Start by sorting the array to streamline triplet selection. Use a binary search or two-pointer approach to find all combinations of indices that satisfy the triangle inequality. This avoids brute-force O(n^3) checks and ensures a reliable, interview-ready solution while covering edge cases with zeros or duplicate elements.
Problem Statement
Given an integer array nums, return the number of triplets that can form a triangle when considered as side lengths. A triplet (i,j,k) is valid if nums[i] + nums[j] > nums[k] for sorted indices i<j<k.
Example: nums = [2,2,3,4]. Valid triangles are (2,2,3), (2,3,4), (2,3,4). The output is 3. Another example: nums = [4,2,3,4], output 4.
Examples
Example 1
Input: nums = [2,2,3,4]
Output: 3
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
Example details omitted.
Constraints
- 1 <= nums.length <= 1000
- 0 <= nums[i] <= 1000
Solution Approach
Sort the Array
Sort nums ascendingly so that triangle conditions can be checked efficiently using indices. Sorting guarantees that nums[i] <= nums[j] <= nums[k], simplifying inequality checks.
Use Two Pointers
Iterate over the array using two pointers for the smaller sides and find the maximum third side k such that nums[i]+nums[j]>nums[k]. Move pointers intelligently to count all valid k without redundant checks.
Binary Search Optimization
For each pair (i,j), perform a binary search to locate the largest index k satisfying the triangle inequality. Add k-j to the count, leveraging the sorted order to reduce time complexity from O(n^3) to O(n^2 log n).
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Sorting takes O(n log n). Two-pointer scanning or binary search for each pair yields O(n^2) or O(n^2 log n) total time. Space is O(1) extra if sorting in place, otherwise O(n).
What Interviewers Usually Probe
- Sorting the array before counting triplets
- Using two pointers or binary search for the third side
- Avoiding O(n^3) brute-force enumeration
Common Pitfalls or Variants
Common pitfalls
- Not sorting the array first, causing incorrect triangle checks
- Ignoring duplicate numbers leading to miscounted triplets
- Miscalculating the range of the third side or using incorrect indices
Follow-up variants
- Count triangles with perimeter below a given threshold
- Find all unique triplets instead of counting them
- Allow negative numbers and determine valid triangle counts accordingly
How GhostInterview Helps
- Automatically identifies the two-pointer or binary search pattern for counting valid triangles
- Provides step-by-step reasoning for why each triplet satisfies the triangle inequality
- Highlights common index mistakes and correct counting strategies for edge cases
Topic Pages
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- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the best approach for Valid Triangle Number?
Sort the array and use either two pointers or binary search to count all triplets satisfying the triangle inequality.
Can zeros or duplicates affect the triangle count?
Yes, zeros cannot form triangles and duplicates must be considered carefully to avoid double counting valid triplets.
What is the time complexity of the optimal solution?
Sorting takes O(n log n) and counting with two pointers or binary search gives O(n^2) or O(n^2 log n).
Why is binary search useful in this problem?
It finds the largest valid third side efficiently, reducing the need for nested loops and avoiding O(n^3) brute-force checks.
How does GhostInterview handle the triangle inequality pattern?
It guides through sorting, pointer selection, and index calculation to ensure all valid triplets are counted correctly with minimal mistakes.
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Open problem page#581 Shortest Unsorted Continuous SubarrayFind the shortest unsorted continuous subarray that, if sorted, would sort the entire array.
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