<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:
bool checkSubarraySum(std::vector<int>& nums, int k) {
    std::unordered_map<int, int> remainder_map;  // A map to store remainders and their corresponding indices.
    remainder_map[0] = -1;  // Initialize the remainder 0 to -1 to handle the case where a valid subarray starts from index 0.
    int sum = 0;  // Initialize sum to keep track of the cumulative sum of the array.

    for (int i = 0; i < nums.size(); ++i) {  // Iterate through the array to compute cumulative sums.
        sum += nums[i];  // Add the current element to the cumulative sum.
        
        if (k != 0) {  // Check if k is non-zero to avoid division by zero.
            sum %= k;  // Take the modulus of the sum to find the remainder when divided by k.
        }
        
        if (remainder_map.find(sum) != remainder_map.end()) {  // Check if the current remainder has been seen before.
            if (i - remainder_map[sum] > 1) {  // Ensure the subarray length is at least 2 (indices are sufficiently far apart).
                return true;  // Return true if a valid subarray is found.
            }
        } else {
            remainder_map[sum] = i;  // If the remainder hasn't been seen before, store the current index in the map.
        }
    }
    
    return false;  // If no valid subarray is found, return false.
}
};
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