LeetCode 1979. Find Greatest Common Divisor of Array Solution in Java, C++, Python & More | Explanation + Code

Description

Given an integer array nums, return the greatest common divisor of the smallest number and largest number in nums.

The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.

 

Example 1:

Input: nums = [2,5,6,9,10]
Output: 2
Explanation:
The smallest number in nums is 2.
The largest number in nums is 10.
The greatest common divisor of 2 and 10 is 2.

Example 2:

Input: nums = [7,5,6,8,3]
Output: 1
Explanation:
The smallest number in nums is 3.
The largest number in nums is 8.
The greatest common divisor of 3 and 8 is 1.

Example 3:

Input: nums = [3,3]
Output: 3
Explanation:
The smallest number in nums is 3.
The largest number in nums is 3.
The greatest common divisor of 3 and 3 is 3.

 

Constraints:

• 2 <= nums.length <= 1000
• 1 <= nums[i] <= 1000

Solutions

Solution 1: Simulation

We can simulate according to the problem description. First, find the maximum and minimum values in the array nums, then find the greatest common divisor of the maximum and minimum values.

The time complexity is O(n), where n is the length of the array nums. The space complexity is O(1).

PythonJavaC++GoTypeScriptRust

class Solution: def findGCD(self, nums: List[int]) -> int: return gcd(max(nums), min(nums))(code-box)

class Solution { public int findGCD(int[] nums) { int a = 1, b = 1000; for (int x : nums) { a = Math.max(a, x); b = Math.min(b, x); } return gcd(a, b); } private int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } }(code-box)

class Solution { public: int findGCD(vector<int>& nums) { auto [min, max] = ranges::minmax_element(nums); return gcd(*min, *max); } };(code-box)

func findGCD(nums []int) int { a, b := slices.Max(nums), slices.Min(nums) return gcd(a, b) } func gcd(a, b int) int { if b == 0 { return a } return gcd(b, a%b) }(code-box)

function findGCD(nums: number[]): number { const min = Math.min(...nums); const max = Math.max(...nums); return gcd(min, max); } function gcd(a: number, b: number): number { if (b == 0) { return a; } return gcd(b, a % b); }(code-box)

impl Solution { pub fn find_gcd(nums: Vec<i32>) -> i32 { let min_val = *nums.iter().min().unwrap(); let max_val = *nums.iter().max().unwrap(); gcd(min_val, max_val) } } fn gcd(mut a: i32, mut b: i32) -> i32 { while b != 0 { let temp = b; b = a % b; a = temp; } a }(code-box)