LeetCode #1977 — HARD

Number of Ways to Separate Numbers

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 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'.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

"327"

Example 2

"094"

Example 3

"0"

Core Insight

What unlocks the optimal approach

If we know the current number has d digits, how many digits can the previous number have?
Is there a quick way of calculating the number of possibilities for the previous number if we know that it must have less than or equal to d digits? Try to do some pre-processing.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1977: Number of Ways to Separate Numbers
class Solution {
    public int numberOfCombinations(String num) {
        final int mod = (int) 1e9 + 7;
        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];
    }
}

// Accepted solution for LeetCode #1977: Number of Ways to Separate Numbers
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]
}

# Accepted solution for LeetCode #1977: Number of Ways to Separate Numbers
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]

// Accepted solution for LeetCode #1977: Number of Ways to Separate Numbers
/**
 * [1977] Number of Ways to Separate Numbers
 *
 * 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.
 *
 *  
 * Constraints:
 *
 * 	1 <= num.length <= 3500
 * 	num consists of digits '0' through '9'.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-ways-to-separate-numbers/
// discuss: https://leetcode.com/problems/number-of-ways-to-separate-numbers/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn number_of_combinations(num: String) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1977_example_1() {
        let num = "327".to_string();

        let result = 2;

        assert_eq!(Solution::number_of_combinations(num), result);
    }

    #[test]
    #[ignore]
    fn test_1977_example_2() {
        let num = "094".to_string();

        let result = 0;

        assert_eq!(Solution::number_of_combinations(num), result);
    }

    #[test]
    #[ignore]
    fn test_1977_example_3() {
        let num = "0".to_string();

        let result = 0;

        assert_eq!(Solution::number_of_combinations(num), result);
    }
}

// Accepted solution for LeetCode #1977: Number of Ways to Separate Numbers
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];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.