Description
https://leetcode.com/problems/4sum/description/
Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
The solution set must not contain duplicate quadruplets.
Example:
Given array nums = [1, 0, -1, 0, -2, 2], and target = 0. A solution set is: [ [-1, 0, 0, 1], [-2, -1, 1, 2], [-2, 0, 0, 2] ]
Explanation
For 4 sum, we need to find 3 other numbers add with nums[i] to equal to the target.
Then it becomes the 3 sum problem.
For 3 sum, we need to find 2 other numbers add with nums[j] to equal to the target – nums[i].
Then it becomes the 2 sum problem.
We sorted the input array. Then we can use two pointers techniques to solve.
Video Tutorial
Java Solution
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length - 3; i++) {
if (i != 0 && nums[i] == nums[i - 1]) {
continue;
}
for (int j = i + 1; j < nums.length - 2; j++) {
if (j != i + 1 && nums[j] == nums[j - 1]) {
continue;
}
int left = j + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[j] + nums[left] + nums[right];
if (sum < target) {
left++;
} else if (sum > target) {
right--;
} else {
List<Integer> quadruplet = new ArrayList<>();
quadruplet.add(nums[i]);
quadruplet.add(nums[j]);
quadruplet.add(nums[left]);
quadruplet.add(nums[right]);
result.add(quadruplet);
left++;
right--;
while (left < right && nums[left] == nums[left - 1]) {
left++;
}
while (left < right && nums[right] == nums[right + 1]) {
right--;
}
}
}
}
}
return result;
// Time Complexity: O(n^3)
// Space Complexity: O(n^3)
}
}