Date: 2026-06-06
Time: 15:57
This file solves LeetCode 2176: Count Equal and Divisible Pairs in an Array. It counts pairs of indices (i, j) where i < j, the elements at those indices are equal, and the product i * j is divisible by k. It follows the repo's standard layout: a Solution class with the LeetCode method signature, plus a module-level wrapper function.
Solution.countPairs(nums, k) -> int — The core algorithm. Takes a list of integers and a divisor k, returns the count of valid (i, j) pairs. The pair validity requires two conditions simultaneously: nums[i] == nums[j] and (i * j) % k == 0.
minmonths(nums, k) -> int — A thin wrapper that delegates to Solution().countPairs. The name minmonths is a misnomer — it doesn't relate to the problem semantics. This is likely an artifact of the code generation pipeline that assigns wrapper names, and the test harness imports through it.
(i, j) pairs with i < j. For this problem's constraints (n ≤ 100), this is the expected approach — there's no need for a more clever solution.Solution class with typed method signature matches LeetCode's submission format.min_months function adapts the class-based interface to a plain function call, consistent with how the repo's test infrastructure invokes solutions.Imports: Only typing.List — no external libraries, no project-internal imports. This is typical for LeetCode solutions that are self-contained.
Imported by: The Imported By list in the provided context is misleading — it lists hundreds of test files from *other* problems. This is almost certainly a tooling artifact (e.g., a shared test runner or import resolution that traces from typing import List globally). The real dependent is count-equal-and-divisible-pairs-in-an-array/testsolution.py, which imports minmonths or Solution to run tests.
1. Initialize count = 0 and get array length n.
2. Outer loop: i from 0 to n-1.
3. Inner loop: j from i+1 to n-1 — ensures i < j without double-counting.
4. For each pair, check both conditions with short-circuit and: value equality first (cheap comparison), then divisibility of the index product.
5. Increment count when both hold. Return the total.
The value-equality check before the modulo is a minor optimization — if the values differ, the modulo is never computed.
i < j for every tested pair, satisfying the problem's strict ordering requirement.None. The function assumes valid inputs per LeetCode constraints (1 ≤ n ≤ 100, 1 ≤ k). No bounds checking, no input validation. An empty nums list would return 0 correctly since neither loop body executes.
count-equal-and-divisible-pairs-in-an-array/test_solution.py — See what test cases exercise this solution and whether edge cases (single element, all equal, k=1) are coveredcount-equal-and-divisible-pairs-in-an-array/review.md — Check if the review flagged the min_months naming issue or suggested optimizationsgrouping-by-value-optimization — A HashMap grouping equal values could reduce to O(n + Σ(gᵢ²)) where gᵢ is group size, skipping pairs with different values entirelynumber-of-good-pairs/solution.py:countPairs — Similar pair-counting pattern without the divisibility constraint, useful comparisonbrute-force-is-optimal-for-constraints — With n ≤ 100, the O(n²) brute force checks at most 4,950 pairs, well within time limits; no algorithmic optimization is necessarywrapper-name-mismatch — The min_months wrapper name has no semantic relationship to the "count equal and divisible pairs" problem, suggesting automated generation rather than manual authoringloop-ordering-guarantees-no-duplicates — The j in range(i + 1, n) inner loop ensures each unordered pair is visited exactly once and i < j is always satisfiedvalue-check-short-circuits-modulo — Python's and short-circuits, so the (i * j) % k == 0 modulo is only evaluated when nums[i] == nums[j]