Number of Ways to Separate Numbers solution and explanation
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The problem
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 <= 3500numconsists of digits'0'through'9'.
Approach overview
Problem Overview: You receive a numeric string and must count how many ways it can be split into a sequence of integers where each number is non‑decreasing compared to the previous one. Leading zeros are not allowed. The challenge is comparing many substrings efficiently while exploring valid partitions.
Approach 1: Recursive Approach with Memoization (Exponential → Pruned with DP)
This approach treats the problem as recursive partitioning of the string. At each index, choose a substring representing the next number and verify two rules: it cannot start with '0', and it must be greater than or equal to the previous number. The recursion continues from the next index. Memoization stores results based on the current index and previous substring length so repeated states are not recomputed. Even with caching, substring comparisons can be expensive, and the number of candidate splits grows quickly. Time complexity is roughly O(n^3) in the worst case due to substring comparisons, and space complexity is O(n^2) for memoization states and recursion stack. This method helps build intuition for how the partitioning works.
Approach 2: Dynamic Programming with Fast String Comparison (Optimal)
The optimized approach uses dynamic programming combined with precomputed substring comparison. Let dp[i][len] represent the number of valid ways to form a sequence ending at index i where the last number has length len. For each candidate number ending at i, you compare it with the previous number of the same or smaller length to ensure the non‑decreasing condition. Direct string comparison would make the solution too slow, so an LCP (longest common prefix) table is precomputed. This idea is closely related to structures used in suffix arrays and enables constant‑time substring comparison.
With LCP values, you can determine whether two substrings are lexicographically ordered without scanning characters repeatedly. Prefix sums over DP states allow fast transitions when the previous number length is smaller. Each position processes possible lengths while enforcing the no‑leading‑zero rule. The final answer is the sum of valid sequences that consume the entire string.
This reduces the overall complexity to O(n^2) time and O(n^2) space. The algorithm relies heavily on string processing and dynamic programming state transitions, which makes it the expected solution for this hard problem.
Recommended for interviews: Start by describing the recursive partition idea to show you understand the constraints and ordering rule. Then move to the dynamic programming solution with LCP‑based substring comparison. Interviewers typically expect the optimized O(n^2) DP approach because it demonstrates strong control over string comparison, state design, and performance optimization.
Approach 1: Dynamic Programming with String Comparison
This approach involves using dynamic programming to solve the problem efficiently. We use a 2D DP table where dp[i][j] indicates the number of ways to partition the string from position i to j. We'll iterate over each possible partitioning point, using string comparison to ensure non-decreasing order without leading zeros.
def numberOfWays(num: str) -> int:
n = len(num)
mod = 1000000007
dp = [0] * (n + 1)
dp[0] = 1
for i in range(1, n + 1):
for j in range(i):
if num[j] != '0':
dp[i] = (dp[i] + dp[j]) % mod
if j != 0:
length = i - j
if j >= length and num[j-length:j] <= num[j:i]:
dp[i] = (dp[i] + dp[j-length]) % mod
return dp[n]
num = "327"
print(numberOfWays(num)) # Output: 2import java.util.*;
public class NumberOfWays {
public static long numberOfWays(String num) {
int n = num.length();
long[] dp = new long[n + 1];
dp[0] = 1;
final int MOD = 1000000007;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
if (num.charAt(j) != '0') {
dp[i] = (dp[i] + dp[j]) % MOD;
if (j > 0) {
int len = i - j;
if (j >= len && num.substring(j - len, j).compareTo(num.substring(j, i)) <= 0) {
dp[i] = (dp[i] + dp[j - len]) % MOD;
}
}
}
}
}
return dp[n];
}
public static void main(String[] args) {
String num = "327";
System.out.println(numberOfWays(num)); // Output: 2
}
}#include <iostream>
#include <vector>
using namespace std;
#define MOD 1000000007
int numberOfWays(string num) {
int n = num.size();
vector<long> dp(n + 1, 0);
dp[0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
if (num[j] != '0') {
dp[i] = (dp[i] + dp[j]) % MOD;
if (j > 0) {
int len = i - j;
if (j >= len && num.substr(j - len, len) <= num.substr(j, len)) {
dp[i] = (dp[i] + dp[j - len]) % MOD;
}
}
}
}
}
return dp[n];
}
int main() {
string num = "327";
cout << numberOfWays(num) << endl; // Output: 2
return 0;
}function numberOfWays(num) {
const mod = 1000000007;
const n = num.length;
const dp = new Array(n + 1).fill(0);
dp[0] = 1;
for (let i = 1; i <= n; i++) {
for (let j = 0; j < i; j++) {
if (num[j] !== '0') {
dp[i] = (dp[i] + dp[j]) % mod;
if (j > 0) {
const len = i - j;
if (j >= len && num.slice(j - len, j) <= num.slice(j, i)) {
dp[i] = (dp[i] + dp[j - len]) % mod;
}
}
}
}
}
return dp[n];
}
console.log(numberOfWays("327")); // Output: 2#include <stdio.h>
#include <string.h>
#define MOD 1000000007
int numberOfWays(char *num) {
int n = strlen(num);
long dp[n + 1];
memset(dp, 0, sizeof(dp));
dp[0] = 1;
long dpCopy[n];
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
if (num[j] != '0') {
strncpy(dpCopy, dp, sizeof(dp));
dp[i] += dpCopy[j];
dp[i] %= MOD;
if (j > 0) {
int len = i - j;
if (j >= len && strncmp(&num[j - len], &num[j], len) <= 0) {
dp[i] = (dp[i] + dpCopy[j - len]) % MOD;
}
}
}
}
}
return dp[n];
}
int main() {
char num[] = "327";
printf("%d\n", numberOfWays(num)); // Output: 2
return 0;
}using System;
public class Solution {
public int NumberOfWays(string num) {
int n = num.Length;
int mod = 1000000007;
long[] dp = new long[n + 1];
dp[0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
if (num[j] != '0') {
dp[i] = (dp[i] + dp[j]) % mod;
if (j > 0) {
int len = i - j;
if (j >= len && string.Compare(num.Substring(j - len, len), num.Substring(j, i - j)) <= 0) {
dp[i] = (dp[i] + dp[j - len]) % mod;
}
}
}
}
}
return (int)dp[n];
}
public static void Main() {
Solution sol = new Solution();
string num = "327";
Console.WriteLine(sol.NumberOfWays(num)); // Output: 2
}
}For Python, this solution initializes a list for the DP array. String slices are efficiently used to check substring conditions, and Python's inherent integer handling ensures modular arithmetic is correctly implemented.
