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| 1 | +// Approach 2: HashMap (Using Map in TypeScript) |
| 2 | +// ⏳ Time Complexity: O(n) ✅ (Efficient for large inputs) |
| 3 | +// 💾 Space Complexity: O(n) ❌ (Extra memory used for Map) |
| 4 | + |
| 5 | +function twoSum(nums: number[], target: number): number[] { |
| 6 | + |
| 7 | + const indices = new Map<number, number>(); |
| 8 | + |
| 9 | + for (let i = 0; i < nums.length; i++) { |
| 10 | + |
| 11 | + const temp = target - nums[i]; |
| 12 | + if (indices.has(temp)) { |
| 13 | + // Ensure correct order |
| 14 | + return [indices.get(temp)!, i]; |
| 15 | + // The exclamation mark (!) in indices.get(temp)! is a non-null assertion operator in TypeScript. |
| 16 | + } |
| 17 | + // If you return [i, indices.get(temp)!], |
| 18 | + // you would be returning the current index first, |
| 19 | + // which is incorrect because the problem statement usually expects indices in the order they were found (earlier index first). |
| 20 | + |
| 21 | + indices.set(nums[i], i); |
| 22 | + } |
| 23 | + return []; |
| 24 | +}; |
| 25 | + |
| 26 | + |
| 27 | +// Approach 1: Brute Force (O(n^2)) |
| 28 | +// ⏳ Time Complexity: O(n^2) ❌ (Inefficient for large inputs) |
| 29 | +// 💾 Space Complexity: O(1) ✅ (Great, no extra memory used!) |
| 30 | + |
| 31 | +// function twoSum(nums: number[], target: number): number[] { |
| 32 | +// for (let i = 0; i < nums.length; i++) { |
| 33 | +// for (let j = i + 1; j < nums.length; j++) { |
| 34 | +// if (nums[i] + nums[j] === target) { |
| 35 | +// return [i, j] |
| 36 | +// } |
| 37 | +// } |
| 38 | +// } |
| 39 | +// return []; |
| 40 | +// }; |
| 41 | + |
| 42 | + |
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