Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 12

Description

Minimum Number of Operations to Make All Array Elements Equal to 1

You are given a 0-indexed array nums consisiting of positive integers. You can do the following operation on the array any number of times:

• Select an index i such that 0 <= i < n - 1 and replace either of nums[i] or nums[i+1] with their gcd value.

Return the minimum number of operations to make all elements of nums equal to 1. If it is impossible, return -1.

The gcd of two integers is the greatest common divisor of the two integers.

Example 1:

Input: nums = [2,6,3,4]
Output: 4
Explanation: We can do the following operations:
- Choose index i = 2 and replace nums[2] with gcd(3,4) = 1. Now we have nums = [2,6,1,4].
- Choose index i = 1 and replace nums[1] with gcd(6,1) = 1. Now we have nums = [2,1,1,4].
- Choose index i = 0 and replace nums[0] with gcd(2,1) = 1. Now we have nums = [1,1,1,4].
- Choose index i = 2 and replace nums[3] with gcd(1,4) = 1. Now we have nums = [1,1,1,1].

Example 2:

Input: nums = [2,10,6,14]
Output: -1
Explanation: It can be shown that it is impossible to make all the elements equal to 1.

Constraints:

• 2 <= nums.length <= 50
• 1 <= nums[i] <= 106

Solution

int gcd(int a, int b) {
  return b ? gcd(b, a % b) : a;
}
class Solution {
public:
  int minOperations(vector<int>& nums) {
    int n = nums.size();
    int answer = INT_MAX;
    int ones = 0;
    for(int i = 0; i < n; ++i) {
      if(nums[i] == 1) ones += 1;
    }
    if(ones) return n - ones;

    for(int i = 0; i < n; ++i) {
      int g = nums[i];
      for(int j = i + 1; j < n; ++j) {
        g = gcd(g, nums[j]);
        if(g != 1) continue;
        answer = min(answer, j - i + n - 1);
        break;
      }
    }

    return answer == INT_MAX ? -1 : answer;
  }
};

// Accepted
// 1068/1068 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 93.06 % of cpp submissions (31.5 MB)