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18. 4Sum 👍

  • Time: $O(n^3)$
  • Space: $O(1)$
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class Solution {
 public:
  vector<vector<int>> fourSum(vector<int>& nums, int target) {
    vector<vector<int>> ans;
    vector<int> path;
    ranges::sort(nums);
    nSum(nums, 4, target, 0, nums.size() - 1, path, ans);
    return ans;
  }

 private:
  // Finds n numbers that add up to the target in [l, r].
  void nSum(const vector<int>& nums, long n, long target, int l, int r,
            vector<int>& path, vector<vector<int>>& ans) {
    if (r - l + 1 < n || target < nums[l] * n || target > nums[r] * n)
      return;
    if (n == 2) {
      // Similar to the sub procedure in 15. 3Sum
      while (l < r) {
        const int sum = nums[l] + nums[r];
        if (sum == target) {
          path.push_back(nums[l]);
          path.push_back(nums[r]);
          ans.push_back(path);
          path.pop_back();
          path.pop_back();
          ++l;
          --r;
          while (l < r && nums[l] == nums[l - 1])
            ++l;
          while (l < r && nums[r] == nums[r + 1])
            --r;
        } else if (sum < target) {
          ++l;
        } else {
          --r;
        }
      }
      return;
    }

    for (int i = l; i <= r; ++i) {
      if (i > l && nums[i] == nums[i - 1])
        continue;
      path.push_back(nums[i]);
      nSum(nums, n - 1, target - nums[i], i + 1, r, path, ans);
      path.pop_back();
    }
  }
};

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class Solution {
  public List<List<Integer>> fourSum(int[] nums, int target) {
    List<List<Integer>> ans = new ArrayList<>();
    Arrays.sort(nums);
    nSum(nums, 4, target, 0, nums.length - 1, new ArrayList<>(), ans);
    return ans;
  }

  // Finds n numbers that add up to the target in [l, r].
  private void nSum(int[] nums, long n, long target, int l, int r, List<Integer> path,
                    List<List<Integer>> ans) {
    if (r - l + 1 < n || target < nums[l] * n || target > nums[r] * n)
      return;
    if (n == 2) {
      // Similar to the sub procedure in 15. 3Sum
      while (l < r) {
        final int sum = nums[l] + nums[r];
        if (sum == target) {
          path.add(nums[l]);
          path.add(nums[r]);
          ans.add(new ArrayList<>(path));
          path.remove(path.size() - 1);
          path.remove(path.size() - 1);
          ++l;
          --r;
          while (l < r && nums[l] == nums[l - 1])
            ++l;
          while (l < r && nums[r] == nums[r + 1])
            --r;
        } else if (sum < target) {
          ++l;
        } else {
          --r;
        }
      }
      return;
    }

    for (int i = l; i <= r; ++i) {
      if (i > l && nums[i] == nums[i - 1])
        continue;
      path.add(nums[i]);
      nSum(nums, n - 1, target - nums[i], i + 1, r, path, ans);
      path.remove(path.size() - 1);
    }
  }
}

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class Solution:
  def fourSum(self, nums: list[int], target: int):
    ans = []

    def nSum(
            l: int, r: int, target: int, n: int, path: list[int],
            ans: list[list[int]]) -> None:
      """Finds n numbers that add up to the target in [l, r]."""
      if r - l + 1 < n or n < 2 or target < nums[l] * n or target > nums[r] * n:
        return
      if n == 2:
        while l < r:
          summ = nums[l] + nums[r]
          if summ == target:
            ans.append(path + [nums[l], nums[r]])
            l += 1
            while nums[l] == nums[l - 1] and l < r:
              l += 1
          elif summ < target:
            l += 1
          else:
            r -= 1
        return

      for i in range(l, r + 1):
        if i > l and nums[i] == nums[i - 1]:
          continue

        nSum(i + 1, r, target - nums[i], n - 1, path + [nums[i]], ans)

    nums.sort()
    nSum(0, len(nums) - 1, target, 4, [], ans)
    return ans