Question

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

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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

Solutions

Solution 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

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;
  }
};

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);
}