Asked in: GOOGLE
Image of the Question
All Testcases Passed ✔
No images available for this passcode.
Solution
class Node:
def _init_(self):
self.h = {}
self.cnt = 0
// ... rest of solution available after purchaseExplanation
```
To approach this problem, start by clearly understanding what a “comparison” means in the context of ranking two rows in a matrix. You are given a matrix M with N rows and D columns, where each row represents a ranking profile. For every unique pair of rows (i, j) where i < j, you compare their elements starting from the first column, moving left to right, until you find the first differing value. The number of elements you need to examine to determine which row ranks higher is counted as the number of comparisons for that pair.
So, the goal is to compute the total number of comparisons needed to determine the ranking for all such pairs. Naively, this would involve iterating over every pair (i, j), and then scanning each row up to D elements, counting how long it takes to distinguish between them. However, this brute-force approach is O(N^2 * D), which might be inefficient when N is large.
To design an efficient strategy, first observe that the number of comparisons between any two rows is simply the number of leading elements they share in common, plus one (the first differing position). If all elements are the same, then the number of comparisons is D. Therefore, for a given pair of rows, the number of comparisons equals the position of the first differing value, or D if all are the same.
This hints at a structure based on common prefixes. Specifically, you can think of grouping the rows based on their shared prefixes. For example, all rows that start with the same first element form a group. Within each group, you can further group by the first two elements, and so on. By recursively dividing the matrix rows based on their shared prefixes, you can count how many comparisons are needed within each group based on how deep the similarity goes.
This leads to a recursive or divide-and-conquer strategy:
1. Initially, group all rows.
2. For each group, count the number of pairs in that group. If the group has K rows, it contributes K*(K-1)/2 pairs.
3. For all those pairs, since they share a prefix of some length (say, level), they have already made ‘level’ comparisons.
4. Then, refine each group further by partitioning them based on the next element in the row (i.e., the next column). This adds one more comparison depth.
5. Continue refining recursively until you reach the last column (D), at which point you stop, as no further distinguishing is possible.
This recursive refinement ensures that for each pair of rows, you correctly count the number of elements compared, without having to brute-force each pair. Since the groupings are refined hierarchically, the total number of comparisons across all levels correctly accumulates as you go deeper into the prefixes.
To implement this logic efficiently, consider using sorting or dictionary-based grouping at each level. For each group at a particular depth, build a map from the value at that column to a list of rows. Then process each list recursively at the next depth level. At each depth, for every group of size K, you know that all pairs within this group required ‘depth + 1’ comparisons. Multiply the number of such pairs (K choose 2) by (depth + 1) and accumulate the result.
Additionally, since you’re working with indices and values, it is important to carry not just the raw row data, but also track which row you are comparing, though this is mostly relevant if exact ranking results were needed. In this case, we only care about the number of comparisons.
This optimized approach essentially partitions the matrix recursively based on shared prefixes, ensuring each pair is only processed once at the level where their commonality ends. This significantly reduces redundant work compared to comparing every pair individually.
To summarize:
1. Recognize that comparisons depend on the first differing position between rows.
2. Use a recursive grouping strategy to avoid checking each pair explicitly.
3. At each level of recursion, compute the number of pairwise comparisons based on group size and current depth.
4. Aggregate all comparisons across all recursive levels.
This method offers both clarity and efficiency and transforms a brute-force N^2 problem into a more tractable recursive grouping problem that leverages the structure of the matrix.
```