2344. Minimum Deletions to Make Array Divisible

Difficulty:
Hard
Related Topics:
Similar Questions:

Problem

You are given two positive integer arrays nums and numsDivide. You can delete any number of elements from nums.

Return **the *minimum* number of deletions such that the smallest element in nums divides all the elements of **numsDivide. If this is not possible, return -1.

Note that an integer x divides y if y % x == 0.

  Example 1:

Input: nums = [2,3,2,4,3], numsDivide = [9,6,9,3,15]
Output: 2
Explanation: 
The smallest element in [2,3,2,4,3] is 2, which does not divide all the elements of numsDivide.
We use 2 deletions to delete the elements in nums that are equal to 2 which makes nums = [3,4,3].
The smallest element in [3,4,3] is 3, which divides all the elements of numsDivide.
It can be shown that 2 is the minimum number of deletions needed.

Example 2:

Input: nums = [4,3,6], numsDivide = [8,2,6,10]
Output: -1
Explanation: 
We want the smallest element in nums to divide all the elements of numsDivide.
There is no way to delete elements from nums to allow this.

  Constraints:

Solution

class Solution {
    public int minOperations(int[] nums, int[] numsDivide) {
        int g = numsDivide[0];
        for (int i : numsDivide) {
            g = gcd(g, i);
        }
        int minOp = 0;
        int smallest = Integer.MAX_VALUE;
        for (int num : nums) {
            if (g % num == 0) {
                smallest = Math.min(smallest, num);
            }
        }
        for (int num : nums) {
            if (num < smallest) {
                ++minOp;
            }
        }
        return smallest == Integer.MAX_VALUE ? -1 : minOp;
    }

    private int gcd(int a, int b) {
        while (b > 0) {
            int tmp = a;
            a = b;
            b = tmp % b;
        }
        return a;
    }
}

Explain:

nope.

Complexity: