Day 1 - Two Sum - Pair with Given Sum.md

Difficulty	Source	Tags
Easy	160 Days of Problem Solving	Arrays Hash

🚀 Day 1. Two Sum - Pair with Given Sum 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given an array arr[] of positive integers and another integer target. Determine if there exist two distinct indices such that the sum of their elements equals target.

🔍 Example Walkthrough:

Input:

arr[] = [1, 4, 45, 6, 10, 8], target = 16

Output:

true

Explanation: arr[3] + arr[4] = 6 + 10 = 16.

Input:

arr[] = [1, 2, 4, 3, 6], target = 11

Output:

false

Explanation: None of the pairs makes a sum of 11.

Constraints

• $1 \leq arr.size \leq 10^5$
• $1 \leq arr[i] \leq 10^5$
• $1 \leq target \leq 2 \times 10^5$

🎯 My Approach:

Hash Set Approach:

1. Use an unordered set (or HashSet in Java / set in Python) to track elements as they are traversed.
2. For each element in the array, compute the complement (target - arr[i]).
3. Check if the complement exists in the set:
   • If it exists, return true.
   • Otherwise, add the current element to the set and continue.
4. If no such pair is found, return false.

Why this approach?

• Using a hash set allows for $O(1)$ average time complexity for insertion and lookup, ensuring optimal performance.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: $O(n)$, as we traverse the array once and perform $O(1)$ operations for insertion and lookup in the hash set.
• Expected Auxiliary Space Complexity: $O(n)$, as we use a hash set to store up to $n$ elements in the worst case.

📝 Solution Code

Code (C++)

class Solution {
  public:
    bool twoSum(vector<int>& arr, int target) {
        unordered_set<int> seen;
        for (int num : arr) {
            if (seen.count(target - num)) return true;
            seen.insert(num);
        }
        return false;
    }
};

Code (Java)

class Solution {
    boolean twoSum(int[] arr, int target) {
        Set<Integer> seen = new HashSet<>();
        for (int num : arr) {
            if (seen.contains(target - num)) return true;
            seen.add(num);
        }
        return false;
    }
}

Code (Python)

class Solution:
    def twoSum(self, arr, target):
        seen = set()
        for num in arr:
            if target - num in seen:
                return True
            seen.add(num)
        return False

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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