Difficulty : Medium
Prerequisites : Segment Tree
Problem : Given n sides,there are two types of queries: Update one of the sides or Find the maximum possible perimeter of a non-degenerate triangle whose sides are in a[l]…a[r]
Approach for subtask-1 and subtask-2 (Brute Force) :
For query of type 1, simply change the value of the element at the required position
For query of type 2, copy the elements a[l]…a[r] (both inclusive) into a temporary array, say S. Sort this array in descending order. Starting from the first element, find the first three consecutive elements which can form a valid triangle i.e. if the three sides are a,b and c, then,
(a+b > c) and (b+c > a) and (a+c > b)
If such a triangle exists, print the sum of the three sides. Else print 0 if no triangle can be formed.
Time Complexity : O(N^2\log_{2}N)
Analysis for subtask-3 :
Here, we will use a segment tree which will appear exactly same as a Merge-Sort tree. We will keep only those elements which form the largest valid triangle and those greater than these sides. First, lets analize the number of elements in each node of the tree.
For a list of sides to not form a valid triangle, the side length must be such that when the sides are sorted, every element is atleast the sum of previous two elements. In the worst case, the values may be of the form:
1, 2, 3, 5, 8, 13, 21, and so on…
This is nothing but the Fibonacci Series. In this series, the 44th number is 1134903170 which is greater than 10^9. This is an important conclusion as this means that if a node contains NO valid triangle, than it will AT MAX hold Just 43 values
Thus, any node in the segment tree will hold at max 45 values where the sides are of the form:
1, 1, 1, 2, 3, 5, 8, 13... and so on till 701408733
Approach :
Building the Segment Tree: Each node of the segment tree will store a vector which is built as follows. Starting from the root, travel through to the leaf. Store the value of the corresponding array element in the leaf. Now for each parent, first create a temporary array and store the elements of its children in sorted merge way.
For example, say there are two vectors, v1 and v2, for the two children. Keep two pointers p1 and p2, one for each vector, initialized to 0. Add the smaller element out of v1[p1] and v2[p2] to our temporary array and increment the corresponding pointer. If one of the list exhausts, add the remaining elements of the other list to the temporary array. It’s important to know that the total number of elements in the temporary array will not be greater than 90.
Now, starting from the third element from the last, travel linearly to find the first set of three consecutive elements which form a valid triangle. Say elements T[i],T[i+1] and T[i+2] form a valid triangle. Store the elements from T[i] till the end of array in the node. It’s safe to say that the side lengths smaller than T[i] will never contribute to the triangle of the largest perimeter.
If no valid triangle exists, store all the elements from the temporary array.
For query of type 1, starting from the root, traverse through to the leaf, change the value of the lone element in the leaf to the new value. Now, when travelling backwards, for each parent, perform sorted merge of the vectors of the children nodes in exactly the same manner as that in building the tree.
For query of type 2, there can be three cases for each node:
The span of current node is completely outside the query range: Return an empty vector.
The span of current node is completely inside the query range: Return the vector present in the node.
The span of current node intersects partially with the query range: Make a call to each of its children. Merge the two vectors it has got from its children in the same manner as dpne during building, and return the Merged-Sorted Vector.
Now, in the main method, linearly travel from the third element from the last in the vector received to find three consecutive elements which can form a valid triangle. The sum of these three elements is our required answer. If no such trio exists, then 0 is our answer.
Time Complexity: For either type of query, in the worst case, we may have to travel to a leaf and return.
So, complexity is O(90\log_{2}N) per query.
Overall Complexity: O(90*Q\log_{2}N) \approx O(Q\log_{2}N)
This is the link to my code.
Hope this editorial helps. I’ll be glad to clear doubts and hear other solutions.
