Leetcode #1977: Number of Ways to Separate Numbers
In this guide, we solve Leetcode #1977 Number of Ways to Separate Numbers in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Dynamic Programming, Suffix Array
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: num = "327"
Output: 2
Explanation: You could have written down the numbers:
3, 27
327
Python Solution
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]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.