Time Complexity: O(n^2)
Space Complexity: O(n)
Approach 2: Recursive Approach with Memoization
This approach leverages recursion with memoization to solve the problem efficiently. We go through the number string recursively, attempting to partition it and storing intermediate results to avoid redundant calculations.
def helper(num, i, mem):
if i == len(num):
return 1
if num[i] == '0':
return 0
if mem[i] != -1:
return mem[i]
result = 0
for j in range(i + 1, len(num) + 1):
if j - i <= i and num[i-(j-i):i] <= num[i:j]:
result = (result + helper(num, j, mem)) % MOD
mem[i] = result
return result
MOD = 1000000007
def numberOfWays(num):
mem = [-1] * (len(num) + 1)
return helper(num, 0, mem)
num = "327"
print(numberOfWays(num)) # Output: 2import java.util.*;
public class NumberOfWays {
static long[] mem;
static final int MOD = 1000000007;
public static long helper(String num, int i) {
if (i == num.length()) return 1;
if (num.charAt(i) == '0') return 0;
if (mem[i] != -1) return mem[i];
long result = 0;
for (int j = i + 1; j <= num.length(); j++) {
if (j - i <= i && num.substring(i - (j - i), i).compareTo(num.substring(i, j)) <= 0) {
result = (result + helper(num, j)) % MOD;
}
}
return mem[i] = result;
}
public static int numberOfWays(String num) {
mem = new long[num.length()];
Arrays.fill(mem, -1);
return (int) helper(num, 0);
}
public static void main(String[] args) {
String num = "327";
System.out.println(numberOfWays(num)); // Output: 2
}
}#include <iostream>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
#define MOD 1000000007
vector<long> mem;
long helper(const string &num, int i) {
if (i == num.size()) return 1;
if (num[i] == '0') return 0;
if (mem[i] != -1) return mem[i];
long result = 0;
for (int j = i + 1; j <= num.size(); j++) {
if (j - i <= i && num.substr(i - (j - i), j - i) <= num.substr(i, j - i)) {
result = (result + helper(num, j)) % MOD;
}
}
return mem[i] = result;
}
int numberOfWays(string num) {
mem.assign(num.size(), -1);
return helper(num, 0);
}
int main() {
string num = "327";
cout << numberOfWays(num) << endl; // Output: 2
return 0;
}function helper(num, i, mem) {
if (i === num.length) return 1;
if (num[i] === '0') return 0;
if (mem[i] !== -1) return mem[i];
let result = 0;
for (let j = i + 1; j <= num.length; j++) {
if (j - i <= i && num.slice(i - (j - i), i) <= num.slice(i, j)) {
result = (result + helper(num, j, mem)) % MOD;
}
}
return mem[i] = result;
}
const MOD = 1000000007;
function numberOfWays(num) {
const mem = new Array(num.length + 1).fill(-1);
return helper(num, 0, mem);
}
console.log(numberOfWays('327')); // Output: 2#include <stdio.h>
#include <string.h>
#define MOD 1000000007
long mem[3501];
long helper(char *num, int i, int n) {
if (i == n) return 1;
if (num[i] == '0') return 0;
if (mem[i] != -1) return mem[i];
long result = 0;
char substr[3501];
for (int j = i + 1; j <= n; j++) {
strncpy(substr, &num[i], j - i);
substr[j - i] = '\0';
if (j - i <= i && strncmp(num + i - (j - i), substr, j - i) <= 0) {
result = (result + helper(num, j, n)) % MOD;
}
}
return mem[i] = result;
}
int numberOfWays(char *num) {
memset(mem, -1, sizeof(mem));
int n = strlen(num);
return helper(num, 0, n);
}
int main() {
char num[] = "327";
printf("%ld\n", numberOfWays(num)); // Output: 2
return 0;
}using System;
using System.Collections.Generic;
public class NumberOfWays {
static long[] mem;
const int MOD = 1000000007;
public static long Helper(string num, int i) {
if (i == num.Length) return 1;
if (num[i] == '0') return 0;
if (mem[i] != -1) return mem[i];
long result = 0;
for (int j = i + 1; j <= num.Length; j++) {
if (j - i <= i && string.Compare(num.Substring(i - (j - i), j - i), num.Substring(i, j - i)) <= 0) {
result = (result + Helper(num, j)) % MOD;
}
}
return mem[i] = result;
}
public static int NumberOfWays(string num) {
mem = new long[num.Length];
Array.Fill(mem, -1);
return (int) Helper(num, 0);
}
public static void Main(string[] args) {
string num = "327";
Console.WriteLine(NumberOfWays(num)); // Output: 2
}
}The Python solution implements the recursive approach with a local helper function. The memoization list `mem` is used to store results and ensure efficient computation.
