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259. 3Sum Smaller

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

• Java
• C++
• Python
• Go
• Javascript
• TypeScript

• 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;
      }
  };
  

• 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
  
  

• 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;
  };
  
  

• function threeSumSmaller(nums: number[], target: number): number {
      nums.sort((a, b) => a - b);
      const n = nums.length;
      let ans = 0;
      for (let i = 0; i < n - 2; ++i) {
          let [j, k] = [i + 1, n - 1];
          while (j < k) {
              const x = nums[i] + nums[j] + nums[k];
              if (x < target) {
                  ans += k - j;
                  ++j;
              } else {
                  --k;
              }
          }
      }
      return ans;
  }
  
  

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