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Solution
#!/bin/python3
import math
import os
// ... rest of solution available after purchaseExplanation
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To solve the problem of counting the number of teams of three coders that form strictly increasing or strictly decreasing sequences in terms of their ratings and satisfy the positional constraint (i < j < k), you need a systematic approach that efficiently explores possible triplets without brute force enumeration, which would be inefficient for large N.
Step 1: Understand the problem and constraints
You have an array of coders’ ratings, all distinct. You want to count triplets (i, j, k) such that their positions satisfy i < j < k, and the ratings satisfy one of the following:
- Strictly increasing: Ri < Rj < Rk
- Strictly decreasing: Ri > Rj > Rk
Each coder can appear in multiple triplets.
Step 2: Naive brute force approach and its inefficiency
A simple solution is to iterate over all possible triplets (i, j, k) where i < j < k and check the rating conditions. This has O(N^3) complexity, which becomes impractical as N grows larger.
Step 3: Optimize by breaking down the problem using dynamic counting
Focus on the middle element j of the triplet. For each position j (1 < j < N), determine:
- How many elements to the left of j have a rating less than Rj (for increasing sequences) and greater than Rj (for decreasing sequences).
- How many elements to the right of j have a rating greater than Rj (for increasing sequences) and less than Rj (for decreasing sequences).
Step 4: Counting increasing triplets via the middle element
For a fixed j, the number of increasing triplets with j in the middle is:
(Number of elements left of j with rating less than Rj) * (Number of elements right of j with rating greater than Rj)
Similarly, for decreasing triplets:
(Number of elements left of j with rating greater than Rj) * (Number of elements right of j with rating less than Rj)
Step 5: How to efficiently find these counts
You need to quickly find counts of elements less or greater than Rj on the left and right sides. Naively counting these for each j would be O(N^2), which may be too slow.
Use data structures or pre-processing to speed this up:
- For counting elements on the left, you can iterate from left to right, and keep track of ratings seen so far.
- For counting elements on the right, you can iterate from right to left similarly.
For example, you might maintain:
- A frequency or presence structure that allows queries like “how many ratings less than Rj have appeared so far” or “how many ratings greater than Rj have appeared so far”.
- Balanced binary search trees, Fenwick trees (Binary Indexed Trees), or Segment Trees can help perform these queries efficiently.
Step 6: Implementation strategy
- Traverse from left to right, for each position j, record how many elements to the left are less than Rj and how many are greater than Rj.
- Traverse from right to left, similarly record how many elements to the right are less than Rj and how many are greater than Rj.
- Combine these counts for each j to compute the number of increasing and decreasing triplets where j is the middle element.
- Sum over all j to get the total count.
Step 7: Why this approach works well
By focusing on the middle element, you reduce the problem from triple nested loops to two passes with data structure support, reducing complexity to roughly O(N log N) with efficient queries.
Step 8: Validating correctness with an example
Using the example:
Ratings = [5, 2, 3, 1, 4]
For j = 2 (rating 2):
- Left smaller: ratings less than 2 on left → none
- Left greater: ratings greater than 2 on left → 5
- Right smaller: ratings less than 2 on right → 1
- Right greater: ratings greater than 2 on right → 3,4
Calculate increasing triplets: left_smaller * right_greater = 0 * 2 = 0
Calculate decreasing triplets: left_greater * right_smaller = 1 * 1 = 1 → (5,2,1)
Similarly for other j, sum results.
Step 9: Edge cases and constraints
- Distinct ratings simplify comparisons.
- Positions always strictly increasing: i < j < k.
- Handle empty sides gracefully (no elements on left or right).
Summary:
1. For each coder position j, count how many coders to the left have ratings less than and greater than Rj.
2. For the same j, count how many coders to the right have ratings less than and greater than Rj.
3. For increasing triplets, multiply left_smaller by right_greater.
4. For decreasing triplets, multiply left_greater by right_smaller.
5. Sum all these products over all j to get the total number of valid triplets.
6. Use data structures for efficient counting of smaller/greater elements on each side.
By following this approach, you efficiently and correctly count all valid teams without explicit triple loops.
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