Asked in: JPMORGAN
Image of the Question
All Testcases Passed ✔
Solution
import heapq
def findTotalWeight(cans):
q = [(i, index) for index,i in enumerate(cans)]
heapq.heapify(q)
// ... rest of solution available after purchaseExplanation
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To approach this problem, start by understanding the process and constraints clearly. You have a line of cans with given weights, and you repeatedly select the lightest can remaining, along with its immediate neighbors, and remove them all. You continue until no cans remain, and you need to sum the weights of all the cans chosen as the "lightest" in each step.
Here is how to think about the problem step-by-step:
1. **Identify the Core Action**
Each selection is centered on the lightest can currently available. This can might be anywhere in the list, and if there are multiple cans with the same minimum weight, the one with the smallest index is chosen. Once that can is picked, the can itself plus its adjacent neighbors (one on the left and one on the right, if they exist) are removed from the line.
2. **Maintaining the Current State of Cans**
Since cans are removed after each selection, the arrangement of cans changes dynamically. You need a data structure or a way to efficiently represent the current lineup of cans after removals. A simple array may become tricky since removing elements shifts indices, and you must keep track of neighbors.
3. **Finding the Minimum Weight Can Efficiently**
The operation of selecting the lightest can is repeated after every removal. You could scan the entire current array to find the minimum each time, but that could be inefficient if done naively (O(n) per step leading to O(n²) overall). To optimize, think about data structures that support quick retrieval of minimum elements along with efficient removal. However, since constraints allow up to 2000 cans, a well-implemented straightforward method may suffice.
4. **Removing the Chosen Can and Its Neighbors**
Once you find the minimum can and its position, remove it and its neighbors:
- If the can is at an edge, fewer than two neighbors might exist.
- Remove the can on the left if it exists.
- Remove the can on the right if it exists.
- Update the current state to reflect these removals.
5. **Handling Index Adjustments Post-Removal**
After removing cans, indices shift in a typical array structure, which can cause confusion when accessing neighbors next time. To avoid complexity, consider using a linked structure or a boolean mask indicating which cans are still present. This approach lets you skip removed cans and find neighbors without shifting the array physically.
6. **Iterative Process**
Repeat the steps:
- Identify the lightest remaining can by checking all cans still present.
- Add its weight to the running sum.
- Remove it and its neighbors.
- Continue until no cans remain.
7. **Edge Cases and Validation**
- Make sure to handle the first and last cans properly since they may have only one neighbor.
- If only one can remains at some point, selecting it removes just itself.
- Multiple cans with the same minimum weight should trigger selection of the one with the smallest index.
8. **Example to Clarify**
Take the example `[5, 4, 1, 3, 2]`:
- Minimum is 1 at index 2. Remove cans at indices 1, 2, and 3 → cans left: `[5, 2]`
- Next minimum is 2 at index 4 (adjusted index after removals). Remove cans at indices 0 and 1 of the reduced array → empty.
- Sum the minimums selected: 1 + 2 = 3.
9. **Data Structures and Implementation Thoughts**
- An array with a parallel boolean array to mark removed cans can simplify neighbor lookup.
- Alternatively, a doubly linked list or using arrays to keep track of "next" and "previous" available cans can help quickly find neighbors after removals.
- Since operations require frequent minimum searches and removals, balance simplicity and efficiency based on constraints.
10. **Summary**
The problem is about simulating a process of repeatedly picking the minimum-weight can (with tie-breaking by index), removing it and its immediate neighbors, and summing the weights of chosen cans. The key challenge lies in efficiently maintaining the state of which cans remain and quickly finding neighbors for removal after each iteration.
By carefully updating the data structure representing the cans and repeating this process until all cans are removed, you can compute the desired sum of weights.
This method is clear, manageable, and within reasonable computational limits given the problem constraints.
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