Formatted question description: https://leetcode.ca/all/1250.html

# 1250. Check If It Is a Good Array

Hard

## Description

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.

53 + 7(-2) = 1

Example 2:

Input: nums = [29,6,10]

Output: true

Explanation: Pick numbers 29, 6 and 10.

291 + 6(-3) + 10*(-1) = 1

Example 3:

Input: nums = [3,6]

Output: false

Constraints:

• 1 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^9

## Solution

This problem is actually a number theory problem. Although marked as Hard difficulty, this problem can be easily solved once the number theory knowledge behind is applied.

For any two positive integers x and y, there exist two integers a and b such that ax+by=1 if and only if the greatest common divisor of a and b is 1.

Therefore, just calculate the greatest common divisor of all the numbers in the given array num. If at one step, the greatest common divisor become 1, then stop calculating further since the greatest common divisor of 1 and any positive integer is 1.

If the greatest common divisor is 1, return True. Otherwise, return False.

• class Solution {
public boolean isGoodArray(int[] nums) {
int length = nums.length;
int gcd = nums;
for (int i = 1; i < length; i++) {
gcd = gcd(gcd, nums[i]);
if (gcd == 1)
return true;
}
return gcd == 1;
}

public int gcd(int num1, int num2) {
if (num1 == 0 && num2 == 0)
return 1;
if (num1 == 1 || num2 == 1)
return 1;
if (num1 > num2) {
int temp = num1;
num1 = num2;
num2 = temp;
}
while (num1 != 0 && num2 != 0) {
num2 %= num1;
int temp = num1;
num1 = num2;
num2 = temp;
}
if (num1 == 0)
return num2;
else
return num1;
}
}

• // OJ: https://leetcode.com/problems/check-if-it-is-a-good-array/
// Time: O(N)
// Space: O(1)
class Solution {
public:
bool isGoodArray(vector<int>& A) {
int d = A;
for (int i = 1; i < A.size(); ++i) {
d = gcd(d, A[i]);
}
return d == 1;
}
};

• print("Todo!")