<p>Given an array of integers <code>nums</code> and an integer <code>k</code>. A continuous subarray is called <strong>nice</strong> if there are <code>k</code> odd numbers on it.</p>

<p>Return <em>the number of <strong>nice</strong> sub-arrays</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,1,2,1,1], k = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,4,6], k = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no odd numbers in the array.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,2,1,2,2,1,2,2,2], k = 2
<strong>Output:</strong> 16
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10^5</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution {
public:
    int numberOfSubarrays(vector<int>& nums, int k) {
        int Result=0;
        int n=nums.size();
        int DP[n];
        for(int i=0;i<n;i++)
        {
            if(i==0)
                DP[i]=nums[i]%2;
            else DP[i]=DP[i-1]+nums[i]%2;
        }
        vector<int> CustomDP;
        vector<pair<int,int>> Start_End;
        for(int i=0;i<n;i++)
        {
            int tempIndex=i;
            while((DP[tempIndex]==DP[i])&&(tempIndex<n))
            {
                tempIndex++;
                if(tempIndex==n)
                    break;
            }
            Start_End.push_back({i,tempIndex-1});
            CustomDP.push_back(DP[i]);
            i=tempIndex-1;
        }
        int Size=Start_End.size();
        for(int i=0;i<n;i++)
        {
            int Index=lower_bound(CustomDP.begin(),CustomDP.end(),k+DP[i]-nums[i]%2)-CustomDP.begin();
            if(Index<Size)
            {
                if(CustomDP[Index]==DP[i]+k-nums[i]%2)
                {
                    Result+=(Start_End[Index].second-Start_End[Index].first+1);
                }
            }
        }
        return Result;
    }
};
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