algoadvance

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
Explanation: Valid combinations are three 4,3,4 and one 2,3,4.

Constraints:

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

Clarifying Questions

1. Can the array contain duplicate elements?
   • Yes, as evident from the example.
2. What should be done if any side length is zero?
   • Zero cannot be a part of any triangle, so we can skip such cases. Given the constraints and problem nature, we assume handling them appropriately.
3. Should the solution handle negative side lengths?
   • The problem guarantees non-negative integers for side lengths, so no need to handle negative values.

Strategy

To solve this problem efficiently, we use the fact that for any three sides to form a triangle, the sum of any two sides must be greater than the third side. We particularly leverage the sorted order of the array to reduce the number of comparisons:

1. Sort the Array: Sorting helps in applying the triangle inequality more straightforwardly.
2. Three-Pointer Approach: Use three nested loops but optimized to avoid unnecessary comparisons:
   • Fix one side (let it be the longest side).
   • Use two pointers to check pairs from the remaining part of the array to see if they form a triangle.

Detailed Steps:

1. Sort the Array: This allows us to focus on the relationship nums[i] + nums[j] > nums[k] more easily.
2. Nested Loop Approach: Use nested loops, but instead of O(n^3) complexity, optimize it by leveraging the sorted nature:
   • Use the innermost loop combined with two pointers technique to directly count valid triangles.

Pseudocode:

• Sort the array.
• Use three nested loops.
• For the innermost loop, start with two pointers, potentially reducing the complexity to O(n^2).

Code

def triangleNumber(nums):
    nums.sort()
    count = 0
    n = len(nums)
    
    for i in range(n-2):
        k = i + 2
        for j in range(i+1, n-1):
            while k < n and nums[i] + nums[j] > nums[k]:
                k += 1
            count += k - j - 1
    
    return count

Time Complexity

• Sorting: O(n log n) where n is the number of elements in nums.
• Nested Loops: The outer two loops go through the array making it O(n^2), and the innermost while-loop runs in constant time on average over all iterations.
• Overall Time Complexity: O(n^2), which is efficient for the given constraints (n <= 1000).

This solution ensures that we consider only those side lengths that can actually form a triangle by sorting and strategically using the triangle inequality rule.

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