Description#
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:
1 <= nums.length, numsDivide.length <= 1051 <= nums[i], numsDivide[i] <= 109
Solutions#
Solution 1#
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| class Solution:
def minOperations(self, nums: List[int], numsDivide: List[int]) -> int:
x = numsDivide[0]
for v in numsDivide[1:]:
x = gcd(x, v)
nums.sort()
for i, v in enumerate(nums):
if x % v == 0:
return i
return -1
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| class Solution {
public int minOperations(int[] nums, int[] numsDivide) {
int x = 0;
for (int v : numsDivide) {
x = gcd(x, v);
}
Arrays.sort(nums);
for (int i = 0; i < nums.length; ++i) {
if (x % nums[i] == 0) {
return i;
}
}
return -1;
}
private int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
}
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| class Solution {
public:
int minOperations(vector<int>& nums, vector<int>& numsDivide) {
int x = 0;
for (int& v : numsDivide) {
x = gcd(x, v);
}
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size(); ++i) {
if (x % nums[i] == 0) {
return i;
}
}
return -1;
}
};
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| func minOperations(nums []int, numsDivide []int) int {
x := 0
for _, v := range numsDivide {
x = gcd(x, v)
}
sort.Ints(nums)
for i, v := range nums {
if x%v == 0 {
return i
}
}
return -1
}
func gcd(a, b int) int {
if b == 0 {
return a
}
return gcd(b, a%b)
}
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Solution 2#
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| class Solution:
def minOperations(self, nums: List[int], numsDivide: List[int]) -> int:
x = gcd(*numsDivide)
nums.sort()
return next((i for i, v in enumerate(nums) if x % v == 0), -1)
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| class Solution {
public int minOperations(int[] nums, int[] numsDivide) {
int x = 0;
for (int v : numsDivide) {
x = gcd(x, v);
}
int y = 1 << 30;
for (int v : nums) {
if (x % v == 0) {
y = Math.min(y, v);
}
}
if (y == 1 << 30) {
return -1;
}
int ans = 0;
for (int v : nums) {
if (v < y) {
++ans;
}
}
return ans;
}
private int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
}
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| class Solution {
public:
int minOperations(vector<int>& nums, vector<int>& numsDivide) {
int x = 0;
for (int& v : numsDivide) {
x = gcd(x, v);
}
int y = 1 << 30;
for (int& v : nums) {
if (x % v == 0) {
y = min(y, v);
}
}
if (y == 1 << 30) {
return -1;
}
int ans = 0;
for (int& v : nums) {
ans += v < y;
}
return ans;
}
};
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| func minOperations(nums []int, numsDivide []int) int {
x := 0
for _, v := range numsDivide {
x = gcd(x, v)
}
y := 1 << 30
for _, v := range nums {
if x%v == 0 {
y = min(y, v)
}
}
if y == 1<<30 {
return -1
}
ans := 0
for _, v := range nums {
if v < y {
ans++
}
}
return ans
}
func gcd(a, b int) int {
if b == 0 {
return a
}
return gcd(b, a%b)
}
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Solution 3#
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| class Solution:
def minOperations(self, nums: List[int], numsDivide: List[int]) -> int:
x = gcd(*numsDivide)
y = min((v for v in nums if x % v == 0), default=0)
return sum(v < y for v in nums) if y else -1
|
