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package leetcode.arrays;
/* Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j,
i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
Example 1:
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
Constraints:
3 <= nums.length <= 3000
-105 <= nums[i] <= 105
Optimal Approach: Sorting + Two Pointers (O(n^2) Time, O(1) Extra Space)
1. Sort the Array → Helps avoid duplicate triplets efficiently.
2. Fix One Element and Use Two-Pointer:
• Fix nums[i] as the first element.
• Use two-pointer (left, right) to find nums[left] + nums[right] = -nums[i].
• Move left or right based on sum.
3. Skip Duplicates:
• If nums[i] == nums[i-1], skip to avoid duplicate triplets.
*/
import java.util.*;
import java.util.stream.Collectors;
public class ThreeSum {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums); // Step 1: Sort the array
for (int i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1])
continue; // Skip duplicate elements
int left = i + 1, right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum == 0) {
result.add(Arrays.asList(nums[i], nums[left], nums[right]));
// Skip duplicate `left` values
while (left < right && nums[left] == nums[left + 1])
left++;
// Skip duplicate `right` values
while (left < right && nums[right] == nums[right - 1])
right--;
left++;
right--;
} else if (sum < 0) {
left++; // Need a larger sum, move `left` right
} else {
right--; // Need a smaller sum, move `right` left
}
}
}
return result;
}
// Using Java 8
public List<List<Integer>> threeSumUsingJava8(int[] nums) {
Arrays.sort(nums);
return Arrays.stream(nums)
.distinct()
.boxed()
.flatMap(i -> Arrays.stream(nums)
.filter(j -> j > i)
.boxed()
.flatMap(j -> Arrays.stream(nums)
.filter(k -> k > j && i + j + k == 0)
.boxed()
.map(k -> Arrays.asList(i, j, k))))
.distinct()
.collect(Collectors.toList());
}
public static void main(String[] args) {
ThreeSum solution = new ThreeSum();
int[] nums1 = { -1, 0, 1, 2, -1, -4 };
System.out.println(solution.threeSum(nums1)); // Output: [[-1, -1, 2], [-1, 0, 1]]
int[] nums2 = { 0, 1, 1 };
System.out.println(solution.threeSum(nums2)); // Output: []
int[] nums3 = { 0, 0, 0 };
System.out.println(solution.threeSum(nums3)); // Output: [[0, 0, 0]]
}
}