18. 4Sum

package Leetcode.TwoPointer;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * @author zhengxingxing
 * @date 2025/02/23
 */
public class FourSum {

    public static List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList<>();
        // Check for invalid input
        if (nums == null || nums.length < 4) {
            return result;
        }

        // Sort array to use two-pointer technique and handle duplicates
        Arrays.sort(nums);
        int n = nums.length;

        // Fix first number
        for (int i = 0; i < n - 3; i++) {
            // Skip duplicates for first number
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }

            // Early termination if smallest possible sum is too large
            if ((long)nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {
                break;
            }
            // Skip if largest possible sum is too small
            if ((long)nums[i] + nums[n - 3] + nums[n - 2] + nums[n - 1] < target) {
                continue;
            }

            // Fix second number
            for (int j = i + 1; j < n - 2; j++) {
                // Skip duplicates for second number
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }

                // Early termination optimizations
                if ((long)nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {
                    break;
                }
                if ((long)nums[i] + nums[j] + nums[n - 2] + nums[n - 1] < target) {
                    continue;
                }

                // Use two pointers to find remaining two numbers
                int left = j + 1, right = n - 1;
                while (left < right) {
                    long sum = (long)nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        // Add valid quadruplet to result
                        result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                        // Skip duplicates for third number
                        while (left < right && nums[left] == nums[left + 1]) {
                            left++;
                        }
                        // Skip duplicates for fourth number
                        while (left < right && nums[right] == nums[right - 1]) {
                            right--;
                        }
                        left++;
                        right--;
                    } else if (sum < target) {
                        left++;  // Sum too small, increase left pointer
                    } else {
                        right--; // Sum too large, decrease right pointer
                    }
                }
            }
        }
        return result;
    }
    
    public static void main(String[] args) {
        // Test Case 1: Standard case with multiple solutions
        int[] nums1 = {1, 0, -1, 0, -2, 2};
        int target1 = 0;
        System.out.println("Test Case 1:");
        System.out.println("Input: nums = [1,0,-1,0,-2,2], target = 0");
        System.out.println("Output: " + fourSum(nums1, target1));
        System.out.println();

        // Test Case 2: Array with all same elements
        int[] nums2 = {2, 2, 2, 2, 2};
        int target2 = 8;
        System.out.println("Test Case 2:");
        System.out.println("Input: nums = [2,2,2,2,2], target = 8");
        System.out.println("Output: " + fourSum(nums2, target2));
        System.out.println();

        // Test Case 3: Large dataset with positive and negative numbers
        int[] nums3 = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5};
        int target3 = 0;
        System.out.println("Test Case 3 (Large Dataset):");
        System.out.println("Output: " + fourSum(nums3, target3));
    }
}