3Sum LeetCode Solution | Easy Approach | C++, Java, Python

Minimum Cost to Merge Stones
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3Sum LeetCode | Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != ji != 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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