<p>A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits <code>s</code> and all we know is that all integers in the array were in the range <code>[1, k]</code> and there are no leading zeros in the array.</p>

<p>Given the string <code>s</code> and the integer <code>k</code>, return <em>the number of the possible arrays that can be printed as </em><code>s</code><em> using the mentioned program</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;1000&quot;, k = 10000
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only possible array is [1000]
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;1000&quot;, k = 10
<strong>Output:</strong> 0
<strong>Explanation:</strong> There cannot be an array that was printed this way and has all integer &gt;= 1 and &lt;= 10.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;1317&quot;, k = 2000
<strong>Output:</strong> 8
<strong>Explanation:</strong> Possible arrays are [1317],[131,7],[13,17],[1,317],[13,1,7],[1,31,7],[1,3,17],[1,3,1,7]
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of only digits and does not contain leading zeros.</li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
    int MOD = 1e9 + 7;
public:
    int numberOfArrays(string s, int k) {
        int n = s.length();
        int klen = to_string(k).length();

        vector<int> dp(n, 0);

        if(k < (s[0] - '0'))
            return 0;
        dp[0] = 1;

        for(int i = 1; i < n; i++)
        {
            if((s[i] - '0') > k)
                return 0;
            
            if((s[i] - '0') != 0)
                dp[i] = dp[i - 1];

            for(int j = (i - 1); j >= 0 && j > (i - klen); j--)
            {
                if(s[j] == '0')
                    continue;

                string temp = s.substr(j, (i - j + 1));
                long num = stol(temp);

                if(num != 0 && num <= (1LL * k))
                    dp[i] = (j == 0) ? ((dp[i] + 1) % MOD): ((dp[i] + dp[j - 1]) % MOD);
            }
        }

        return dp[n - 1];
    }
};
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