1. Two Sum

Problem Statement

Given an array of integers nums and an integer target, return 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.

You can return the answer in any order.

Examples

Example 1:

Input: nums = [2,7,11,15], target = 9 Output: [0,1] Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6 Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6 Output: [0,1]

Difficulty

Easy

Constraints

• 2 ≤ nums.length ≤ 10⁴
• -10⁹ ≤ nums[i] ≤ 10⁹
• -10⁹ ≤ target ≤ 10⁹
• Only one valid answer exists

Solution 1: Using Hash Map

Approach:

1. Use a hash map to store value → index
2. Iterate through array
3. Check complement (target - current value)
4. If found, return answer
5. Else store value

Java

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            int need = target - nums[i];
            if (map.containsKey(need)) {
                return new int[]{map.get(need), i};
            }
            map.put(nums[i], i);
        }
        return new int[]{};
    }
}

C++

vector twoSum(vector& nums, int target) {
    unordered_map m;
    for (int i = 0; i < nums.size(); i++) {
        int need = target - nums[i];
        if (m.count(need)) return {m[need], i};
        m[nums[i]] = i;
    }
    return {};
}

Python

def twoSum(nums, target):
    seen = {}
    for i, num in enumerate(nums):
        need = target - num
        if need in seen:
            return [seen[need], i]
        seen[num] = i

Complexity

Time: O(n)

Space: O(n)

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