#1977

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Number of Ways to Separate Numbers

Calculate the number of valid non-decreasing integer sequences from a string using state transition dynamic programming efficiently.

string dynamic programming suffix array

Problem Statement

You are given a string num representing a sequence of digits with no separators. The original sequence consisted of positive integers written in non-decreasing order, but commas were omitted. No integer had leading zeros, and each number is at least 1 digit long.

Return the total number of possible sequences of integers that could produce the string num. Since the result may be large, return it modulo 10^9 + 7. For example, given num = "327", the possible sequences are [3,27] and [327].

Examples

Example 1

Input: num = "327"

Output: 2

You could have written down the numbers: 3, 27 327

Example 2

Input: num = "094"

Output: 0

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

Example 3

Input: num = "0"

Output: 0

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

Constraints

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

Solution Approach

Dynamic Programming Table Setup

Define dp[i] as the number of ways to split the prefix num[0..i]. Iterate through possible previous splits and maintain valid non-decreasing conditions using string comparisons.

Handling Leading Zeros and Lengths

Skip splits where a number starts with '0' to avoid invalid sequences. Use length-based checks to ensure the current number is at least as large as the previous number in value.

Optimizing with Prefix Comparisons

Use precomputed suffix arrays or LCP-like techniques to compare substrings efficiently. This reduces repeated string comparisons and ensures the solution scales up to num.length of 3500.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time and space complexity depend on the DP implementation. Naive DP can reach O(n^2) for both, while optimizations using suffix arrays or prefix comparisons reduce constant factors without changing the asymptotic bounds.

What Interviewers Usually Probe

Are you considering how to represent previous number states efficiently?
Can you handle cases with leading zeros correctly?
Think about reducing repeated string comparisons using suffix information.

Common Pitfalls or Variants

Common pitfalls

Not skipping numbers starting with zero leads to invalid sequences.
Comparing numbers incorrectly without handling lengths can yield wrong DP transitions.
Inefficient repeated substring comparisons can cause timeouts on long inputs.

Follow-up variants

Count sequences under strictly increasing integer constraints instead of non-decreasing.
Allow leading zeros and compute all possible sequences including them.
Return all valid sequences as actual lists instead of only the count.

FAQ

What is the main challenge in Number of Ways to Separate Numbers?

The main challenge is efficiently counting all non-decreasing integer sequences without leading zeros using state transition DP.

How does the dynamic programming approach work for this problem?

DP tracks the number of ways to split the string up to each position while validating non-decreasing order and avoiding leading zeros.

Can suffix arrays improve performance?

Yes, precomputing suffix comparisons or LCP arrays reduces repeated string comparisons and speeds up DP transitions.

What should I watch out for regarding leading zeros?

Any number starting with '0' is invalid except for single-digit '0', so these splits must be skipped.

Is this problem suitable for practicing state transition dynamic programming?

Absolutely, it is a classic example of state transition DP combined with string manipulation and substring comparisons.

terminal

Solution

Solution 1: Dynamic Programming + Prefix Sum

Define $dp[i][j]$ to represent the number of ways to partition the first $i$ characters of the string `num` such that the length of the last number is $j$. Clearly, the answer is $\sum_{j=0}^{n} dp[n][j]$. The initial value is $dp[0][0] = 1$.

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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]

Continue Practicing

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