How to Solve the Two Sum Challenge on LeetCode

 

🧩 Two Sum – Explained simply

🚀 Difficulty: Easy | 💼 Asked in Many Interviews

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📘 The Problem (from LeetCode)

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 must:

• Return exactly one solution (you can assume there is always one)

• Not use the same element twice

• Return the indexes (positions) of the numbers, not the numbers themselves

• The order of the result doesn’t matter

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💡 Let’s Understand With an Example:

nums = [2, 7, 11, 15]
target = 9

We need to find two numbers from the array whose sum is 9.

Let’s try:

• 2 + 7 = ✅ 9 → Found at index 0 and 1

Answer: [0, 1]

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🎓 Real-Life Analogy:

Imagine you're shopping. Your friend gives you ₹9 and a list of item prices. You need to pick 2 items that exactly total ₹9.

Item prices = [2, 7, 11, 15]
You pick ₹2 (item at index 0) and ₹7 (index 1). That’s the answer → [0, 1].

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❗ Common Misunderstanding

Some people only check each element and the next one (i and i+1). But what if the answer is in some other pair (like index 1 and 3)? That logic will miss some valid answers.

We need to try every pair, not just adjacent ones.

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🧐 Solution 1: Brute Force (Beginner Friendly)

Try every possible pair of numbers. If their sum is the target, return their indexes.

🧮 Logic:

• Loop over each number with index i

• Inside that loop, loop over every number after i (index j)

• Check if nums[i] + nums[j] == target

• If yes, return [i, j]

🕐 Time Complexity:

• O(n^2) (nested loops)

• 🧠 Pseudocode:

  Loop i from 0 to length-1
      Loop j from i+1 to length
          if nums[i] + nums[j] == target
              return [i, j]

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💻 PHP Code

class Solution {
    function twoSum($nums, $target) {
        $n = count($nums); // Get the size of the array
        for ($i = 0; $i < $n; $i++) {
            for ($j = $i + 1; $j < $n; $j++) {
                // Check if this pair sums up to the target
                if ($nums[$i] + $nums[$j] == $target) {
                    return [$i, $j]; // Return their positions
                }
            }
        }
        return []; // Return empty if no pair found
    }
}

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☕ Java Code

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums[i] + nums[j] == target) {
                    return new int[] { i, j };
                }
            }
        }
        return new int[] {};
    }
}

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🐍 Python Code

class Solution:
    def twoSum(self, nums, target):
        n = len(nums)
        for i in range(n):
            for j in range(i + 1, n):
                if nums[i] + nums[j] == target:
                    return [i, j]
        return []

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🌐 JavaScript Code

function twoSum(nums, target) {
    for (let i = 0; i < nums.length; i++) {
        for (let j = i + 1; j < nums.length; j++) {
            if (nums[i] + nums[j] === target) {
                return [i, j];
            }
        }
    }
    return [];
}

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💡 Solution 2: Optimized Using Hash Map (For Faster Performance)

Instead of checking every pair, we can:

• Use a hash map to remember numbers we’ve already seen

• For each number, check if target - current number already exists in the map

✅ Time: O(n), Space: O(n)

Python (Optimized):

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

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✅ Summary

Approach	Time Complexity	Space	Easy to Understand
Brute Force	O(n²)	O(1)	✅
Hash Map	O(n)	O(n)	❌ (slightly tricky)

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📒 Final Tips

• Master brute force first → it's simple and helps understand the problem.

• Later, try to learn the hash map method — it's fast and often asked in interviews.

• Practice writing in multiple languages (like we just did!) — great for polyglot devs.

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