# 1819. Number of Different Subsequences GCDs

## Description

You are given an array nums that consists of positive integers.

The GCD of a sequence of numbers is defined as the greatest integer that divides all the numbers in the sequence evenly.

• For example, the GCD of the sequence [4,6,16] is 2.

A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array.

• For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10].

Return the number of different GCDs among all non-empty subsequences of nums.

Example 1:

Input: nums = [6,10,3]
Output: 5
Explanation: The figure shows all the non-empty subsequences and their GCDs.
The different GCDs are 6, 10, 3, 2, and 1.


Example 2:

Input: nums = [5,15,40,5,6]
Output: 7


Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 2 * 105

## Solutions

Solution 1: Enumeration + Mathematics

For all sub-sequences of the array $nums$, their greatest common divisor (GCD) will not exceed the maximum value $mx$ in the array.

Therefore, we can enumerate each number $x$ in $[1,.. mx]$, and determine whether $x$ is the GCD of a sub-sequence of the array $nums$. If it is, then we increment the answer by one.

So the problem is transformed into: determining whether $x$ is the GCD of a sub-sequence of the array $nums$. We can do this by enumerating the multiples $y$ of $x$, and checking whether $y$ exists in the array $nums$. If $y$ exists in the array $nums$, then we calculate the GCD $g$ of $y$. If $g = x$ occurs, then $x$ is the GCD of a sub-sequence of the array $nums$.

The time complexity is $O(n + M \times \log M)$, and the space complexity is $O(M)$. Here, $n$ and $M$ are the length of the array $nums$ and the maximum value in the array $nums$, respectively.

• class Solution {
public int countDifferentSubsequenceGCDs(int[] nums) {
int mx = Arrays.stream(nums).max().getAsInt();
boolean[] vis = new boolean[mx + 1];
for (int x : nums) {
vis[x] = true;
}
int ans = 0;
for (int x = 1; x <= mx; ++x) {
int g = 0;
for (int y = x; y <= mx; y += x) {
if (vis[y]) {
g = gcd(g, y);
if (x == g) {
++ans;
break;
}
}
}
}
return ans;
}

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

• class Solution {
public:
int countDifferentSubsequenceGCDs(vector<int>& nums) {
int mx = *max_element(nums.begin(), nums.end());
vector<bool> vis(mx + 1);
for (int& x : nums) {
vis[x] = true;
}
int ans = 0;
for (int x = 1; x <= mx; ++x) {
int g = 0;
for (int y = x; y <= mx; y += x) {
if (vis[y]) {
g = gcd(g, y);
if (g == x) {
++ans;
break;
}
}
}
}
return ans;
}
};

• class Solution:
def countDifferentSubsequenceGCDs(self, nums: List[int]) -> int:
mx = max(nums)
vis = set(nums)
ans = 0
for x in range(1, mx + 1):
g = 0
for y in range(x, mx + 1, x):
if y in vis:
g = gcd(g, y)
if g == x:
ans += 1
break
return ans


• func countDifferentSubsequenceGCDs(nums []int) (ans int) {
mx := slices.Max(nums)
vis := make([]bool, mx+1)
for _, x := range nums {
vis[x] = true
}
for x := 1; x <= mx; x++ {
g := 0
for y := x; y <= mx; y += x {
if vis[y] {
g = gcd(g, y)
if g == x {
ans++
break
}
}
}
}
return
}

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