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Leetcode #2197: Replace Non-Coprime Numbers in Array

In this guide, we solve Leetcode #2197 Replace Non-Coprime Numbers in Array in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given an array of integers nums. Perform the following steps: Find any two adjacent numbers in nums that are non-coprime.

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Stack, Array, Math, Number Theory

Intuition

The problem has a natural nested or last-in-first-out structure.

A stack lets us resolve matches in the correct order as we scan.

Approach

Push items as they appear and pop when you can finalize a decision.

The stack captures the unresolved part of the input.

Steps:

  • Push elements as you scan.
  • Pop when a rule or match is satisfied.
  • Use the stack to compute results.

Example

Input: nums = [6,4,3,2,7,6,2] Output: [12,7,6] Explanation: - (6, 4) are non-coprime with LCM(6, 4) = 12. Now, nums = [12,3,2,7,6,2]. - (12, 3) are non-coprime with LCM(12, 3) = 12. Now, nums = [12,2,7,6,2]. - (12, 2) are non-coprime with LCM(12, 2) = 12. Now, nums = [12,7,6,2]. - (6, 2) are non-coprime with LCM(6, 2) = 6. Now, nums = [12,7,6]. There are no more adjacent non-coprime numbers in nums. Thus, the final modified array is [12,7,6]. Note that there are other ways to obtain the same resultant array.

Python Solution

class Solution: def replaceNonCoprimes(self, nums: List[int]) -> List[int]: stk = [] for x in nums: stk.append(x) while len(stk) > 1: x, y = stk[-2:] g = gcd(x, y) if g == 1: break stk.pop() stk[-1] = x * y // g return stk

Complexity

The time complexity is O(n×log⁡M)O(n \times \log M)O(n×logM), and the space complexity is O(n)O(n)O(n). The space complexity is O(n)O(n)O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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