Given an array nums of positive integers. Your task is to select some subset of nums, multiply each element by an integer and add all these numbers. The array is said to be good if you can obtain a sum of 1 from the array by any possible subset and multiplicand.
Return True if the array is good otherwise return False.
Example 1:
Input: nums = [12,5,7,23]
Output: true
Explanation: Pick numbers 5 and 7.
5*3 + 7*(-2) = 1
Example 2:
Input: nums = [29,6,10]
Output: true
Explanation: Pick numbers 29, 6 and 10.
29*1 + 6*(-3) + 10*(-1) = 1
Example 3:
Input: nums = [3,6]
Output: false
Constraints:
1 <= nums.length <= 10^51 <= nums[i] <= 10^9
This problem is an application of the Bezout's lemma: if gcd(A1, A2,...., An) = d, then there are integers X1, X2,..., Xn such that d = A1 * X1 + A2 * X2 + ... + An * Xn. d is the smallest positive integer of this form. Every number of this form is a multiple of d. This lemma is also a generalization of the extended euclid's algorithm: we can always get integer x and y such that a * x + b * y = gcd(a, b).
Applying the above lemma, we know that we just need to check if there exists two numbers X and Y that are coprime, gcd(X, Y) == 1. Here X and Y can be a single array number or a series of numbers' gcd. So we can just compute the entire array's gcd. If at any point the gcd becomes 1, we know that either we have a number of 1 or we have 2 co-prime numbers and we can find use these numbers to get a sum of 1 for sure.
Euclid's GCD algorithms takes O(log(min(a, b))). So the runtime is O(N * log(min(a, b))).
class Solution { public boolean isGoodArray(int[] nums) { int g = nums[0]; for(int i = 1; i < nums.length; i++) { g = gcd(g, nums[i]); if(g == 1) { break; } } return g == 1; } private int gcd(int a, int b) { if(b == 0) { return a; } return gcd(b, a % b); } }