- Translate each rule into one explicit state update.
- Maintain the invariant after every processed item.
- Return the accumulated state once all relevant input has been handled.
Code notes
- 33 lines of Python from the credited upstream file 2800.py.
- The implementation visibly relies on sequence storage.
- No explicit loop blocks detected.
Complexity
Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.
Check the problem constraints before deciding whether this complexity will pass.
Use this to learn the idea, then write your own version.
1class Solution:2 def minimumString(self, a: str, b: str, c: str) -> str:3 def merge(a: str, b: str) -> str:4 """Merges a and b."""5 if a in b: 6 return b7 for i in range(len(a)):8 aSuffix = a[i:]9 bPrefix = b[:len(aSuffix)]10 if aSuffix == bPrefix:11 return a + b[len(bPrefix):]12 return a + b13 14 abc = merge(a, merge(b, c))15 acb = merge(a, merge(c, b))16 bac = merge(b, merge(a, c))17 bca = merge(b, merge(c, a))18 cab = merge(c, merge(a, b))19 cba = merge(c, merge(b, a))20 return self._getMin([abc, acb, bac, bca, cab, cba])21 22 def _getMin(self, words: list[str]) -> str:23 """Returns the lexicographically smallest string."""24 25 def getMin(a: str, b: str) -> str:26 """Returns the lexicographically smaller string."""27 return a if len(a) < len(b) or (len(a) == len(b) and a < b) else b28 29 res = words[0]30 for i in range(1, len(words)):31 res = getMin(res, words[i])32 return res33