Complexity:
- Time: O(n) — Each element is processed once using a hash map lookup.
- Space: O(n) — Worst case: the map stores all elements.
Parameters:
| Name | Type | Description |
|---|---|---|
nums |
Array.<number>
|
The list of integers to search through. Must be a finite array. |
target |
number
|
The value representing the desired sum of two elements within |
- Since:
- 1.0.0
- See:
-
threeSum- For finding three numbers that sum to a targetfourSum- For finding four numbers that sum to a target- LeetCode #1: Two Sum
- Subset Sum Problem (Wikipedia)
If nums is not an array or if elements are not valid numbers.
TypeError
A 2-element array [index1, index2]
where nums[index1] + nums[index2] === target.
Returns an empty array [] if no valid pair exists.
Array.<number>
Examples
Basic usage with positive numbers
twoSum([2, 7, 11, 15], 9);
// → [0, 1] because nums[0] + nums[1] = 2 + 7 = 9Works with negative numbers
twoSum([-4, 8, 5, -1], 4);
// → [0, 1] because nums[0] + nums[1] = -4 + 8 = 4Handles duplicate values
twoSum([3, 3], 6);
// → [0, 1] because nums[0] + nums[1] = 3 + 3 = 6Returns empty array when no solution exists
twoSum([1, 2, 3], 10);
// → [] because no two numbers sum to 10Real-world usage: Finding items within budget
const prices = [10, 20, 30, 40];
const budget = 50;
const [first, second] = twoSum(prices, budget);
if (first !== undefined) {
console.log(`Buy items ${first} and ${second} for $${budget}`);
}