Welcome to Subscribe On Youtube

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)
}

All Problems

All Solutions