actually for both of the steps 1 & 2 the complexity will be O(n log(b)).

Here n is the length of the array and b is the smaller integer of a pair(a,b) for which we calculate the GCD/LCM.

For simplicity we call it O(n log(n)) instead of O(n log(b)).

Explanation: To calculate the GCD or LCM for the whole array we need to run a loop up to the length of the array which is n and inside that loop we need to calculate GCD/LCM which will take log(n). like below.

previousGCD = array[0];
for(int i = 1; i<n; i++) {// n
	previousGCD = GCD(previousGCD, array[i]);//log(n) 
}

so the complexity is O(n*log(n)). Same goes for LCM also.