Problem Overview & Intuition

By sorting the array, we fix two pointers (i, j) in nested loops and use two pointers (left, right) for the remaining pair. This reduces the search from O(N⁴) brute force to O(N³). Duplicate skipping ensures uniqueness.

Step-by-Step Approach

1. Sort nums in ascending order.
2. Outer loop i from 0 to n - 4, skipping duplicates.
3. Inner loop j from i + 1 to n - 3, skipping duplicates.
4. Run two pointers left = j + 1, right = n - 1.
5. When sum === target, record quadruplet and skip inner duplicates.
6. Return result.

Optimal Implementation (JavaScript)

function fourSum(nums, target) {
  nums.sort((a, b) => a - b);
  const result = [];
  const n = nums.length;
  for (let i = 0; i < n - 3; i++) {
    if (i > 0 && nums[i] === nums[i - 1]) continue;
    for (let j = i + 1; j < n - 2; j++) {
      if (j > i + 1 && nums[j] === nums[j - 1]) continue;
      let left = j + 1, right = n - 1;
      while (left < right) {
        const sum = nums[i] + nums[j] + nums[left] + nums[right];
        if (sum === target) {
          result.push([nums[i], nums[j], nums[left], nums[right]]);
          while (left < right && nums[left] === nums[left + 1]) left++;
          while (left < right && nums[right] === nums[right - 1]) right--;
          left++;
          right--;
        } else if (sum < target) left++;
        else right--;
      }
    }
  }
  return result;
}

Complexity Analysis