Problem Description

The problem presents us with an integer array nums and asks us to find the count of all possible triplets in this array that can form a valid triangle. A set of three numbers forms the sides of a triangle if and only if the sum of any two sides is greater than the third side. This is known as the triangle inequality theorem.

Intuition

To solve this problem, we can start by sorting the array. Sorting helps us to apply the triangle inequality more effectively by having an ordered list to work with. In this way, when we pick two numbers, the smaller two sides of a potential triangle, we already know that any number following them in the array will be greater or equal than these two.

Once we have the sorted array, the intuition behind the solution is as follows:

• The outer loop picks the first side a of the potential triangle.
• The second, inner loop picks the second side b, which is just the next number in the sorted array after the first side.
• Using the triangle inequality theorem, we know that in order to form a triangle, the third side c must be less than the sum of a and b but also greater than their difference. Since the array is sorted and we only need the sum condition ("less than the sum"), we can use binary search to quickly find the largest index k such that nums[k] is less than the sum of nums[i] plus nums[j] (where nums[i] is the first side and nums[j] is the second side).

With this approach, for each pair of a and b, binary search helps us find the count of possible third sides. We then subtract the index of the second side j from k to get the number of valid third sides c, and add this to our answer ans. We repeat this process until we have considered all possible pairs of sides a and b.

To sum up, the solution uses a combination of sorting, iterating through pairs of potential sides, and binary search to efficiently calculate the number of valid triangles that can be formed from the array.

Learn more about Greedy, Two Pointers, Binary Search and Sorting patterns.

Solution Approach

The solution begins with sorting the array nums. Sorting is a critical step because it allows us to take advantage of the triangle inequality in a sorted sequence, which makes it easier to reason about the relationship between the sides. With sorting, we guarantee that if nums[i] and nums[j] (where i < j) are two sides of a potential triangle, then any nums[k] (where k > j) is at least as large as nums[j]. This means that for the triangle inequality to be satisfied (nums[i] + nums[j] > nums[k]), we only need to find the first k where the inequality does not hold and then all indices before k will form valid triangles with nums[i] and nums[j].

Here is how the solution is implemented:

• After sorting, a for loop is used to iterate through the array as the outer loop. This loop starts from the first element and stops at the third-to-last element, with i representing the index of the first side of a potential triangle.

• Inside the outer loop, a nested for loop starts from i+1, iterates over the remaining elements, and stops at the second-to-last element, with j representing the index of the second side of a potential triangle.

• For each pair of sides represented by nums[i] and nums[j], binary search is used to find the right-most element which could be a candidate for the third side of the triangle. The bisect_left function from the Python bisect module is used to find the index k, where nums[k] is the smallest number greater than or equal to the sum of nums[i] and nums[j]. Since nums[k] itself cannot form a triangle with nums[i] and nums[j] (it's not strictly less than the sum), we subtract one to get the index of the largest valid third side.

• Now with k-1 being the largest index of a valid third side and j the index of the second side, all numbers in the range (j, k) will satisfy the triangle inequality. Thus, ans (the count of valid triangles) is incremented by k - j - 1.

• By the end of these loops, ans will contain the total number of triplets that can form valid triangles.

The solution applies the sorting algorithm, iteration through nested loops, and binary search to solve the problem efficiently.

Example Walkthrough

Let's apply our solution approach on a small example array, nums = [4, 2, 3, 4].

First, we sort the array to get nums = [2, 3, 4, 4]. Sorting helps us apply the triangle inequality theorem effectively.

Next, we begin our outer loop with i = 0, where nums[i] = 2.

Inside our outer loop, we start our inner loop with j = i + 1, which makes j = 1, and nums[j] = 3.

For nums[i] = 2 and nums[j] = 3, we perform a binary search to find the largest index k where nums[k] is less than the sum of nums[i] and nums[j], which is 2 + 3 = 5. In this sorted array, nums[k] could be either 4 or 4, both are less than 5. Binary search finds that k = 3 (the last index of 4 in the sorted array). Since the actual element at k cannot be part of the triangle, to get the largest valid k, we subtract one from it, making k = 2.

Now we calculate the number of valid c sides between j and k, which is k - j - 1 = 2 - 1 - 1 = 0. There is no third side that can form a triangle with nums[i] = 2 and nums[j] = 3.

Next, we increment j to 2, making nums[j] = 4, and repeat the binary search. This finds that k = 3 (as 4 + 4 is not less than any element in the array). Subtracting one from k gives us k = 2. The number of valid third sides c is k - j - 1 = 2 - 2 - 1 = -1. No valid triangle can be formed in this case as well.

Then we increment i to 1 and j to i + 1 which is 2, and now we have nums[i] = 3 and nums[j] = 4. Repeating the binary search for the sum 3 + 4 = 7, we find that no such k exists since all existing elements are less than 7. Thus, k is the length of the array, k = 4. And the number of valid c sides is k - j - 1 = 4 - 2 - 1 = 1, so we can form one triangle using the indices (1, 2, 3).

Our next pair would be nums[i] = 3 and nums[j] = 4 (again, but with j = 3), but since there's no element after nums[j] when j = 3, we cannot form a triangle, and the process stops here.

Therefore, through our example, we found only one valid triplet (3, 4, 4) which can form a triangle. Our answer ans would be 1 for this example.

Solution Implementation

Time and Space Complexity

The given Python code sorts an array and then uses a double loop to find every triplet that can form a valid triangle, using binary search to optimize the finding of the third side.

Time Complexity

The time complexity of the code can be broken down as follows:

1. nums.sort() has a time complexity of O(n log n) where n is the number of elements in nums.
2. The outer for-loop runs (n - 2) times.
3. The inner for-loop runs (n - i - 2) times per iteration of the outer loop, summing up to roughly (1/2) * n^2 in the order of growth.
4. The bisect_left call within the inner loop does a binary search, which has a time complexity of O(log n).

Thus, the total time complexity of the nested loop (excluding the sort) would be O((n^2/2) * log n), as each iteration of j could potentially result in a binary search. When adding the sort, the overall time complexity is O(n log n + (n^2/2) * log n) which simplifies to O(n^2 log n) when n is large.

Space Complexity

Space complexity is easier to analyze since the only extra space used is for sorting, which in Python is O(n) because Timsort (the algorithm behind Python's sort) can require this much space. No other significant extra space is used, as the variables i, j, k, and ans use constant space. Therefore, the space complexity of the algorithm is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.