LeetCode 611: Valid Triangle Number Solution in Python – A Step-by-Step Guide

What Is LeetCode 611: Valid Triangle Number?

In LeetCode 611: Valid Triangle Number, you’re given an array of non-negative integers representing stick lengths, and your task is to count how many unique triplets (sets of three distinct elements) can form a valid triangle. A triplet [a, b, c] is valid if a + b > c, a + c > b, and b + c > a. For example, in [2, 3, 4, 5], triplets like [2, 3, 4] and [3, 4, 5] work, but [2, 3, 5] doesn’t (2+3=5, not >5). This problem tests your ability to optimize triplet counting with the triangle inequality theorem.

Problem Statement

• Input: A list of non-negative integers nums.
• Output: An integer—the number of valid triangle triplets.
• Rules:
• Valid triangle: a + b > c, a + c > b, b + c > a.
• Triplets are unordered (e.g., [2,3,4] same as [4,3,2]).
• Use distinct elements (no repeats within a triplet).

Constraints

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

Examples

• Input: [2, 3, 4, 5]
• Output: 3 (Valid: [2,3,4], [2,4,5], [3,4,5])
• Input: [1, 2, 3, 4]
• Output: 3 (Valid: [1,2,3], [1,2,4], [2,3,4])
• Input: [1, 1, 1, 1]
• Output: 4 (Valid: any 3 of the 1s, e.g., [1,1,1] four ways)

These examples hint at a pattern—let’s solve it!

Understanding the Problem: Counting Triangle Triplets

To solve LeetCode 611: Valid Triangle Number in Python, we need to count triplets in an array that satisfy the triangle inequality. A brute-force approach—checking every triplet with three nested loops—is O(n³), far too slow for 1000 elements. Instead, we’ll use:

• Best Solution (Two-Pointer Technique): O(n²) time, O(1) space—fast and elegant.
• Alternative Solution (Binary Search): O(n² log n) time, O(1) space—smart but more complex.

Let’s start with the two-pointer solution, breaking it down for beginners.

Best Solution: Two-Pointer Technique After Sorting

Why This Is the Best Solution

The two-pointer technique is the top choice for LeetCode 611 because it’s efficient—O(n²) time after an O(n log n) sort—and uses minimal extra space (O(1)). By sorting the array first, it simplifies the inequality check to just a + b > c (where a ≤ b < c), and two pointers optimize the counting process. It’s like lining up sticks and sliding pointers to count valid sets!

How It Works

Think of this as a stick-sorting game:

• Step 1: Sort the Array:
• Arrange nums in ascending order.
• Step 2: Fix the Largest Side:
• For each c (largest side), use two pointers for a and b.
• Step 3: Use Two Pointers:
• left starts at 0, right at c-1.
• If nums[left] + nums[right] > nums[c], all pairs from left to right work (since right is the smallest possible b).
• Count right - left valid pairs, move right left.
• If not, move left right.
• Step 4: Sum Counts:
• Add up all valid triplets.

It’s like a sliding window of triangle possibilities!

Step-by-Step Example

Example: [2, 3, 4, 5]

• Sort: [2, 3, 4, 5] (already sorted).
• c = 5 (index 3):
• left = 0 (2), right = 2 (4):
• 2 + 4 = 6 > 5, count = 2 - 0 = 2 ([2,4,5], [3,4,5]).
• right = 1 (3): 2 + 3 = 5, not > 5, left += 1.
• left = 1 (3), right = 1: Stop (left ≥ right).
• c = 4 (index 2):
• left = 0 (2), right = 1 (3):
• 2 + 3 = 5 > 4, count += 1 - 0 = 1 ([2,3,4]).
• right = 0: Stop.
• c = 3 (index 1):
• No pairs possible (need 2 smaller numbers).
• Total: 2 + 1 = 3.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def triangleNumber(self, nums: List[int]) -> int:
        # Step 1: Sort the array
        nums.sort()
        n = len(nums)
        count = 0

        # Step 2: Iterate over largest side
        for c in range(2, n):
            left = 0
            right = c - 1

            # Step 3: Two-pointer technique
            while left < right:
                if nums[left] + nums[right] > nums[c]:
                    count += right - left  # All pairs from left to right work
                    right -= 1
                else:
                    left += 1

        # Step 4: Return total count
        return count

• Line 4: Sort nums ascending.
• Lines 8-18: Process each c:
• left and right define range of smaller sides.
• If left + right > c, count all valid bs from left to right.
• Adjust pointers based on inequality.
• Time Complexity: O(n²)—sort is O(n log n), two-pointer is O(n²).
• Space Complexity: O(1)—no extra space.

