Otter.Ai: Fewest Adjacent-GCD Operations to Turn Every Array Element into 1

Quick Overview

Given an array of positive integers, find the fewest operations that make every element 1, where each operation replaces one of two neighboring values with their greatest common divisor, or return -1 if it cannot be done. Tests GCD properties, lower-bound reasoning and careful case analysis on small arrays.

Fewest Adjacent-GCD Operations to Turn Every Array Element into 1

Company: Otter.Ai

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Onsite

You are given an array `nums` of positive integers. In one operation, you choose an index `i` with `0 <= i < len(nums) - 1` and replace either `nums[i]` or `nums[i + 1]`, but not both, with the greatest common divisor (GCD) of those two values. Return the minimum number of operations needed to make every element of `nums` equal to `1`. If that is impossible, return `-1`. ### Function Signature ```python def min_operations_to_all_ones(nums: list[int]) -> int: ``` ### Rules - `gcd(a, b)` is the largest positive integer that divides both `a` and `b`; for example, `gcd(12, 18) = 6` and `gcd(4, 9) = 1`. - Operations are applied one at a time, and each one sees the values left by the previous operations. - Return `0` if every element is already `1`. ### Constraints - `2 <= len(nums) <= 50` - `1 <= nums[i] <= 10^6` ### Examples **Example 1** ```text Input: nums = [6, 10, 15] Output: 4 ``` One optimal sequence: replace `10` with `gcd(10, 15) = 5`, giving `[6, 5, 15]`; replace `6` with `gcd(6, 5) = 1`, giving `[1, 5, 15]`; replace `5` with `gcd(1, 5) = 1`, giving `[1, 1, 15]`; replace `15` with `gcd(1, 15) = 1`, giving `[1, 1, 1]`. No sequence of three operations works. **Example 2** ```text Input: nums = [3, 1, 9, 1] Output: 2 ``` Replace `3` with `gcd(3, 1) = 1`, then replace `9` with `gcd(9, 1) = 1`. **Example 3** ```text Input: nums = [4, 6, 8] Output: -1 ``` Every value is even, so every GCD that can be formed is even and no element can ever become `1`.

Overview: Given an array of positive integers, find the fewest operations that make every element 1, where each operation replaces one of two neighboring values with their greatest common divisor, or return -1 if it cannot be done. Tests GCD properties, lower-bound reasoning and careful case analysis on small arrays.

You are given an array `nums` of positive integers. In one operation you choose an index `i` with `0 <= i < len(nums) - 1` and replace either `nums[i]` or `nums[i + 1]`, but not both, with the greatest common divisor (GCD) of those two values. Return the minimum number of operations needed to make every element of `nums` equal to `1`. If that is impossible, return `-1`. Return `0` if every element is already `1`. Rules: - `gcd(a, b)` is the largest positive integer that divides both `a` and `b`; for example, `gcd(12, 18) = 6` and `gcd(4, 9) = 1`. - Operations are applied one at a time, and each one sees the values left by the previous operations. Implement `min_operations_to_all_ones(nums)`, which returns an integer. Constraints: - `2 <= len(nums) <= 50` - `1 <= nums[i] <= 10^6` - No input value, GCD or answer exceeds 2^31 - 1, so 32-bit integers suffice in every language. Example 1: Input: nums = [6, 10, 15] Output: 4 Explanation: Replace 10 with gcd(10, 15) = 5, giving [6, 5, 15]; replace 6 with gcd(6, 5) = 1, giving [1, 5, 15]; replace 5 with gcd(1, 5) = 1, giving [1, 1, 15]; replace 15 with gcd(1, 15) = 1, giving [1, 1, 1]. No sequence of three operations works. Example 2: Input: nums = [3, 1, 9, 1] Output: 2 Explanation: Replace 3 with gcd(3, 1) = 1, then replace 9 with gcd(9, 1) = 1.

Constraints

2 <= len(nums) <= 50
1 <= nums[i] <= 10^6

Examples

Input: ([6, 10, 15],)

Expected Output: 4

Explanation: Source example 1: no adjacent pair is coprime; the shortest GCD-1 window is all three elements (2 operations to create a 1), then 2 more.

Input: ([3, 1, 9, 1],)

Expected Output: 2

Explanation: Source example 2: two 1s already exist, so each of the two non-1 elements needs one operation.

Hints

Once some element equals 1, how many operations does each remaining non-1 element need, and can any of them be skipped?
Before any 1 exists, every value in the array is the GCD of some contiguous stretch of the original array. Which stretches could ever produce a 1?
If the GCD of the whole array is greater than 1, can a 1 ever appear?

Approach

Let n = len(nums) and c = the number of elements already equal to 1. Each operation changes exactly one element, and every element that is not 1 must change at least once, so at least n - c operations are needed. If c > 0, that many suffice: repeatedly pair a 1 with a non-1 neighbor and overwrite the neighbor with gcd(1, x) = 1, spreading the 1s outward. So the answer is n - c (which is 0 when every element is already 1).

If c = 0, a 1 must first be created. Invariant: at any moment, each element equals the GCD of some contiguous segment of the original array that contains its position, and one operation can grow that segment by at most one neighboring element. So the first 1 needs a contiguous window of the original array whose GCD is 1, and building it from a window of length L takes at least L - 1 operations; folding GCDs left to right across the shortest such window achieves exactly L - 1. After that, the other n - 1 elements each need one more operation, and none of them can have been fixed earlier, because no 1 existed before. The answer is (L - 1) + (n - 1). If no window has GCD 1 (equivalently, the GCD of the whole array exceeds 1), every reachable value keeps a common factor greater than 1 and the answer is -1.

Implementation: for each start i, keep a running GCD while extending j to the right, stop at the first j where it reaches 1, and track the minimum j - i. Edge cases: an all-ones array returns 0; a single non-coprime pair returns -1; an existing 1 makes the window search unnecessary. All values stay at or below 10^6, so 32-bit integers suffice.

Time complexity:

O(n^2 log M), where M is the largest value in nums

Space complexity:

O(1)