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Problem solution · C++

Lexicographically Smallest String After Reverse

Lexicographically Smallest String After Reverse: a C++ 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
130 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, 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

  • 130 lines of C++ from the credited upstream file lexicographically-smallest-string-after-reverse.cpp.
  • The implementation visibly relies on sequence storage.
  • 8 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 · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(nlogn)// Space: O(n) // rolling hash, binary searchclass Solution {public:    string lexSmallest(string s) {        static const int64_t MOD = 1e9 + 7;        static const int64_t B = 29;        const int n = size(s);         const auto& binary_search = [](auto left, auto right, const auto& check) {            while (left <= right) {                const auto mid = left + (right - left) / 2;                if (check(mid)) {                    right = mid - 1;                } else {                    left = mid + 1;                }            }            return left;        };         vector<int64_t> prefix(n + 1);        for (int i = 0; i + 1 < size(prefix); ++i) {            prefix[i + 1] = (prefix[i] * B + s[i]) % MOD;        }        vector<int64_t> suffix(n + 1);        for (int i = size(suffix) - 2; i >= 0; --i) {            suffix[i] = (suffix[i + 1] * B + s[i]) % MOD;        }        vector<int64_t> base(n + 1, 1);        for (int i = 0; i + 1 < size(base); ++i) {            base[i + 1] = (base[i] * B) % MOD;        }         const auto& get_prefix_hash = [&](int l, int r) {            if (l > r) {                return static_cast<int64_t>(0);            }            return (prefix[r + 1] - prefix[l] * base[r - l + 1] % MOD + MOD) % MOD;        };         const auto& get_suffix_hash = [&](int l, int r) {            if (l > r) {                return static_cast<int64_t>(0);            }            return (suffix[l] - suffix[r + 1] * base[r - l + 1] % MOD + MOD) % MOD;        };         const auto& get_total_hash = [&](int k, int t, int l) {            if (!t) {                return l <= k                    ? get_suffix_hash(k - l, k - 1)                    : ((get_suffix_hash(0, k - 1) * base[l - k]) % MOD + get_prefix_hash(k, l - 1)) % MOD;            }            const auto& nk = n - k;            return l <= nk                ? get_prefix_hash(0, l - 1)                : ((get_prefix_hash(0, nk - 1) * base[l - nk]) % MOD + get_suffix_hash(n - (l - nk), n - 1)) % MOD;        };         const auto& get_char = [&](int k, int t, int idx) {            if (!t) {                return idx < k                    ? s[(k - 1) - idx]                    : s[idx];            }            return idx < n - k                ? s[idx]                : s[(n - 1) - (idx - (n - k))];        };            int best_k = 1, best_i = 0;        auto is_less = [&](int k, int i) {            const auto& idx = binary_search(0, n - 1, [&](int x) {                return get_total_hash(k, i, x + 1) != get_total_hash(best_k, best_i, x + 1);            });            return idx != n && get_char(k, i, idx) < get_char(best_k, best_i, idx);        };        const auto& mn = ranges::min(s);        for (int k = 1; k <= n; ++k) {            if (s[k - 1] != mn) {                continue;            }            if (is_less(k, 0)) {                best_k = k;                best_i = 0;            }        }        for (int k = 1; k <= n; ++k) {            if (!(s[size(s) - k] >= s.back())) {                continue;            }            if (is_less(k, 1)) {                best_k = k;                best_i = 1;            }        }        string result(s);        if (!best_i) {            reverse(begin(result), begin(result) + best_k);        } else {            reverse(begin(result) + (size(result) - best_k), end(result));        }        return result;    }}; // Time:  O(n^2)// Space: O(1)// brute forceclass Solution2 {public:    string lexSmallest(string s) {        string result = s;        for (int k = 2; k <= size(s); ++k) {            string t = s;            reverse(begin(t), begin(t) + k);            result = min(result, t);        }        for (int k = 2; k <= size(s); ++k) {            string t = s;            reverse(begin(t) + (size(t) - k), end(t));            result = min(result, t);        }        return result;    }};

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