Solution 1 - GCD of Min and Max

GCD of all numbers equals gcd(min(nums), max(nums)).

TimeComplexity: O(n + \log M)

SpaceComplexity: O(1)

""" 1979.1 - Find Greatest Common Divisor of Array - Solution 1 - Min/Max GCD """

from math import gcd
from typing import List

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

function gcd(a:number,b:number):number{while(b!==0){[a,b]=[b,a%b];}return a;}
function findGCD(nums:number[]):number{
 let lo=Infinity, hi=-Infinity;
 for(const x of nums){ if(x<lo)lo=x; if(x>hi)hi=x; }
 return gcd(lo,hi);
}

package main

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

func findGCD(nums []int) int {
	lo, hi := nums[0], nums[0]
	for _, x := range nums {
		if x < lo { lo = x }
		if x > hi { hi = x }
	}
	return gcd(lo, hi)
}

func main() {}

fn gcd(mut a:i32, mut b:i32) -> i32 {
	while b != 0 { let t=a%b; a=b; b=t; }
	a
}

fn find_gcd(nums: &[i32]) -> i32 {
	let lo = *nums.iter().min().unwrap();
	let hi = *nums.iter().max().unwrap();
	gcd(lo, hi)
}

fn main() {}

1979 - Find Greatest Common Divisor of Array

Solution 1 - GCD of Min and Max