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611. Valid Triangle Number

Description

Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.

 

Example 1:

Input: nums = [2,2,3,4]
Output: 3
Explanation: Valid combinations are: 
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3

Example 2:

Input: nums = [4,2,3,4]
Output: 4

 

Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 1000

Solutions

First enumerate two edges, and then use binary search to locate the third edge.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int triangleNumber(int[] nums) {
          Arrays.sort(nums);
          int n = nums.length;
          int res = 0;
          for (int i = n - 1; i >= 2; --i) {
              int l = 0, r = i - 1;
              while (l < r) {
                  if (nums[l] + nums[r] > nums[i]) {
                      res += r - l;
                      --r;
                  } else {
                      ++l;
                  }
              }
          }
          return res;
      }
  }
  
  

• class Solution {
  public:
      int triangleNumber(vector<int>& nums) {
          sort(nums.begin(), nums.end());
          int ans = 0, n = nums.size();
          for (int i = 0; i < n - 2; ++i) {
              for (int j = i + 1; j < n - 1; ++j) {
                  int k = lower_bound(nums.begin() + j + 1, nums.end(), nums[i] + nums[j]) - nums.begin() - 1;
                  ans += k - j;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def triangleNumber(self, nums: List[int]) -> int:
          nums.sort()
          ans, n = 0, len(nums)
          for i in range(n - 2):
              for j in range(i + 1, n - 1):
                  k = bisect_left(nums, nums[i] + nums[j], lo=j + 1) - 1
                  ans += k - j
          return ans
  
  

• func triangleNumber(nums []int) int {
  	sort.Ints(nums)
  	ans := 0
  	for i, n := 0, len(nums); i < n-2; i++ {
  		for j := i + 1; j < n-1; j++ {
  			left, right := j+1, n
  			for left < right {
  				mid := (left + right) >> 1
  				if nums[mid] >= nums[i]+nums[j] {
  					right = mid
  				} else {
  					left = mid + 1
  				}
  			}
  			ans += left - j - 1
  		}
  	}
  	return ans
  }
  

• function triangleNumber(nums: number[]): number {
      nums.sort((a, b) => a - b);
      let n = nums.length;
      let ans = 0;
      for (let i = n - 1; i >= 2; i--) {
          let left = 0,
              right = i - 1;
          while (left < right) {
              if (nums[left] + nums[right] > nums[i]) {
                  ans += right - left;
                  right--;
              } else {
                  left++;
              }
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn triangle_number(mut nums: Vec<i32>) -> i32 {
          nums.sort();
          let n = nums.len();
          let mut res = 0;
          for i in (2..n).rev() {
              let mut left = 0;
              let mut right = i - 1;
              while left < right {
                  if nums[left] + nums[right] > nums[i] {
                      res += right - left;
                      right -= 1;
                  } else {
                      left += 1;
                  }
              }
          }
          res as i32
      }
  }
  
  

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