2654

2654. Minimum Number of Operations to Make All Array Elements Equal to 1 (Medium)

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

 

Follow-up:

The O(n) time complexity solution works, but could you find an O(1) constant time complexity solution?

Related Topics:
Array, Math, Number Theory

Solution 1.

• If GCD of all numbers is not 1, impossible.
• If some A[i] is already 1, assume there are k 1s, we need N-k operations to make all 1s.
• If GCD of two adjacent numbers is 1, we do one operation here to get a one, then take N-1 operations to make all 1s.
• If no GCD of two adjacent numbers is 1, we find the minimum subarray whose GCD is 1. If this subarray is of length K, we take K-1 operations to get a one, then take N-1 operations to make all 1s.

// OJ: https://leetcode.com/problems/minimum-number-of-operations-to-make-all-array-elements-equal-to-1
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    int minOperations(vector<int>& A) {
        int N = A.size(), mnLen = N, one = 0, all = A[0];
        for (int n : A) {
            one += n == 1;
            all = gcd(all, n);
        }
        if (one) return N - one;
        if (all > 1) return - 1;
        for (int i = N - 1; i >= 0; --i) {
            int val = A[i], j = i - 1;
            for (; j >= 0 && val > 1; --j) {
                val = gcd(val, A[j]);
            }
            if (val == 1) mnLen = min(mnLen, i - j);
            else break;
        }
        return mnLen + N - 2;
    }
};