How do you solve Valid Triangle Number on LeetCode?

Valid Triangle Number (LeetCode #611) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Sorting helps reduce the number of checks needed.

What is the brute force solution for Valid Triangle Number?

The brute force approach for Valid Triangle Number is Brute Force, with O(n³) time complexity and O(1) space complexity. The brute-force approach checks all possible triplets in the array to see if they can form a triangle. This is straightforward but inefficient as it examines every combination. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Valid Triangle Number?

Valid Triangle Number uses the following concepts: Array, Two Pointers, Binary Search, Greedy, Sorting. The recommended patterns to study are: Two Pointers, Sorting.

What are the interview tips for Valid Triangle Number?

Always consider edge cases, such as arrays with fewer than three elements. Explain your thought process clearly while coding. Practice identifying patterns in problems to improve your problem-solving speed.

What are common mistakes when solving Valid Triangle Number?

Not sorting the array before applying the two-pointer technique. Misunderstanding the triangle inequality conditions.

#611

Valid Triangle Number

Medium

ArrayTwo PointersBinary SearchGreedySortingTwo PointersSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n³)	O(n²)
Space	O(1)	O(1)

💡

Intuition

Time O(n²)Space O(1)

By sorting the array first, we can use a two-pointer technique to efficiently count valid triplets. This reduces the number of checks needed by leveraging the properties of sorted arrays.

⚙️

Algorithm

4 steps

1Step 1: Sort the array in non-decreasing order.
2Step 2: Iterate through the array with index 'k' starting from the third element.
3Step 3: For each 'k', use two pointers 'i' (starting from 0) and 'j' (starting from k-1) to find valid pairs (i, j) such that nums[i] + nums[j] > nums[k].
4Step 4: If a valid pair is found, all elements between 'i' and 'j' can form valid triangles with nums[k]. Increment the count accordingly and move the pointer 'i' forward.

solution.py13 lines

1def triangleNumber(nums):
2    nums.sort()
3    count = 0
4    n = len(nums)
5    for k in range(2, n):
6        i, j = 0, k - 1
7        while i < j:
8            if nums[i] + nums[j] > nums[k]:
9                count += (j - i)
10                j -= 1
11            else:
12                i += 1
13    return count

ℹ

Complexity note: The time complexity is O(n²) due to the two-pointer technique inside a single loop, while the sorting step is O(n log n). The space complexity is O(1) as we are using a constant amount of extra space.

1Sorting helps reduce the number of checks needed.
2Using two pointers can significantly improve efficiency.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.