<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 {
private: 
    int GetSubArraySum(const vector<int>& nums, int beginIdx, int endIdx)
    {
        int sum = 0;
        
        for (int i = beginIdx; i <= endIdx; i++)
        {
            sum += nums[i];
        }

        return sum;
    }

public:
    bool checkSubarraySum(vector<int>& nums, int k) 
    {
        // 1. find all subarraies -> length 1  x  N
        //                           length 2  x (N-1)
        //                           length 3  x (N-2)
        //                           ...
        //                           length N  x  1

        //                                       (1 + N) * N / 2 -> O(n^2)
        // 2. get sum of each subarray and see if it's "good subarray" -> O(n)


        int numsSize = nums.size();

        // good subarray lenght is at least 2
        if (numsSize <= 1)
        {
            return false;
        }

        // for (int subArraySize = 2; subArraySize <= numsSize; subArraySize++)
        // {
        //     for (int beginIdx = 0; beginIdx + subArraySize - 1 < numsSize; beginIdx++)
        //     {
        //         int subArraySum = GetSubArraySum(nums, beginIdx, beginIdx + subArraySize - 1);
        //         if (subArraySum % k == 0)
        //         {
        //             return true;
        //         }
        //     }
        // }

        // return false;


        vector<int> prefixSum;
        prefixSum.resize(numsSize);

        prefixSum[0] = nums[0];
        for (int i = 1; i < numsSize; i++)
        {
            prefixSum[i] = prefixSum[i-1] + nums[i];
        }

        unordered_map<int, int> hashMap; // prefixMod -> index
        hashMap[0] = -1;

        for (int i = 0; i < numsSize; i++)
        {
            int prefixMod = prefixSum[i] % k;

            if (hashMap.find(prefixMod) != hashMap.end())
            {
                if (i - hashMap[prefixMod] >= 2)
                {
                    return true;
                }
            }
            else
            {
                hashMap[prefixMod] = i;
            }
        }

        return false;
    }
};
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