This is like a triangle counter—fast and sleek!

Alternative Solution: Binary Search Approach

Why an Alternative Approach?

The binary search approach also sorts the array but uses binary search to find valid third sides—O(n² log n) time, O(1) space. It’s clever, leveraging sorted order to pinpoint counts, but the log factor makes it slower than two-pointers. It’s like searching a sorted stick pile for matches!

How It Works

Picture this as a stick hunt:

• Step 1: Sort nums.
• Step 2: Fix two sides a and b.
• Step 3: Binary search for smallest c where a + b > c.
• Step 4: Count valid cs beyond b.
• Step 5: Sum all counts.

It’s like a precise triangle locator!

Step-by-Step Example

Example: [2, 3, 4, 5]

• Sort: [2, 3, 4, 5].
• a = 2, b = 3:
• 2 + 3 = 5, find c > 3 and 5 > c: None (4, 5 fail), count = 0.
• a = 2, b = 4:
• 2 + 4 = 6 > 5, count = 1 ([2,4,5]).
• a = 2, b = 5:
• No c possible.
• a = 3, b = 4:
• 3 + 4 = 7 > 5, count = 1 ([3,4,5]).
• a = 3, b = 5:
• No c.
• a = 4, b = 5:
• No c.
• Total: 0 + 1 + 1 = 2 (misses [2,3,4] due to implementation, corrected below).

Code for Binary Search Approach

Copy codeclass Solution:
    def triangleNumber(self, nums: List[int]) -> int:
        nums.sort()
        n = len(nums)
        count = 0

        for i in range(n-2):
            for j in range(i+1, n-1):
                # Binary search for smallest k where nums[i] + nums[j] > nums[k]
                left, right = j + 1, n - 1
                while left <= right:
                    mid = (left + right) // 2
                    if nums[i] + nums[j] > nums[mid]:
                        left = mid + 1
                    else:
                        right = mid - 1
                count += left - j - 1  # Number of valid k’s

        return count

• Lines 7-16: Fix a and b, search for c:
• Binary search finds first invalid c.
• Count valid cs between j and that point.
• Time Complexity: O(n² log n)—n² pairs, log n search.
• Space Complexity: O(1)—no extra space.

It’s a precise triangle finder—smart but slower!

Comparing the Two Solutions

• Two-Pointer (Best):
• Pros: O(n²) time, O(1) space, fast and elegant.
• Cons: Requires sorting intuition.
• Binary Search (Alternative):
• Pros: O(n² log n) time, O(1) space, clever search.
• Cons: Slower due to log factor.

Two-pointer wins for speed.

Additional Examples and Edge Cases

• Input: [0, 0, 0]
• Output: 0 (0+0=0, not >0).
• Input: [1, 2, 2]
• Output: 1 ([1,2,2]).

Both handle these well.

Complexity Breakdown

• Two-Pointer: Time O(n²), Space O(1).
• Binary Search: Time O(n² log n), Space O(1).

Two-pointer is optimal.

Key Takeaways

• Two-Pointer: Sliding efficiency—smart!
• Binary Search: Search precision—clear!
• Math: Triangles are fun.
• Python Tip: Sorting rocks—see Python Basics.

Final Thoughts: Count Those Triangles

LeetCode 611: Valid Triangle Number in Python is a neat triplet-counting challenge. The two-pointer technique offers speed and simplicity, while binary search adds a clever twist. Want more? Try LeetCode 15: 3Sum or LeetCode 16: 3Sum Closest. Ready to build? Head to Solve LeetCode 611 on LeetCode and count those triangles today!