Asked in: Walmart
Image of the Question
All Testcases Passed ✔
Solution
from functools import lru_cache
import bisect
n,q = map(int, input().split())
arr = list(map(int, input().split()))
// ... rest of solution available after purchaseExplanation
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To approach this problem, start by carefully understanding the movement rules for Jatin and Mano in the Skyline Street scenario. Both can only move to the right, and they can only move onto buildings taller than the one they currently stand on. They continue moving until no taller building exists ahead. Our goal is to find the earliest (leftmost) building where both friends could end up, starting from their respective initial buildings, following these rules.
Key insights and steps to think through:
1. **Understanding Movement Constraints:**
Both Jatin and Mano start from specific buildings. From any building, they can only move right to a building that is strictly taller than their current one. This means their path is strictly increasing in height. Importantly, at each step, they must find the next taller building to the right.
2. **Implications of Strictly Increasing Movement:**
Because they can only jump to a taller building, their paths form a chain of increasing building heights to the right. This is similar to "jumping" to the next greater element to the right. Hence, from any building, their movement can be represented as a sequence of buildings where each next building is taller than the current one.
3. **Preprocessing Next Taller Buildings:**
To efficiently answer queries for multiple starting positions, precompute for every building the "next taller building to the right." This means for each building at position i, find the closest building j > i where height[j] > height[i]. If no such building exists, denote it as -1.
This preprocessing step can be achieved efficiently using a stack and scanning the buildings from right to left (a common approach for Next Greater Element problems).
4. **Constructing Jump Chains:**
After computing the next taller building for each position, each building can be thought of as a node with a pointer to the next building in the sequence. Thus, starting from any building, the path Jatin or Mano will take is simply following these "next taller building" pointers until reaching the end (no taller building to the right).
5. **Query Handling:**
For each query, given starting positions J (Jatin) and M (Mano), simulate or find the sequences of buildings they will visit by following the next taller pointers starting at J and M respectively.
6. **Efficiently Finding Meeting Points:**
The naive way would be to fully list out both sequences and check for common buildings, then pick the leftmost common building. However, this can be inefficient if sequences are long and there are many queries.
Instead, think about using a data structure or approach to quickly find intersection points:
- Since buildings only move forward, the sequences are strictly increasing in indices.
- Traverse both sequences simultaneously like merging two sorted lists, since both are strictly increasing sequences of building indices.
- Move forward along the sequences, advancing the pointer with the smaller current building index.
- When both pointers point to the same building index, that is a meeting point.
- The first such match encountered is the leftmost meeting point.
7. **Handling No Meeting Case:**
If the sequences do not intersect (no common building index), return -1.
8. **Optimizations:**
For multiple queries, repeating the traversal each time may be inefficient.
Consider:
- Precomputing and storing the full jump chain for each building may use too much memory.
- Alternatively, use caching or memoization to speed up repeated traversals.
- Use efficient pointer advancement since the sequences are sorted lists of indices.
- Since buildings point to only one next taller building, sequences won't loop, ensuring safe traversal.
9. **Edge Cases:**
- Starting points might be the same building; then meeting point is trivial (that building).
- If starting building has no taller building to the right, the path contains only that building.
- Buildings could be in descending order, leading to no movement possible beyond starting buildings.
10. **Summarizing the Approach:**
- Preprocess next taller building indices for all buildings.
- For each query:
- Retrieve the sequences of buildings that Jatin and Mano would visit by following the next taller building pointers.
- Use a two-pointer approach on these sequences to find the first common building.
- Return that building's index or -1 if none found.
This approach leverages classic next greater element computation, efficient traversal of sequences, and careful merging to find the meeting point, enabling the solution to handle large inputs and many queries efficiently.
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