Leetcode #2767: Partition String Into Minimum Beautiful Substrings
In this guide, we solve Leetcode #2767 Partition String Into Minimum Beautiful Substrings 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, partition the string into one or more substrings such that each substring is beautiful. A string is beautiful if: It doesn't contain leading zeros.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Hash Table, String, Dynamic Programming, Backtracking
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: s = "1011"
Output: 2
Explanation: We can paritition the given string into ["101", "1"].
- The string "101" does not contain leading zeros and is the binary representation of integer 51 = 5.
- The string "1" does not contain leading zeros and is the binary representation of integer 50 = 1.
It can be shown that 2 is the minimum number of beautiful substrings that s can be partitioned into.
Python Solution
class Solution:
def minimumBeautifulSubstrings(self, s: str) -> int:
def dfs(i: int) -> int:
if i >= n:
return 0
if s[i] == "0":
return inf
x = 0
ans = inf
for j in range(i, n):
x = x << 1 | int(s[j])
if x in ss:
ans = min(ans, 1 + dfs(j + 1))
return ans
n = len(s)
x = 1
ss = {x}
for i in range(n):
x *= 5
ss.add(x)
ans = dfs(0)
return -1 if ans == inf else ans
Complexity
The time complexity is O(n). The space complexity is O(n).
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.