Determine if possible to get Array by breaking rod segments into two halves in C++

In computer-based problem solving, there are some problems that we can only solve by working with basic things like sticks or groups of items that are similar. There is one problem like this: given a particular series of these basic things (an array), can we split a stick (that we start with) so that we have exactly those pieces. We can break the stick as many times as we need to, but when we do break it, it has to be into two equal parts.

This task is interesting because its setup is similar to some situations we come across in real life as well as in other computer programs where we have to follow certain steps repeatedly until something ends.

When dealing with this task, we first have to understand what it means to break a stick according to the rules, unlike what we would normally do. If a stick is divided into two equal parts at each break, what are the different lengths it can take?

• If we say the original stick is L, L/2, L/4, L/8,…and so on until the length becomes very small are the only possible lengths that can be gotten from breaking the stick in this way.
• For example, it is possible to divide a rod of length eight into two equal parts because 8/2=4. From what we know already, the lengths 8 and 4 are allowed. At a later step, if we divide one of the sticks of length four into two equal parts, the lengths we would get are 4, 2, and 8/4=2.
• The last length mentioned, 2, is the smallest possible length obtainable when working with sticks of any positive integer length.

Understanding the Problem Statement:

Formally, the problem is stated as follows:

We are presented with an integer L representing the length of an initial rod. Also given is an array array[] containing the various lengths of segments for which we desire to make cuts. Our job is to specify whether we can generate all these segment lengths by consecutively halving the rod. The only important condition is that in each operation, the rod must be halved into equal pieces. It entails that we cannot make arbitrary cuts at any position. In order to understand this problem, consider an example, L=16 and arr={8,4,2,2}. Here, the rod starts out as one segment with a length of 16. The first valid break is clearly splitting into two segments of lengths of eight each. After that, we can take one of those 8’s and split it into two 4’s, and then with one 4, we can obtain two 2’s, thus forming the entire array. This case returns YES because we can obtain all pieces through valid halving operations. In this case, not everything is that simple. Take L=16 and arr={9,4,2,1}. Segment 9 cannot be obtained because cutting 16 gives segments 8, 4, 2, and 1 only that keep splitting themselves into halves. It means 9 never would appear from any such kind of operation. Hence, the answer becomes NO. In addition, it gives a very important observation: all numbers would have to belong to a power of 2 because only powers of 2 can ever come up using a rod of length L.

In order to set the terms of feasibility conditions in formal terms, we must check for the following:

The rod length, L, must be enough to cover all elements in arr[]. If the sum over arr[] exceeds L, it is impossible to obtain all segments. Every number in arr[] must be a power of 2. Any number that is not a power of 2 may not be formed through valid halving operations. The breakdown process does not require a different order of natural halving. It means we can decide which half to break down but we cannot force divisions that do not follow the natural law of halving like a power of 2.

Approach:

In order to check whether we can form the desired array by breaking the rod in half, we will resort to a naive but efficient approach. We will first check whether all segments in arr[] are powers of 2. Since the rod can be broken down only into halves, any segment that is not a power of 2 cannot possibly be formed, so an instant NO is returned. A number can efficiently be checked whether it is a power of 2 by using the bitwise operation x & (x - 1) == 0.

Afterwards, just need to simulate the breaking process. For that, we can use a priority queue with a max heap and always break the largest one first. We start with a rod of length L and keep breaking the largest one available. Whenever we divide a section, we enqueue the two parts again. If we can reconstruct the original list from the elements present in the segments any number of times possible, output YES. If not, output NO.

Alternatively, another person can use a multiset to keep track of segments available at any point in time. They can model this process of dividing into segments (not necessarily distinct) starting with {L} and check if it is possible to choose a segment of maximum length suData Structuresch that after breaking it from some point, the remaining pieces can be used to represent all elements from the array.

If we sort arr[] by value in non-increasing order, it allows us to simplify the process of checking whether breaking the rod at a particular point will give us all of the necessary pieces. We just need to see if it is possible to select all of the other required pieces from the ones that are bigger.

Code Implementation:

Output:

YES, the given array can be obtained by breaking the rod.
NO, it is not possible to obtain the given array by breaking the rod.   

Code Explanation:

• Check if every number in arr[] is a power of 2. If not, return NO.
• Sort arr[] in descending order to process bigger requirements first.
• Use a max heap (priority queue) to always process the largest available segment.
• Keep breaking the largest segment unless we fulfil all requirements, or it is no longer possible to break segments without violating one requirement.
• We use a hash map to keep track of how many more times each requirement needs to be fulfilled. It helps us combine small segments greedily without going over the required size for any requirement.
• If we have met all requirements at the i-th index, we push that index into the answer vector and update our hash map for that requirement.
• In the end, if we have fulfilled all requirements, print YES; otherwise, NO.

Conclusion:

In conclusion, the issue of whether it is possible to get a given array by breaking a rod into two equal halves is interesting because its solution relies on recursion and priority-based processing. We can use powers of 2 to solve some cases quickly. In order to solve this efficiently, we can use a priority queue (max heap) to always pick the biggest segment to break. We also use a hash map to keep track of how many times each segment length should occur so that we know when we have all the segments of a certain length.

If we sort the array in descending order, it will help us give out segments with a minimum number of breaks, hence reducing logic. The time complexity O(n log n + n log L) for this approach, where L is the maximum segment size, ensures that this solution is practical even for large constraints. There are real-world applications for this kind of problem, including cutting things into certain lengths using material most efficiently, representing even- and odd-sized things with binary trees, and dividing computer memory into sections that are either accessible or not accessible.

We have now arrived at a solution where every segment is accounted for this is the best possible answer. If we have made all the segments that were asked for, the answer is YES. If something went wrong along the way (for example, we made too many copies of one segment), the answer is NO.