<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(vector<int>& nums, int k) {
        
        unordered_set<int> st;
        
        int total = 0, preSum;
        
        for(auto &x: nums)
        {
            preSum = total;
            
            total += x;
            
            if(st.count(total % k))
            {
                return true;
            }
                
            st.insert(preSum % k);
            
        }
        
        return false;
        
        the code has a time complexity of O(n) and a space complexity of O(min(n, k)), which makes it efficient for most practical scenarios.
       
       int n = nums.size();
        
        for(int i=0; i<=n-2; i++)
        {
            int sum = 0;
            
            for(int j=i; j<i+2; j++) //...
            {
                sum += nums[j];
                
                if(sum/ k == 1)
                {
                    return true;
                }
            }
        }
        
        return false;
       
    }
};*/

/*class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) 
    {
      unordered_map<int,int>mp;
        
      mp[0] = 0;
        
      int sum = 0;
        
      for(int i=0;i<nums.size();i++)
      {
          sum+=nums[i];
          
          if(mp.find(sum%k) == mp.end())
          {
              mp[sum%k] = i+1;
          }
          
          else if(mp[sum%k]<i)
          {
              return true;
          }
          
      }
        return false;
    }
};*/

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int sum = 0;
        unordered_map<int, int> mp; // Map to store remainder and index

        mp[0] = -1;  // Initialize to handle the case when sum % k == 0 for the whole subarray
        
        for (int i = 0; i < nums.size(); i++) {
            sum += nums[i];
            
            // If k is not zero, we want to handle the modulo operation
            if (k != 0) {
                sum %= k;
            }
            
            // If the same remainder is found again, check the distance between indices
            if (mp.find(sum) != mp.end()) {
                if (i - mp[sum] > 1) {
                    return true;
                }
            } else {
                mp[sum] = i;  // Store the first occurrence of this remainder
            }
        }
        
        return false;
    }
};

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