Two Sum — Find Pairs with a Target Sum

📌 Definition:

📖 Detailed Explanation:

Brute force checks every pair — O(n^2). The optimal one-pass approach keeps a hash map of value -> index; for each element x, check whether (target - x) has already been seen. O(n) time, O(n) space. Clarify whether indices or values are required and how duplicates are handled.

💻 Solutions:

• Java
• JavaScript
• Python

public static int[] twoSum(int[] a, int target) {
    Map<Integer,Integer> seen = new HashMap<>();
    for (int i = 0; i < a.length; i++) {
        int need = target - a[i];
        if (seen.containsKey(need)) return new int[]{seen.get(need), i};
        seen.put(a[i], i);
    }
    return new int[]{-1, -1};
}

function twoSum(a, target) {
  const seen = new Map();
  for (let i = 0; i < a.length; i++) {
    const need = target - a[i];
    if (seen.has(need)) return [seen.get(need), i];
    seen.set(a[i], i);
  }
  return [-1, -1];
}

def two_sum(a, target):
    seen = {}
    for i, x in enumerate(a):
        if target - x in seen:
            return [seen[target - x], i]
        seen[x] = i
    return [-1, -1]

🔑 Key Points:

• Hash map of value -> index
• For each x look up complement target - x
• One pass, O(n) beats O(n^2)
• Clarify indices vs values, duplicates

🌍 Real-World Example: