# 1977. Number of Ways to Separate Numbers

## 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 109 + 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.


Constraints:

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

## Solutions

• class Solution {
private static final int MOD = (int) 1e9 + 7;

public int numberOfCombinations(String num) {
int n = num.length();
int[][] lcp = new int[n + 1][n + 1];
for (int i = n - 1; i >= 0; --i) {
for (int j = n - 1; j >= 0; --j) {
if (num.charAt(i) == num.charAt(j)) {
lcp[i][j] = 1 + lcp[i + 1][j + 1];
}
}
}
int[][] dp = new int[n + 1][n + 1];
dp[0][0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= i; ++j) {
int v = 0;
if (num.charAt(i - j) != '0') {
if (i - j - j >= 0) {
int x = lcp[i - j][i - j - j];
if (x >= j || num.charAt(i - j + x) >= num.charAt(i - j - j + x)) {
v = dp[i - j][j];
}
}
if (v == 0) {
v = dp[i - j][Math.min(j - 1, i - j)];
}
}
dp[i][j] = (dp[i][j - 1] + v) % MOD;
}
}
return dp[n][n];
}
}

• class Solution {
public:
const int mod = 1e9 + 7;

int numberOfCombinations(string num) {
int n = num.size();
vector<vector<int>> lcp(n + 1, vector<int>(n + 1));
for (int i = n - 1; i >= 0; --i) {
for (int j = n - 1; j >= 0; --j) {
if (num[i] == num[j]) {
lcp[i][j] = 1 + lcp[i + 1][j + 1];
}
}
}
auto cmp = [&](int i, int j, int k) {
int x = lcp[i][j];
return x >= k || num[i + x] >= num[j + x];
};
vector<vector<int>> dp(n + 1, vector<int>(n + 1));
dp[0][0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= i; ++j) {
int v = 0;
if (num[i - j] != '0') {
if (i - j - j >= 0 && 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];
}
};

• 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]


• func numberOfCombinations(num string) int {
n := len(num)
lcp := make([][]int, n+1)
dp := make([][]int, n+1)
for i := range lcp {
lcp[i] = make([]int, n+1)
dp[i] = make([]int, n+1)
}
for i := n - 1; i >= 0; i-- {
for j := n - 1; j >= 0; j-- {
if num[i] == num[j] {
lcp[i][j] = 1 + lcp[i+1][j+1]
}
}
}
cmp := func(i, j, k int) bool {
x := lcp[i][j]
return x >= k || num[i+x] >= num[j+x]
}
dp[0][0] = 1
var mod int = 1e9 + 7
for i := 1; i <= n; i++ {
for j := 1; j <= i; j++ {
v := 0
if num[i-j] != '0' {
if i-j-j >= 0 && 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]
}