What is the time complexity of Valid Triangle Number?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Valid Triangle Number?

This problem uses Array, Two Pointers, Binary Search, Greedy, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Valid Triangle Number?

Insight 1: Sorting the input array allows us to efficiently check the triangle inequality condition. If we have a c. Insight 2: Using a two-pointer approach after sorting allows us to count valid combinations in O(n^2) time.

Which companies ask LeetCode 611. Valid Triangle Number in interviews?

This problem is frequently asked at Expedia, LinkedIn.

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Valid Triangle Number - LeetCode 611 Solution

Valid Triangle Number - Complete Solution Guide

Valid Triangle Number is LeetCode problem 611, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem requires us to find the number of triplets in a given array of integers that can form valid triangles. A triangle is valid if the sum of any two sides is greater than the third side. In other words, for sides a, b, and c, the conditions a + b > c, a + c > b, and b + c > a must all be true. The input is an array of integers, and the output is the number of valid triangle triplets that can be formed from the array elements.

Solution Approach

The solution first sorts the input array. Then, it iterates through the array from the largest element to the third element. For each element (nums[k]), it uses two pointers (left and right) to find pairs of elements (nums[left] and nums[right]) such that nums[left] + nums[right] > nums[k]. If this condition is met, then all pairs between 'left' and 'right' will also satisfy this condition, so we increment the count by (right - left) and decrement 'right'. Otherwise, we increment 'left'. This two-pointer approach efficiently finds all valid combinations for each potential largest side.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Initialize a counter `count` to 0.
Step 3: Iterate from the end of the array (largest element) to the third element (index 2). Let the index be `k`.
Step 4: Initialize two pointers, `left` to 0 and `right` to `k - 1`.
Step 5: While `left` is less than `right`:
Step 6: If `nums[left] + nums[right] > nums[k]`:
Step 7: Increment `count` by `right - left` because all elements between `left` and `right` will satisfy the triangle inequality.
Step 8: Decrement `right` to look for smaller values that satisfy the inequality.
Step 9: Else (if `nums[left] + nums[right] <= nums[k]`):
Step 10: Increment `left` to find a larger value.
Step 11: Return `count`.

Key Insights

Insight 1: Sorting the input array allows us to efficiently check the triangle inequality condition. If we have a <= b <= c, we only need to check if a + b > c.
Insight 2: Using a two-pointer approach after sorting allows us to count valid combinations in O(n^2) time.
Insight 3: Iterating from the largest potential side (k) down to smaller sides improves efficiency because if nums[left] + nums[right] > nums[k], then all combinations nums[i] + nums[right] where left <= i < right will also be greater than nums[k].

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Binary Search, Greedy, Sorting.

Companies

Asked at: Expedia, LinkedIn.

Valid Triangle Number | LeetCode 611

Medium

Array Two Pointers Binary Search Greedy Sorting

Time: O(n^2)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int triangleNumber(int[] nums) {
        int n = nums.length;
        if (n < 3) {
            return 0;
        }

        Arrays.sort(nums);
        
        int count = 0;
        for (int k = n - 1; k > 1; k--) {
            int left = 0;
            int right = k - 1;
            while (left < right) {
                if (nums[left] + nums[right] > nums[k]) {
                    count += right - left;
                    right--;
                } else {
                    left++;
                }
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 611: Valid Triangle Number - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Greedy Sorting

Companies

ExpediaLinkedIn

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:611

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Sort the input array `nums` in ascending order.
2
Step 2: Initialize a counter `count` to 0.
3
Step 3: Iterate from the end of the array (largest element) to the third element (index 2). Let the index be `k`.
4
Step 4: Initialize two pointers, `left` to 0 and `right` to `k - 1`.
5
Step 5: While `left` is less than `right`:
6
Step 6: If `nums[left] + nums[right] > nums[k]`:
7
Step 7: Increment `count` by `right - left` because all elements between `left` and `right` will satisfy the triangle inequality.
8
Step 8: Decrement `right` to look for smaller values that satisfy the inequality.
9
Step 9: Else (if `nums[left] + nums[right] <= nums[k]`):
10
Step 10: Increment `left` to find a larger value.
11
Step 11: Return `count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The sorting algorithm typically used has O(1) auxiliary space. The two pointers (`left` and `right`) and the `count` variable take constant space. The nested loops contribute to the time complexity but not to the space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search

Instead of using two pointers, for each pair (i, j), we can use binary search to find the largest index k such that nums[i] + nums[j] > nums[k]. Then, all indices between j+1 and k are valid third sides.

Time: O(n^2 log n)Space: O(1)

Trade-offs: The binary search approach has a slightly worse time complexity (O(n^2 log n)) compared to the two-pointer approach (O(n^2)). The two-pointer approach is generally preferred due to its simplicity and better constant factors.

✨

Key Insights

1Insight 1: Sorting the input array allows us to efficiently check the triangle inequality condition. If we have a <= b <= c, we only need to check if a + b > c.
2Insight 2: Using a two-pointer approach after sorting allows us to count valid combinations in O(n^2) time.
3Insight 3: Iterating from the largest potential side (k) down to smaller sides improves efficiency because if nums[left] + nums[right] > nums[k], then all combinations nums[i] + nums[right] where left <= i < right will also be greater than nums[k].

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `nums` is empty or has fewer than 3 elements. The code handles this by returning 0 because no triangle can be formed.
Edge case 2: `nums` contains zero or negative numbers. Since the problem statement indicates that 0 <= nums[i] <= 1000, we don't have to explicitly handle negative numbers. If negative values were allowed, the triangle inequality would need extra checks, and the sorting-based approach might require modification.
Edge case 3: Duplicate numbers in the input array. The algorithm correctly handles duplicate numbers because it considers all combinations of triplets, even if some numbers are the same.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the input array. Without sorting, it's significantly harder to efficiently check the triangle inequality condition for all possible triplets.

✗Common mistake 2: Incorrectly implementing the triangle inequality check. It's essential to ensure that the sum of any two sides is strictly greater than the third side.

✗Common mistake 3: Off-by-one errors in the two-pointer logic. Carefully managing the `left` and `right` pointers is crucial to correctly count valid triplets.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Determining if a set of measurements can form a physical triangle, such as in surveying or construction.

🏦Real-world scenario 2: In computer graphics, checking if three points form a valid triangle for rendering or collision detection. The triangle inequality ensures that the triangle doesn't collapse into a line.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 11, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Valid Triangle Number"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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