What is the time complexity of Number of Ways to Separate Numbers?

The optimal solution has a time complexity of O(n^2).

What is the space complexity of Number of Ways to Separate Numbers?

The space complexity is O(n^2).

What data structures are used to solve Number of Ways to Separate Numbers?

This problem uses String, Dynamic Programming, Suffix Array. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Ways to Separate Numbers?

Insight 1: Dynamic programming is suitable for this problem because we need to count the number of valid combinations, and the subproblems overlap. We can build a table of results for smaller substrings. Insight 2: The core idea is to iterate through all possible split positions and determine if the resulting sequence is non-decreasing and contains no leading zeros. The non-decreasing constraint needs to be handled carefully by comparing substrings.

Which companies ask LeetCode 1977. Number of Ways to Separate Numbers in interviews?

This problem is frequently asked at Walmart Labs.

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Number of Ways to Separate Numbers - LeetCode 1977 Solution

Number of Ways to Separate Numbers - Complete Solution Guide

Number of Ways to Separate Numbers is LeetCode problem 1977, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 numb

Detailed Explanation

The problem asks us to find the number of ways a given string of digits, `num`, can be split into a non-decreasing sequence of positive integers. The integers formed after splitting cannot have leading zeros. The result needs to be returned modulo 10^9 + 7. For example, if `num` is "327", the possible splits are [3, 27] and [327].

Solution Approach

The solution uses dynamic programming. `dp[i][j]` stores the number of ways to split the first `i` characters of `num` such that the last number in the sequence has length `j`. `S[i][j]` is a prefix sum array used to efficiently calculate the number of valid splits ending at position `i` with a length less than or equal to `j`. The LCP (Longest Common Prefix) array `lcp[i][j]` stores the length of the longest common prefix between the substrings of `num` starting at indices `i` and `j`. LCP is used to optimize string comparisons when checking if the resulting sequence is non-decreasing. The approach builds the `dp` table iteratively by considering all possible lengths of the last number and checking if adding it results in a valid non-decreasing sequence.

Step-by-Step Algorithm

Step 1: Initialize `lcp` array. `lcp[i][j]` represents the longest common prefix length between num[i:] and num[j:].
Step 2: Initialize `dp` array. `dp[i][j]` represents the number of ways to split num[:i] such that the last number has length j.
Step 3: Initialize `S` array. `S[i][j]` represents the sum of `dp[i][k]` for all `k` from 1 to `j`. It is used for prefix sum optimization.
Step 4: Iterate through the input string. The outer loop iterates from `i = 1` to `n` (the length of `num`).
Step 5: The inner loop iterates from `j = 1` to `i`, representing the length of the last number in the sequence.
Step 6: Check for leading zeros: If the last number starts with '0', skip this split.
Step 7: If the current split is at the beginning of the string (prev_len == 0), set `dp[i][j] = 1` because this represents a valid split consisting of only one number.
Step 8: Calculate the maximum allowed length of the previous number `limit_k`. It's the minimum between current number's length `j` and length of remaining part `prev_len`.
Step 9: Calculate the number of ways to split `num[:i]` by using precomputed sum of ways until current index `ways=S[prev_len][limit_k]`.
Step 10: Check if the current number is greater than or equal to previous number if possible.
Step 11: If current number is not greater than equal to previous number, subtract the number of ways `dp[prev_len][j]`.
Step 12: Update `dp[i][j]` with the number of ways and precompute the prefix sum using `dp` table.

Key Insights

Insight 1: Dynamic programming is suitable for this problem because we need to count the number of valid combinations, and the subproblems overlap. We can build a table of results for smaller substrings.
Insight 2: The core idea is to iterate through all possible split positions and determine if the resulting sequence is non-decreasing and contains no leading zeros. The non-decreasing constraint needs to be handled carefully by comparing substrings.
Insight 3: Longest Common Prefix (LCP) optimization helps to compare numbers effectively and avoid redundant string comparisons. It's a key performance optimization.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: String, Dynamic Programming, Suffix Array.

Companies

Asked at: Walmart Labs.

