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Problem solution · Python

Lexicographically Smallest String After Reverse II

Lexicographically Smallest String After Reverse II: a Python solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
65 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Lexicographically Smallest String After Reverse II, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 65 lines of Python from the credited upstream file lexicographically-smallest-string-after-reverse-ii.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeLexicographically Smallest String After Reverse II · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlogn)# Space: O(n) # rolling hash, binary searchclass Solution(object):    def lexSmallest(self, s):        """        :type s: str        :rtype: str        """        MOD = 10**9+7        B = 29        def binary_search(left, right, check):            while left <= right:                mid = left+(right-left)//2                if check(mid):                    right = mid-1                else:                    left = mid+1            return left         def get_prefix_hash(l, r):            return (prefix[r+1]-prefix[l]*base[r-l+1])%MOD if l <= r else 0         def get_suffix_hash(l, r):            return (suffix[l]-suffix[r+1]*base[r-l+1])%MOD if l <= r else 0                def get_total_hash(k, t, l):            if not t:                return get_suffix_hash(k-l, k-1) if l <= k else ((get_suffix_hash(0, k-1))*base[l-k]+get_prefix_hash(k, l-1))%MOD            nk = len(s)-k            return get_prefix_hash(0, l-1) if l <= nk else ((get_prefix_hash(0, nk-1))*base[l-nk]+get_suffix_hash(len(s)-(l-nk), len(s)-1))%MOD         def get_char(k, t, idx):            if not t:                return s[(k-1)-idx] if idx < k else s[idx]            return s[idx] if idx < len(s)-k else s[(len(s)-1)-(idx-(len(s)-k))]         def is_less(k, i):            idx = binary_search(0, len(s)-1, lambda x: get_total_hash(k, i, x+1) != get_total_hash(best_k, best_i, x+1))            return idx != len(s) and get_char(k, i, idx) < get_char(best_k, best_i, idx)         prefix = [0]*(len(s)+1)        for i in xrange(len(prefix)-1):            prefix[i+1] = (prefix[i]*B+ord(s[i]))%MOD        suffix = [0]*(len(s)+1)        for i in reversed(xrange(len(suffix)-1)):            suffix[i] = (suffix[i+1]*B+ord(s[i]))%MOD        base = [1]*(len(s)+1)        for i in xrange(len(base)-1):            base[i+1] = (base[i]*B)%MOD        best_k, best_i = 1, 0        mn = min(s)        for k in xrange(1, len(s)+1):            if s[k-1] != mn:                continue            if is_less(k, 0):                best_k, best_i = k, 0        for k in xrange(1, len(s)+1):            if not s[-k] >= s[-1]:                continue            if is_less(k, 1):                best_k, best_i = k, 1        return s[:best_k][::-1]+s[best_k:] if not best_i else s[:-best_k]+s[-best_k:][::-1]

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