Number of Ways to Separate Numbers
<p>You wrote down many <strong>positive</strong> integers in a string called <code>num</code>. However, you realized that you forgot to add commas to seperate the different numbers. You remember that the list of integers was <strong>non-decreasing</strong> and that <strong>no</strong> integer had leading zeros.</p>

<p>Return <em>the <strong>number of possible lists of integers</strong> that you could have written down to get the string </em><code>num</code>. Since the answer may be 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> num = &quot;327&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You could have written down the numbers:
3, 27
327
</pre>

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

<pre>
<strong>Input:</strong> num = &quot;094&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> No numbers can have leading zeros and all numbers must be positive.
</pre>

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

<pre>
<strong>Input:</strong> num = &quot;0&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> No numbers can have leading zeros and all numbers must be positive.
</pre>

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

<ul>
	<li><code>1 &lt;= num.length &lt;= 3500</code></li>
	<li><code>num</code> consists of digits <code>&#39;0&#39;</code> through <code>&#39;9&#39;</code>.</li>
</ul>
String; Dynamic Programming; Suffix Array
class Solution:
    def numberOfCombinations(self, num: str) -> int:
        def cmp(i, j, k):
            x = lcp[i][j]
            return x >= k or num[i + x] >= num[j + x]

        mod = 10**9 + 7
        n = len(num)
        lcp = [[0] * (n + 1) for _ in range(n + 1)]
        for i in range(n - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                if num[i] == num[j]:
                    lcp[i][j] = 1 + lcp[i + 1][j + 1]

        dp = [[0] * (n + 1) for _ in range(n + 1)]
        dp[0][0] = 1
        for i in range(1, n + 1):
            for j in range(1, i + 1):
                v = 0
                if num[i - j] != '0':
                    if i - j - j >= 0 and cmp(i - j, i - j - j, j):
                        v = dp[i - j][j]
                    else:
                        v = dp[i - j][min(j - 1, i - j)]
                dp[i][j] = (dp[i][j - 1] + v) % mod
        return dp[n][n]
