comments | difficulty | edit_url | rating | source | tags | |||||
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true |
Hard |
1640 |
Weekly Contest 302 Q4 |
|
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 <= 105
1 <= nums[i], numsDivide[i] <= 109
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
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);
}
}
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;
}
};
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)
}
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)
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);
}
}
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;
}
};
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)
}
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