Lesson 33 of 250
18. 4Sum
NeetCode: 4Sum · LeetCode: 18. 4Sum · Difficulty: Medium
Given an array nums of n integers and an integer target, return all unique quadruplets [a, b, c, d] (values, not indices) with a + b + c + d == target. The result must not contain duplicate quadruplets, and may be returned in any order. Sums can exceed the int range, so the code adds in long.
Example
Input: nums = [1, 0, -1, 0, -2, 2], target = 0 Output: [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]
Input: nums = [2, 2, 2, 2, 2], target = 8 Output: [[2, 2, 2, 2]]Brute force: four nested loops plus a set
Try every choice of four distinct indices i < j < k < l. When the sum matches, sort the four values and insert them into a TreeSet<List<Integer>> with a comparator that orders lists element by element; the set removes duplicate quadruplets and gives a deterministic order.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.TreeSet;
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
int n = nums.length;
// Lists compare by contents; this comparator orders them lexicographically.
TreeSet<List<Integer>> found = new TreeSet<>((a, b) -> {
for (int i = 0; i < 4; i++) {
int c = Integer.compare(a.get(i), b.get(i));
if (c != 0) {
return c;
}
}
return 0;
});
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
for (int k = j + 1; k < n; k++) {
for (int l = k + 1; l < n; l++) {
long sum = (long) nums[i] + nums[j] + nums[k] + nums[l];
if (sum == target) {
int[] q = {nums[i], nums[j], nums[k], nums[l]};
Arrays.sort(q);
found.add(Arrays.asList(q[0], q[1], q[2], q[3]));
}
}
}
}
}
return new ArrayList<>(found);
}
}
public class Main {
public static void main(String[] args) {
Solution s = new Solution();
System.out.println(s.fourSum(new int[]{1, 0, -1, 0, -2, 2}, 0));
System.out.println(s.fourSum(new int[]{2, 2, 2, 2, 2}, 8));
System.out.println(s.fourSum(new int[]{1, 2, 3}, 6));
System.out.println(s.fourSum(new int[]{1_000_000_000, 1_000_000_000, 1_000_000_000, 1_000_000_000}, -294_967_296));
}
}[[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]
[[2, 2, 2, 2]]
[]
[]Time: O(n^4) for the loops (plus O(log q) per hit for the tree set, where q is the number of distinct quadruplets). Space: O(q) for the set.
Optimal: sort, fix two numbers, sweep with two pointers
This extends 3Sum with one more outer loop. Sort the array first. Fix the first two numbers nums[i] and nums[j], then the remaining two must sum to target - nums[i] - nums[j]. Because the rest of the array is sorted, a left pointer l = j + 1 and a right pointer r = n - 1 find that pair in one pass: if the sum is too small move l right, if too big move r left.
Duplicates are avoided by skipping equal values instead of using a set: skip nums[i] if it equals the previous value at the same loop level, do the same for j (only when j > i + 1), and after recording a match advance l and r past equal neighbours. All sums are computed in long: (long) a[i] + a[j] + ... widens the first operand, so the whole addition is done in long. Adding four int values first and widening afterwards would wrap before the cast.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
int[] a = nums.clone();
Arrays.sort(a);
int n = a.length;
List<List<Integer>> res = new ArrayList<>();
for (int i = 0; i < n - 3; i++) {
if (i > 0 && a[i] == a[i - 1]) {
continue;
}
for (int j = i + 1; j < n - 2; j++) {
if (j > i + 1 && a[j] == a[j - 1]) {
continue;
}
int l = j + 1, r = n - 1;
while (l < r) {
long sum = (long) a[i] + a[j] + a[l] + a[r];
if (sum < target) {
l++;
} else if (sum > target) {
r--;
} else {
res.add(Arrays.asList(a[i], a[j], a[l], a[r]));
l++;
r--;
while (l < r && a[l] == a[l - 1]) {
l++;
}
while (l < r && a[r] == a[r + 1]) {
r--;
}
}
}
}
}
return res;
}
}
public class Main {
public static void main(String[] args) {
Solution s = new Solution();
System.out.println(s.fourSum(new int[]{1, 0, -1, 0, -2, 2}, 0));
System.out.println(s.fourSum(new int[]{2, 2, 2, 2, 2}, 8));
System.out.println(s.fourSum(new int[]{1, 2, 3}, 6));
System.out.println(s.fourSum(new int[]{1_000_000_000, 1_000_000_000, 1_000_000_000, 1_000_000_000}, -294_967_296));
}
}[[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]
[[2, 2, 2, 2]]
[]
[]Time: O(n^3) (sorting is O(n log n), two nested loops times an O(n) sweep). Space: O(1) extra besides the sorted copy (nums.clone(), so the input is not changed) and the output.
The last test uses four values of 1_000_000_000. Their true sum, 4_000_000_000, does not fit in int, and the target would match only after an int overflow wrap, so the correct answer is no match.