Overview

Given an array of integers nums and an integer target, return the indices of the two numbers such that they add up to target.

You may assume that each input has exactly one solution, and you may not use the same element twice. Return the indices in ascending order.

Implement twoSum(nums, target).

Constraints

• 2 ≤ nums.length ≤ 10000
• -1000000000 ≤ nums[i] ≤ 1000000000
• Exactly one valid pair exists.
• Return indices sorted ascending: [i, j] where i < j.
• Aim for $O(n)$ time using a hash map.

Examples

twoSum([2, 7, 11, 15], 9);
// → [0, 1]  — nums[0] + nums[1] = 2 + 7 = 9

twoSum([3, 2, 4], 6);
// → [1, 2]  — nums[1] + nums[2] = 2 + 4 = 6

twoSum([3, 3], 6);
// → [0, 1]

Notes

• Use a hash map to store each number's index as you iterate. For each element, check if the complement (target - nums[i]) exists in the map.
• Single-pass $O(n)$ solution.

Solution

function twoSum(nums, target) {
  const map = new Map();

  for (let i = 0; i < nums.length; i++) {
    const complement = target - nums[i];
    if (map.has(complement)) {
      return [map.get(complement), i];
    }
    map.set(nums[i], i);
  }
}

Resources

• Two Sum — LeetCode
• Hash Table — Wikipedia