<p>Given an integer array nums and an integer k, return <code>true</code> <em>if </em><code>nums</code><em> has a <strong>good subarray</strong> or </em><code>false</code><em> otherwise</em>.</p>

<p>A <strong>good subarray</strong> is a subarray where:</p>

<ul>
	<li>its length is <strong>at least two</strong>, and</li>
	<li>the sum of the elements of the subarray is a multiple of <code>k</code>.</li>
</ul>

<p><strong>Note</strong> that:</p>

<ul>
	<li>A <strong>subarray</strong> is a contiguous part of the array.</li>
	<li>An integer <code>x</code> is a multiple of <code>k</code> if there exists an integer <code>n</code> such that <code>x = n * k</code>. <code>0</code> is <strong>always</strong> a multiple of <code>k</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [23,<u>2,4</u>,6,7], k = 6
<strong>Output:</strong> true
<strong>Explanation:</strong> [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [<u>23,2,6,4,7</u>], k = 6
<strong>Output:</strong> true
<strong>Explanation:</strong> [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [23,2,6,4,7], k = 13
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= sum(nums[i]) &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>1 &lt;= k &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    inline static int hash_map(vector<int>& nums, int k, int n){
        unordered_set<int> mod_k;// 1 <= k <= 2^31 - 1
        int prefix=0, last=0;
        mod_k.reserve(n);
        for(int i=0; i<n; i++){
            prefix+=nums[i];
            prefix%=k;
            if (mod_k.count(prefix))
                return 1;
            mod_k.insert(last);
            last=prefix;
        }
        return 0;
    }
    inline static int array(vector<int>& nums, int k, int n){
        bitset<1000000> mod_k=0;// mod k=0,1,...,k-1
        int prefix=0, last=0;
        for(int x: nums){
            prefix+=x;
            prefix%=k;
            if (mod_k[prefix])
                return 1;
            mod_k[last]=1;
            last=prefix;
        }
        return 0;
    }
    static bool checkSubarraySum(vector<int>& nums, int k) {
        int n=nums.size();
        if (n<2) return 0; //its length is at least two
        if (k>=1e6) return hash_map(nums, k, n);
        return array(nums, k, n);
    }     
};
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