Number of Ways to Separate Numbers | LeetCode 1977

Hard

String Dynamic Programming Suffix Array

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int numberOfCombinations(String num) {
        int n = num.length();
        int MOD = (int)1e9 + 7;

        if (num.charAt(0) == '0') {
            return 0;
        }

        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];
        int[][] S = new int[n + 1][n + 1];

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= i; j++) {
                int start_idx = i - j;
                if (num.charAt(start_idx) == '0') {
                    continue;
                }

                int prev_len = i - j;
                if (prev_len == 0) {
                    dp[i][j] = 1;
                    continue;
                }
                
                int limit_k = Math.min(j, prev_len);
                long ways = S[prev_len][limit_k];
                
                if (j <= prev_len) {
                    int prev_num_start = prev_len - j;
                    int common_len = lcp[prev_num_start][start_idx];
                    
                    if (common_len < j && num.charAt(prev_num_start + common_len) > num.charAt(start_idx + common_len)) {
                        ways = (ways - dp[prev_len][j] + MOD) % MOD;
                    }
                }
                
                dp[i][j] = (int)ways;
            }
            
            for (int j = 1; j <= n; j++) {
                S[i][j] = (S[i][j - 1] + dp[i][j]) % MOD;
            }
        }

        return S[n][n];
    }
}

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Problem Info

Description

Problem 1977: Number of Ways to Separate Numbers - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Suffix Array

Companies

Walmart Labs

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1977

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Initialize `lcp` array. `lcp[i][j]` represents the longest common prefix length between num[i:] and num[j:].
2
Step 2: Initialize `dp` array. `dp[i][j]` represents the number of ways to split num[:i] such that the last number has length j.
3
Step 3: Initialize `S` array. `S[i][j]` represents the sum of `dp[i][k]` for all `k` from 1 to `j`. It is used for prefix sum optimization.
4
Step 4: Iterate through the input string. The outer loop iterates from `i = 1` to `n` (the length of `num`).
5
Step 5: The inner loop iterates from `j = 1` to `i`, representing the length of the last number in the sequence.
6
Step 6: Check for leading zeros: If the last number starts with '0', skip this split.
7
Step 7: If the current split is at the beginning of the string (prev_len == 0), set `dp[i][j] = 1` because this represents a valid split consisting of only one number.
8
Step 8: Calculate the maximum allowed length of the previous number `limit_k`. It's the minimum between current number's length `j` and length of remaining part `prev_len`.
9
Step 9: Calculate the number of ways to split `num[:i]` by using precomputed sum of ways until current index `ways=S[prev_len][limit_k]`.
10
Step 10: Check if the current number is greater than or equal to previous number if possible.
11
Step 11: If current number is not greater than equal to previous number, subtract the number of ways `dp[prev_len][j]`.
12
Step 12: Update `dp[i][j]` with the number of ways and precompute the prefix sum using `dp` table.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is determined by the nested loops that iterate through all possible substrings to calculate the DP values and LCP values. The space complexity is due to the creation of three 2D arrays `dp`, `S`, and `lcp`, each of size (n+1) x (n+1).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach with memoization can also be used to solve this problem. The idea is to define a recursive function that takes the current index as input and returns the number of ways to split the string from that index onwards. Memoization can be used to avoid recomputing the same subproblems.

Time: O(n^2)Space: O(n^2)

Trade-offs: Recursion with memoization can be more intuitive for some, but the iterative DP approach is generally more efficient in practice because it avoids the overhead of function calls.

✨

Key Insights

1Insight 1: Dynamic programming is suitable for this problem because we need to count the number of valid combinations, and the subproblems overlap. We can build a table of results for smaller substrings.
2Insight 2: The core idea is to iterate through all possible split positions and determine if the resulting sequence is non-decreasing and contains no leading zeros. The non-decreasing constraint needs to be handled carefully by comparing substrings.
3Insight 3: Longest Common Prefix (LCP) optimization helps to compare numbers effectively and avoid redundant string comparisons. It's a key performance optimization.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Input string starts with '0'. The function should return 0, as no number can have leading zeros. The code handles this case at the beginning by checking `if num[0] == '0': return 0`.
Edge case 2: Input contains only zero. The code returns zero since there are no valid splits with positive non-zero numbers.
Edge case 3: Non-decreasing sequence check. Correctly implementing the non-decreasing sequence logic using LCP optimization is crucial. The algorithm explicitly checks `if common_len < j and num[prev_num_start + common_len] > num[start_idx + common_len]` to ensure the non-decreasing order.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the non-decreasing sequence constraint. It's essential to compare the numbers correctly, considering their lengths and using LCP to optimize the comparison.

✗Common mistake 2: Failing to handle leading zeros. The code needs to explicitly check for leading zeros and avoid splits that result in numbers with leading zeros.

✗Common mistake 3: Integer overflow. The number of ways can be very large, so it's important to use the modulo operator to prevent integer overflow.

✗Common mistake 4: Not using prefix sums to optimize DP transitions which could lead to higher time complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression. Similar dynamic programming techniques can be used in data compression algorithms to find optimal ways to encode data.

🏦Real-world scenario 2: Financial modeling. Decomposing a time series of stock prices into non-decreasing or non-increasing segments can be useful for identifying trends and patterns in financial markets.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Ways to Separate Numbers"?

For this hard-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?