Two Sum Problem Using Hash Map

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. Each input will have exactly one solution, and you may not use the same element twice.

Difficulty:

JavaScript Solution

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

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

Explanation: We store each number with its index in a hash map. For every element, we check if its complement (target - current number) already exists. This allows constant-time lookup.

⏱ O(n) | 💾 O(n)

Java Solution

class Solution {
  public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();

    for (int i = 0; i < nums.length; i++) {
      int diff = target - nums[i];
      if (map.containsKey(diff)) {
        return new int[]{map.get(diff), i};
      }
      map.put(nums[i], i);
    }
    return new int[]{};
  }
}

Explanation: HashMap stores visited numbers and their indices. For each number, we check if its complement exists in the map.

⏱ O(n) | 💾 O(n)

Python Solution

def two_sum(nums, target):
    lookup = {}
    for i, num in enumerate(nums):
        diff = target - num
        if diff in lookup:
            return [lookup[diff], i]
        lookup[num] = i

Explanation: We iterate through the array and store elements in a dictionary for constant-time lookup of complements.

⏱ O(n) | 💾 O(n)

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Two Sum Problem Using Hash Map