3Sum LeetCode | 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]]
Example 2:
Input: nums = [] Output: []
Example 3:
Input: nums = [0] Output: []
Constraints:
0 <= nums.length <= 3000
-105 <= nums[i] <= 105
3Sum LeetCode Solutions
✅Time: O(n*n)
✅Space: O(|ans|)
C++
class Solution {
public:
string intToRoman(int num) {
const vector<pair<int, string>> valueSymbols{
{1000, "M"}, {900, "CM"}, {500, "D"}, {400, "CD"}, {100, "C"},
{90, "XC"}, {50, "L"}, {40, "XL"}, {10, "X"}, {9, "IX"},
{5, "V"}, {4, "IV"}, {1, "I"}};
string ans;
for (const auto& [value, symbol] : valueSymbols) {
if (num == 0)
break;
while (num >= value) {
num -= value;
ans += symbol;
}
}
return ans;
}
};
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
if (nums.size() < 3)
return {};
vector<vector<int>> ans;
sort(begin(nums), end(nums));
for (int i = 0; i + 2 < nums.size(); ++i) {
if (i > 0 && nums[i] == nums[i - 1])
continue;
// choose nums[i] as the first num in the triplet,
// and search the remaining nums in [i + 1, n - 1]
int l = i + 1;
int r = nums.size() - 1;
while (l < r) {
const int sum = nums[i] + nums[l] + nums[r];
if (sum == 0) {
ans.push_back({nums[i], nums[l++], nums[r--]});
while (l < r && nums[l] == nums[l - 1])
++l;
while (l < r && nums[r] == nums[r + 1])
--r;
} else if (sum < 0) {
++l;
} else {
--r;
}
}
}
return ans;
}
};
class Solution {
public:
string intToRoman(int num) {
const vector<pair<int, string>> valueSymbols{
{1000, "M"}, {900, "CM"}, {500, "D"}, {400, "CD"}, {100, "C"},
{90, "XC"}, {50, "L"}, {40, "XL"}, {10, "X"}, {9, "IX"},
{5, "V"}, {4, "IV"}, {1, "I"}};
string ans;
for (const auto& [value, symbol] : valueSymbols) {
if (num == 0)
break;
while (num >= value) {
num -= value;
ans += symbol;
}
}
return ans;
}
};
Java
class Solution {
public String intToRoman(int num) {
final int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
final String[] symbols = {"M", "CM", "D", "CD", "C", "XC", "L",
"XL", "X", "IX", "V", "IV", "I"};
StringBuilder sb = new StringBuilder();
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
if (nums.length < 3)
return new ArrayList<>();
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i + 2 < nums.length; ++i) {
if (i > 0 && nums[i] == nums[i - 1])
continue;
// choose nums[i] as the first num in the triplet,
// and search the remaining nums in [i + 1, n - 1]
int l = i + 1;
int r = nums.length - 1;
while (l < r) {
final int sum = nums[i] + nums[l] + nums[r];
if (sum == 0) {
ans.add(Arrays.asList(nums[i], nums[l++], nums[r--]));
while (l < r && nums[l] == nums[l - 1])
++l;
while (l < r && nums[r] == nums[r + 1])
--r;
} else if (sum < 0) {
++l;
} else {
--r;
}
}
}
return ans;
}
}
Python
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
if len(nums) < 3:
return []
ans = []
nums.sort()
for i in range(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
l = i + 1
r = len(nums) - 1
while l < r:
sum = nums[i] + nums[l] + nums[r]
if sum == 0:
ans.append((nums[i], nums[l], nums[r]))
l += 1
r -= 1
while nums[l] == nums[l - 1] and l < r:
l += 1
while nums[r] == nums[r + 1] and l < r:
r -= 1
elif sum < 0:
l += 1
else:
r -= 1
return ans
3Sum LeetCode Tutorial
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