Time Complexity: O(n^2)
Space Complexity: O(n)
Approach 3: 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.
For dp[i][j], the end of the previous number should be i-j. We can enumerate dp[i-j][k], where k \le j. For the part where k < j, i.e., the number of ways with a length less than j can be directly added to dp[i][j], i.e., dp[i][j] = sum_{k=0}^{j-1} dp[i-j][k]. Because the previous number is shorter, it means it is smaller than the current number. Here, prefix sum can be used for optimization.
However, when k = j, we need to compare the sizes of the two numbers of the same length. If the previous number is larger than the current number, this situation is invalid, and we should not add it to dp[i][j]. Otherwise, we can add it to dp[i][j]. Here, we can preprocess the "longest common prefix" in O(n^2) time, and then compare the sizes of two numbers of the same length in O(1) time.
The time complexity is O(n^2), and the space complexity is O(n^2). Where n is the length of the string num.
function numberOfCombinations(num: string): number {
const n: number = num.length;
const mod: number = 1_000_000_007;
const lcp: number[][] = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
const dp: number[][] = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
for (let i = n - 1; i >= 0; i--) {
for (let j = n - 1; j >= 0; j--) {
if (num[i] === num[j]) {
lcp[i][j] = 1 + lcp[i + 1][j + 1];
}
}
}
function cmp(i: number, j: number, k: number): boolean {
const x: number = lcp[i][j];
return x >= k || num[i + x] >= num[j + x];
}
dp[0][0] = 1;
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= i; j++) {
let v: number = 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][Math.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]
}Complexity comparison
| Approach | Complexity |
|---|---|
| Dynamic Programming with String Comparison | Time Complexity: O(n^2) |
| Recursive Approach with Memoization | Time Complexity: O(n^2) |
| Dynamic Programming + Prefix Sum | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Recursive with Memoization | O(n^3) | O(n^2) | Useful for understanding the partition logic and constraints before optimizing. |
| Dynamic Programming with LCP String Comparison | O(n^2) | O(n^2) | Best choice for large inputs; avoids repeated substring comparisons using precomputed LCP. |
Frequently asked questions
Is Number of Ways to Separate Numbers easy or hard?
Number of Ways to Separate Numbers is classified as Hard on LeetCode. The difficulty comes from designing a DP state that enforces non‑decreasing numeric order while comparing substrings efficiently across many partitions.
Number of Ways to Separate Numbers Python/Java solution
Most implementations use a DP table plus an LCP preprocessing step. The algorithm works well in Python, Java, C++, and JavaScript because the complexity is O(n^2). The key is efficient substring comparison rather than language‑specific features.
How to solve Number of Ways to Separate Numbers in O(n^2)?
Use dynamic programming where dp[i][len] represents the number of valid ways ending at index i with a last number of length len. Precompute an LCP table so two substrings can be compared in constant time. This enables quick validation of the non‑decreasing constraint while iterating over candidate lengths.
What is the best approach for Number of Ways to Separate Numbers?
The optimal solution uses dynamic programming combined with fast substring comparison via an LCP (Longest Common Prefix) table. DP tracks how many valid sequences end at each position with a specific number length. LCP allows constant‑time lexicographic comparison between substrings, reducing the complexity to O(n^2) time and O(n^2) space.
Is Number of Ways to Separate Numbers asked at Google/Amazon/Meta?
Hard dynamic programming problems involving string comparison frequently appear in interviews at companies like Google, Amazon, and Meta. This specific problem tests DP state design, substring comparison optimization, and careful handling of constraints such as leading zeros.
What data structure is used in Number of Ways to Separate Numbers?
The core data structure is a dynamic programming table that tracks valid sequences by position and substring length. An auxiliary LCP table (related to suffix array concepts) is used for constant‑time substring comparison between candidate numbers.
What is the time complexity of Number of Ways to Separate Numbers?
The optimized dynamic programming solution runs in O(n^2) time with O(n^2) space. A preprocessing step builds an LCP table for substring comparison, which also costs O(n^2). Naive recursive solutions can reach O(n^3) or worse due to repeated substring comparisons.