You are given a 0-indexed integer array nums, and you are allowed to traverse between its indices. You can traverse between index i and index j, i != j, if and only if gcd(nums[i], nums[j]) > 1, where gcd is the greatest common divisor.
Your task is to determine if for every pair of indices i and j in nums, where i < j, there exists a sequence of traversals that can take us from i to j.
Return true if it is possible to traverse between all such pairs of indices, or false otherwise.
\n
Example 1:
\n\n\nInput: nums = [2,3,6]\nOutput: true\nExplanation: In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).\nTo go from index 0 to index 1, we can use the sequence of traversals 0 -> 2 -> 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 > 1.\nTo go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 > 1.\n\n\n
Example 2:
\n\n\nInput: nums = [3,9,5]\nOutput: false\nExplanation: No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.\n\n\n
Example 3:
\n\n\nInput: nums = [4,3,12,8]\nOutput: true\nExplanation: There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 105Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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