Algorithm Notes
Summary: Permutation In String — notes not yet curated.
Time: Estimate via loops/recurrences; common classes: O(1), O(log n), O(n), O(n log n), O(n^2)
Space: Count auxiliary structures and recursion depth.
Tip: See the Big-O Guide for how to derive bounds and compare trade-offs.
Big-O Guide
Source
"""
Permutation In String
TODO: Add problem description
"""
from collections import Counter
from src.interview_workbook.leetcode._registry import register_problem
from src.interview_workbook.leetcode._types import Category, Difficulty
class Solution:
def solve(self, *args):
"""Return True if any permutation of s1 is a substring of s2."""
s1, s2 = args
if len(s1) > len(s2):
return False
s1_count = Counter(s1)
window = Counter()
l = 0
for r, ch in enumerate(s2):
window[ch] += 1
if r - l + 1 > len(s1):
window[s2[l]] -= 1
if window[s2[l]] == 0:
del window[s2[l]]
l += 1
if window == s1_count:
return True
return False
def demo():
"""Run a simple demonstration for Permutation In String."""
s1 = "ab"
s2 = "eidbaooo"
result = Solution().solve(s1, s2)
return f"Input: s1={s1}, s2={s2} -> Contains permutation? {result}"
register_problem(
id=567,
slug="permutation_in_string",
title="Permutation In String",
category=Category.SLIDING_WINDOW,
difficulty=Difficulty.MEDIUM,
tags=["hash table", "two pointers", "string", "sliding window"],
url="https://leetcode.com/problems/permutation-in-string/",
notes="Sliding window with character counts compared to target counts.",
)