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Valid Triangle Number

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

Solution Explanation for Valid Triangle Number

The problem asks to find the number of triplets in an array nums that can form the sides of a triangle. A triangle is valid if the sum of the lengths of any two sides is greater than the length of the third side.

Approach:

The most efficient approach involves sorting the array and then using a two-pointer technique or binary search for each pair of potential sides. Here's a breakdown:

  1. Sorting: First, sort the input array nums in ascending order. This is crucial for efficiently finding valid triplets. Sorting takes O(n log n) time.

  2. Iterating Through Potential Sides: Iterate through the array using a nested loop. The outer loop iterates from i = 0 to n - 3 (where n is the length of nums). The inner loop iterates from j = i + 1 to n - 2. Each pair (nums[i], nums[j]) represents two potential sides of a triangle.

  3. Finding the Third Side (Two-Pointer or Binary Search): For each pair (nums[i], nums[j]), we need to find how many numbers in the array are greater than nums[i] + nums[j] and are also greater than nums[j] (to ensure the third side is larger than the second side and obeys the triangle inequality). There are two methods to achieve this efficiently:

    • Two-Pointer Approach: This method is faster. Use a left pointer starting at j + 1 and a right pointer at n - 1. Move the pointers until the condition nums[left] + nums[j] > nums[i] is satisfied. The number of valid triangles is then given by right - left. This approach has a time complexity of O(n) for the inner loop and O(n^2) for the outer loops.

    • Binary Search Approach: We can use binary search to find the largest index k such that nums[k] < nums[i] + nums[j]. This binary search takes O(log n) time. The number of valid triangles for this pair is then k - j. This approach has a time complexity of O(n^2 log n).

Time Complexity Analysis:

  • Sorting: O(n log n)
  • Two-Pointer Approach: O(n^2) (Outer loop is O(n^2), inner loop is O(n))
  • Binary Search Approach: O(n^2 log n) (Outer loop is O(n^2), binary search is O(log n))

Therefore, the overall time complexity of the solution using the two-pointer approach is O(n^2), dominated by the nested loops. The solution using binary search is O(n^2 log n). The space complexity is O(1) (constant) because we're only using a few extra variables, not dependent on the input size.

Space Complexity Analysis: O(1)

Code Examples (Two-Pointer Approach):

The provided Java, C++, Go, and TypeScript solutions all use the two-pointer approach, which is more efficient than binary search in this case. The Python solution uses bisect_left which is essentially a binary search, leading to slightly higher time complexity.

This detailed explanation, including complexity analysis and code examples in multiple languages, gives a comprehensive understanding of how to solve the Valid Triangle Number problem efficiently. The two-pointer approach is generally preferred for its better time complexity in this scenario.