LeetCode-in-Java

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?

Solution

public class Solution {
    private int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    public int minOperations(int[] nums) {
        int cnt1 = 0;
        int minsubarray = Integer.MAX_VALUE;
        for (int x : nums) {
            if (x == 1) {
                cnt1++;
            }
        }

        if (cnt1 > 0) {
            return nums.length - cnt1;
        }

        for (int i = 0; i < nums.length; i++) {
            int curgcd = nums[i];
            for (int j = i + 1; j < nums.length; j++) {
                curgcd = (gcd(curgcd, nums[j]));
                if (curgcd == 1) {
                    minsubarray = Math.min(minsubarray, j - i);
                    break;
                }
            }
        }
        if (minsubarray != Integer.MAX_VALUE) {
            return minsubarray + nums.length - 1;
        }
        return -1;
    }
}

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