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| 1 | +### Using Sorting and Two Pointers (Time Complexity (O(n^2))) |
| 2 | +This approach is more efficient and first sorts the array. For every number in the array, it reduces the problem to finding pairs (similar to the two-pointer method). |
| 3 | + |
| 4 | +Algorithm: |
| 5 | +Sort the input array. |
| 6 | +Fix one number (arr[i) and use two pointers (left and right) to find two other numbers that sum up to targetSum - arr[i]. |
| 7 | + |
| 8 | +```js |
| 9 | +function findTripletsTwoPointers(arr, targetSum) { |
| 10 | + const triplets = []; |
| 11 | + arr.sort((a, b) => a - b); // Sort the array |
| 12 | + |
| 13 | + for (let i = 0; i < arr.length - 2; i++) { |
| 14 | + // Skip duplicate values to prevent repetitive triplets |
| 15 | + if (i > 0 && arr[i] === arr[i - 1]) continue; |
| 16 | + |
| 17 | + let left = i + 1; |
| 18 | + let right = arr.length - 1; |
| 19 | + |
| 20 | + while (left < right) { |
| 21 | + const sum = arr[i] + arr[left] + arr[right]; |
| 22 | + |
| 23 | + if (sum === targetSum) { |
| 24 | + triplets.push([arr[i], arr[left], arr[right]]); |
| 25 | + |
| 26 | + // Skip duplicate values to prevent repetitive triplets |
| 27 | + while (left < right && arr[left] === arr[left + 1]) left++; |
| 28 | + while (left < right && arr[right] === arr[right - 1]) right--; |
| 29 | + |
| 30 | + left++; |
| 31 | + right--; |
| 32 | + } else if (sum < targetSum) { |
| 33 | + left++; // Increase the sum |
| 34 | + } else { |
| 35 | + right--; // Decrease the sum |
| 36 | + } |
| 37 | + } |
| 38 | + } |
| 39 | + |
| 40 | + return triplets; |
| 41 | +} |
| 42 | + |
| 43 | +// Example usage |
| 44 | +const array = [1, 2, 3, 4, 5, 6, 1]; |
| 45 | +const target = 10; |
| 46 | +console.log(findTripletsTwoPointers(array, target)); // Output: [[1, 3, 6], [1, 4, 5], [2, 3, 5], [1, 3, 6]] |
| 47 | +``` |
| 48 | + |
| 49 | +### Using Hashing (Time Complexity (O(n^2)), Space Complexity (O(n))) |
| 50 | +This approach uses a hash table to track numbers for each fixed element. It avoids sorting and directly checks for valid pairs that sum to targetSum - arr[i]. |
| 51 | + |
| 52 | +Algorithm: |
| 53 | +Fix one number and use a hash table to find the remaining two numbers. |
| 54 | +For each pair of numbers, check if a third number exists in the hash table to complete the triplet. |
| 55 | + |
| 56 | +```js |
| 57 | +function findTripletsUsingHashing(arr, targetSum) { |
| 58 | + const triplets = []; |
| 59 | + const n = arr.length; |
| 60 | + |
| 61 | + for (let i = 0; i < n - 2; i++) { |
| 62 | + const seen = new Set(); // Hash table to store complements |
| 63 | + const currentTarget = targetSum - arr[i]; |
| 64 | + |
| 65 | + for (let j = i + 1; j < n; j++) { |
| 66 | + const complement = currentTarget - arr[j]; |
| 67 | + if (seen.has(complement)) { |
| 68 | + triplets.push([arr[i], arr[j], complement]); |
| 69 | + } |
| 70 | + seen.add(arr[j]); // Add the current number to the hash table |
| 71 | + } |
| 72 | + } |
| 73 | + |
| 74 | + return triplets; |
| 75 | +} |
| 76 | + |
| 77 | +// Example usage |
| 78 | +const array = [1, 2, 3, 4, 5, 6]; |
| 79 | +const target = 10; |
| 80 | +console.log(findTripletsUsingHashing(array, target)); // Output: [[1, 4, 5], [2, 3, 5]] |
| 81 | +``` |
| 82 | + |
| 83 | +| Approach | Time Complexity | Space Complexity | Notes | |
| 84 | +| :-- | :-- | :-- | :-- | |
| 85 | +| Brute Force | (O(n^3)) | (O(1)) | Very inefficient for large arrays. Avoid using this in real-world apps. | |
| 86 | +| Two Pointers (Sorting) | (O(n^2)) | (O(1)) | Efficient and handles duplicate values. Requires sorting the array. | |
| 87 | +| Hashing | (O(n^2)) | (O(n)) | Efficient but uses extra space for hash table. No need to sort array. | |
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