#15. 3Sum
- Solved
Description
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
My Solution
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
public class Solution {
public IList<IList<int>> ThreeSum(int[] nums) {
Array.Sort(nums);
List<IList<int>> result = new List<IList<int>>();
for(int i = 0; i < nums.Length -2; i++) {
if(i > 0 && nums[i] == nums[i-1]) {
continue;
}
for(int j = i + 1; j < nums.Length -1; j++) {
if(j > i + 1 && nums[j] == nums[j-1]) {
continue;
}
for(int k = j + 1; k < nums.Length; k++) {
if(k > j + 1 && nums[k] == nums[k-1])
continue;
if(nums[i] + nums[j] + nums[k] == 0) {
result.Add(new List<int>() {nums[i], nums[j], nums[k]});
}
}
}
}
return result;
}
}
Runtime
Time Limit Exceeded
Memory
Time Limit Exceeded
Big O Notation
Time complexity: O(n^3)
Space complexity: O(n)
Reflection
Because of Time Complexity of O(n^3), I could not submit this solution. This question is categorized with ‘Two Pointers’. I have to find the way use just 2 pointers not 3 pointers.
Best Solution
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
public class Solution {
public IList<IList<int>> ThreeSum(int[] nums) {
List<IList<int>> result = new List<IList<int>>();
Array.Sort(nums);
for (int i = 0; i < nums.Length; i++) {
if (i > 0 && nums[i] == nums[i-1]) {
continue;
}
int j = i + 1;
int k = nums.Length - 1;
while (j < k) {
int total = nums[i] + nums[j] + nums[k];
if (total > 0) {
k--;
} else if (total < 0) {
j++;
} else {
result.Add(new List<int>() {nums[i], nums[j], nums[k]});
j++;
while (nums[j] == nums[j-1] && j < k) {
j++;
}
}
}
}
return result;
}
}
Runtime
26 ms / Beats 94.09%
Memory
74.92 MB / Beats 86.48%
Big O Notation
Time complexity: O(n^2)
Space complexity: O(n)