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LeetCode Solutions

259. 3Sum Smaller

Time: $O(n^2)$

Space: $O(1)$

			


class Solution {
 public:
  int threeSumSmaller(vector<int>& nums, int target) {
    if (nums.size() < 3)
      return 0;

    int ans = 0;

    sort(begin(nums), end(nums));

    for (int i = 0; i + 2 < nums.size(); ++i) {
      int l = i + 1;
      int r = nums.size() - 1;
      while (l < r)
        if (nums[i] + nums[l] + nums[r] < target) {
          // (nums[i], nums[l], nums[r])
          // (nums[i], nums[l], nums[r - 1])
          // ...,
          // (nums[i], nums[l], nums[l + 1])
          ans += r - l;
          ++l;
        } else {
          --r;
        }
    }

    return ans;
  }
};




			


class Solution {
  public int threeSumSmaller(int[] nums, int target) {
    if (nums.length < 3)
      return 0;

    int ans = 0;

    Arrays.sort(nums);

    for (int i = 0; i + 2 < nums.length; ++i) {
      int l = i + 1;
      int r = nums.length - 1;
      while (l < r)
        if (nums[i] + nums[l] + nums[r] < target) {
          // (nums[i], nums[l], nums[r])
          // (nums[i], nums[l], nums[r - 1])
          // ...,
          // (nums[i], nums[l], nums[l + 1])
          ans += r - l;
          ++l;
        } else {
          --r;
        }
    }

    return ans;
  }
}




			


class Solution:
  def threeSumSmaller(self, nums: List[int], target: int) -> int:
    if len(nums) < 3:
      return 0

    ans = 0

    nums.sort()

    for i in range(len(nums) - 2):
      l = i + 1
      r = len(nums) - 1
      while l < r:
        if nums[i] + nums[l] + nums[r] < target:
          # (nums[i], nums[l], nums[r])
          # (nums[i], nums[l], nums[r - 1])
          # ...,
          # (nums[i], nums[l], nums[l + 1])
          ans += r - l
          l += 1
        else:
          r -= 1

    return ans