Question

259. 3Sum Smaller

中文文档

Description

Given an array of n integers nums and an integer target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

 

Example 1:

Input: nums = [-2,0,1,3], target = 2
Output: 2
Explanation: Because there are two triplets which sums are less than 2:
[-2,0,1]
[-2,0,3]

Example 2:

Input: nums = [], target = 0
Output: 0

Example 3:

Input: nums = [0], target = 0
Output: 0

 

Constraints:

• n == nums.length
• 0 <= n <= 3500
• -100 <= nums[i] <= 100
• -100 <= target <= 100

Solutions

Solution 1

class Solution:
  def threeSumSmaller(self, nums: List[int], target: int) -> int:
    nums.sort()
    ans, n = 0, len(nums)
    for i in range(n):
      j, k = i + 1, n - 1
      while j < k:
        s = nums[i] + nums[j] + nums[k]
        if s >= target:
          k -= 1
        else:
          ans += k - j
          j += 1
    return ans

class Solution {
  public int threeSumSmaller(int[] nums, int target) {
    Arrays.sort(nums);
    int ans = 0;
    for (int i = 0, n = nums.length; i < n; ++i) {
      int j = i + 1;
      int k = n - 1;
      while (j < k) {
        int s = nums[i] + nums[j] + nums[k];
        if (s >= target) {
          --k;
        } else {
          ans += k - j;
          ++j;
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  int threeSumSmaller(vector<int>& nums, int target) {
    sort(nums.begin(), nums.end());
    int ans = 0;
    for (int i = 0, n = nums.size(); i < n; ++i) {
      int j = i + 1, k = n - 1;
      while (j < k) {
        int s = nums[i] + nums[j] + nums[k];
        if (s >= target)
          --k;
        else {
          ans += k - j;
          ++j;
        }
      }
    }
    return ans;
  }
};

func threeSumSmaller(nums []int, target int) int {
  sort.Ints(nums)
  ans := 0
  for i, n := 0, len(nums); i < n; i++ {
    j, k := i+1, n-1
    for j < k {
      s := nums[i] + nums[j] + nums[k]
      if s >= target {
        k--
      } else {
        ans += k - j
        j++
      }
    }
  }
  return ans
}

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var threeSumSmaller = function (nums, target) {
  nums.sort((a, b) => a - b);
  let ans = 0;
  for (let i = 0, n = nums.length; i < n; ++i) {
    let j = i + 1;
    let k = n - 1;
    while (j < k) {
      s = nums[i] + nums[j] + nums[k];
      if (s >= target) {
        --k;
      } else {
        ans += k - j;
        ++j;
      }
    }
  }
  return ans;
};