611. Valid Triangle Number

MediumLeetCode

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

How to solve the problem

• Two Pointers
• Brute Force (Time Limit Exceeded)

# Approach 1: Two Pointers
class Solution:
    def triangleNumber(self, nums: List[int]) -> int:
        n = len(nums)
        nums.sort()
        result = 0
        for k in range(2, n):  # k is the longest side starting from the third element in nums
            left = 0
            right = k - 1
            while left < right:
                if nums[left] + nums[right] > nums[k]:
                    result += right - left # all the left to right are available
                    right -= 1
                else:
                    left += 1
        return result

# Approach 2: Brute Force (Time Limit Exceeded)
class Solution:
    def triangleNumber(self, nums: List[int]) -> int:
        n = len(nums)
        result = 0
        nums.sort()
        for i in range(n):
            for j in range(i + 1, n):
                for k in range(j + 1, n):
                    if nums[i] + nums[j] > nums[k]:
                        result += 1
                        continue
        return result

Complexity

• Approach 1 (Two Pointers): Time complexity: O(n²), Space complexity: O(1)
• Approach 2 (Brute Force): Time complexity: O(n³), Space complexity: O(1)