Solving the Two Sum Problem in JavaScript

Problem: Given an array of integers and a target number, return indices of the two numbers that add up to the target.

🔢 Input:

let arr = [1, 2, 6, 9, 3, 2, 8, 7];
let resultNumber = 10;

🧠 Brute Force Approach:

This solution checks all possible pairs (i, j) in the array.

let indexes = [];
for(let i = 0; i < arr.length; i++) {
  for(let j = i + 1; j < arr.length; j++) {
    if(arr[i] + arr[j] === resultNumber) {
      indexes.push([i, j]);
    }
  }
}
console.log(indexes);

📈 Time Complexity:

O(n²) – because we check each pair of elements once.

🧩 Sample Output:

[ [ 0, 3 ], [ 1, 6 ], [ 4, 7 ], [ 5, 6 ] ]

💡 Optimization Tip:

We can reduce time complexity to O(n) using a hash map. Stay tuned for the optimized version in my next post!

✅ Summary:

• Clear understanding of nested loops
• Simple yet effective for small input sizes
• Ideal for beginners practicing DSA

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