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

Description

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?

Solutions

  • class Solution {
    public int minOperations(int[] nums) {
        int n = nums.length;
        int cnt = 0;
        for (int x : nums) {
            if (x == 1) {
                ++cnt;
            }
        }
        if (cnt > 0) {
            return n - cnt;
        }
        int mi = n + 1;
        for (int i = 0; i < n; ++i) {
            int g = 0;
            for (int j = i; j < n; ++j) {
                g = gcd(g, nums[j]);
                if (g == 1) {
                    mi = Math.min(mi, j - i + 1);
                }
            }
        }
        return mi > n ? -1 : n - 1 + mi - 1;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

  • class Solution {
public:
    int minOperations(vector<int>& nums) {
        int n = nums.size();
        int cnt = 0;
        for (int x : nums) {
            if (x == 1) {
                ++cnt;
            }
        }
        if (cnt) {
            return n - cnt;
        }
        int mi = n + 1;
        for (int i = 0; i < n; ++i) {
            int g = 0;
            for (int j = i; j < n; ++j) {
                g = gcd(g, nums[j]);
                if (g == 1) {
                    mi = min(mi, j - i + 1);
                }
            }
        }
        return mi > n ? -1 : n - 1 + mi - 1;
    }
};

  • class Solution:
    def minOperations(self, nums: List[int]) -> int:
        n = len(nums)
        cnt = nums.count(1)
        if cnt:
            return n - cnt
        mi = n + 1
        for i in range(n):
            g = 0
            for j in range(i, n):
                g = gcd(g, nums[j])
                if g == 1:
                    mi = min(mi, j - i + 1)
        return -1 if mi > n else n - 1 + mi - 1


  • func minOperations(nums []int) int {
	n := len(nums)
	cnt := 0
	for _, x := range nums {
		if x == 1 {
			cnt++
		}
	}
	if cnt > 0 {
		return n - cnt
	}
	mi := n + 1
	for i := 0; i < n; i++ {
		g := 0
		for j := i; j < n; j++ {
			g = gcd(g, nums[j])
			if g == 1 {
				mi = min(mi, j-i+1)
			}
		}
	}
	if mi > n {
		return -1
	}
	return n - 1 + mi - 1
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

  • function minOperations(nums: number[]): number {
    const n = nums.length;
    let cnt = 0;
    for (const x of nums) {
        if (x === 1) {
            ++cnt;
        }
    }
    if (cnt > 0) {
        return n - cnt;
    }
    let mi = n + 1;
    for (let i = 0; i < n; ++i) {
        let g = 0;
        for (let j = i; j < n; ++j) {
            g = gcd(g, nums[j]);
            if (g === 1) {
                mi = Math.min(mi, j - i + 1);
            }
        }
    }
    return mi > n ? -1 : n - 1 + mi - 1;
}

function gcd(a: number, b: number): number {
    return b === 0 ? a : gcd(b, a % b);
}

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