Leetcode #1573: Number of Ways to Split a String
In this guide, we solve Leetcode #1573 Number of Ways to Split a String 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
Given a binary string s, you can split s into 3 non-empty strings s1, s2, and s3 where s1 + s2 + s3 = s. Return the number of ways s can be split such that the number of ones is the same in s1, s2, and s3.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Math, String
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: s = "10101"
Output: 4
Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'.
"1|010|1"
"1|01|01"
"10|10|1"
"10|1|01"
Python Solution
class Solution:
def numWays(self, s: str) -> int:
def find(x):
t = 0
for i, c in enumerate(s):
t += int(c == '1')
if t == x:
return i
cnt, m = divmod(sum(c == '1' for c in s), 3)
if m:
return 0
n = len(s)
mod = 10**9 + 7
if cnt == 0:
return ((n - 1) * (n - 2) // 2) % mod
i1, i2 = find(cnt), find(cnt + 1)
j1, j2 = find(cnt * 2), find(cnt * 2 + 1)
return (i2 - i1) * (j2 - j1) % (10**9 + 7)
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.