Formatted question description: https://leetcode.ca/all/1977.html

# 1977. Number of Ways to Separate Numbers

Hard

## Description

You wrote down many positive integers in a string called num. However, you realized that you forgot to add commas to seperate the different numbers. You remember that the list of integers was non-decreasing and that no integer had leading zeros.

Return the number of possible lists of integers that you could have written down to get the string num. Since the answer may be large, return it modulo 10^9 + 7.

Example 1:

Input: num = “327”

Output: 2

Explanation: You could have written down the numbers:

3, 27

327

Example 2:

Input: num = “094”

Output: 0

Explanation: No numbers can have leading zeros and all numbers must be positive.

Example 3:

Input: num = “0”

Output: 0

Explanation: No numbers can have leading zeros and all numbers must be positive.

Example 4:

Input: num = “9999999999999”

Output: 101

Constraints:

• 1 <= num.length <= 3500
• num consists of digits '0' through '9'.

## Solution

Use dynamic programming with precalculation. First, precalculate the longest common prefixes of each pair of substrings. Next, use dynamic programming to count the number of ways.

class Solution {
public int numberOfCombinations(String num) {
final int MODULO = 1000000007;
if (num.charAt(0) == '0')
return 0;
int length = num.length();
int[][] lcp = new int[length][length];
for (int i = length - 1; i >= 0; i--) {
lcp[i][length - 1] = num.charAt(i) == num.charAt(length - 1) ? 1 : 0;
for (int j = i + 1; j < length - 1; j++)
lcp[i][j] = num.charAt(i) == num.charAt(j) ? lcp[i + 1][j + 1] + 1 : 0;
}
int[][] dp = new int[length][length];
for (int i = 0; i < length; i++)
dp[i] = 1;
for (int i = 1; i < length; i++) {
if (num.charAt(i) == '0')
continue;
int presum = 0;
for (int j = i; j < length; ++j) {
int range = j - i + 1;
dp[i][j] = presum;
if (i - range >= 0) {
if (lcp[i - range][i] >= range || num.charAt(i - range + lcp[i - range][i]) < num.charAt(i + lcp[i - range][i]))
dp[i][j] = (dp[i][j] + dp[i - range][i - 1]) % MODULO;
presum = (presum + dp[i - range][i - 1]) % MODULO;
}
}
}
int ways = 0;
for (int i = 0; i < length; i++)
ways = (ways + dp[i][length - 1]) % MODULO;
return ways;
}
}