CPMagical CP Library

Problem solution · C++

Total Waviness of Numbers in Range II

Total Waviness of Numbers in Range II: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Dynamic programming
Source
Kamyu LeetCode Solutions
Length
244 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

Dynamic programming

For Total Waviness of Numbers in Range II, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 244 lines of C++ from the credited upstream file total-waviness-of-numbers-in-range-ii.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, cached states.
  • 22 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeTotal Waviness of Numbers in Range II · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(logn * 11 * 11 * 2 * 2 * 10)// Space: O(logn * 11 * 11 * 2 * 2) // memoization by listclass Solution {public:    long long totalWaviness(long long num1, long long num2) {        const auto& count = [&](long long x) {            auto encode = [&](int i, int prev, int prev2, bool zero, bool tight) {                long long key = i;                key = key * (10 + 1) + (prev + 1);                key = key * (10 + 1) + (prev2 + 1);                key = key * 2 + (zero ? 1 : 0);                key = key * 2 + (tight ? 1 : 0);                return key;            };             const auto& s = to_string(x);            vector<pair<long long, long long>> lookup(size(s) * (10 + 1) * (10 + 1) * 2 * 2, {-1, -1});            const auto dp = [&](this auto&& dp, int i, int prev, int prev2, bool zero, bool tight) -> pair<long long, long long> {                if (i == size(s)) {                    return {1, 0};                }                long long key = encode(i, prev, prev2, zero, tight);                if (lookup[key].first == -1) {                    long long cnt = 0, w = 0;                    const auto& mx = tight ? s[i] - '0' : 9;                    for (int d = 0; d <= mx; ++d) {                        const auto& new_tight = tight && (d == s[i] - '0');                        const auto& new_zero = zero && (d == 0);                        const auto& new_prev2 = prev;                        const auto& new_prev = !new_zero ? d : -1;                        const auto& [new_cnt, nw] = dp(i + 1, new_prev, new_prev2, new_zero, new_tight);                        cnt += new_cnt;                        if (!zero && prev2 != -1 && ((prev2 < prev && prev > d) || (prev2 > prev && prev < d))) {                            w += new_cnt;                        }                        w += nw;                    }                    lookup[key] = {cnt, w};                }                return lookup[key];            };             return dp(0, -1, -1, true, true).second;        };         return count(num2) - count(num1 - 1);    }}; // Time:  O(logn * 11 * 11 * 2 * 2 * 10)// Space: O(11 * 11 * 2 * 2)// dp by listclass Solution2 {public:    long long totalWaviness(long long num1, long long num2) {        const auto& count = [&](long long x) {            auto encode = [&](int prev, int prev2, int zero, int tight) {                long long key = prev + 1;                key = key * (10 + 1) + (prev2 + 1);                key = key * 2 + zero;                key = key * 2 + tight;                return key;            };             const auto& s = to_string(x);            const int state_size = (10 + 1) * (10 + 1) * 2 * 2;            vector<pair<long long, long long>> dp(state_size, {-1, -1});            vector<pair<long long, long long>> new_dp(state_size, {-1, -1});            for (int prev = -1; prev <= 9; ++prev) {                for (int prev2 = -1; prev2 <= 9; ++prev2) {                    for (int zero = 0; zero <= 1; ++zero) {                        for (int tight = 0; tight <= 1; ++tight) {                            const auto& key = encode(prev, prev2, zero, tight);                            dp[key] = {1, 0};                        }                    }                }            }            for (int i = size(s) - 1; i >= 0; --i) {                fill(begin(new_dp), end(new_dp), make_pair(-1LL, -1LL));                for (int prev = -1; prev <= 9; ++prev) {                    for (int prev2 = -1; prev2 <= 9; ++prev2) {                        for (int zero = 0; zero <= 1; ++zero) {                            for (int tight = 0; tight <= 1; ++tight) {                                long long cnt = 0, w = 0;                                const auto& mx = tight ? s[i] - '0' : 9;                                for (int d = 0; d <= mx; ++d) {                                    const auto& new_tight = tight && (d == s[i] - '0');                                    const auto& new_zero = zero && (d == 0);                                    const auto& new_prev2 = prev;                                    const auto& new_prev = !new_zero ? d : -1;                                    const auto& key = encode(new_prev, new_prev2, new_zero, new_tight);                                    if (dp[key].first != -1) {                                        const auto& [new_cnt, nw] = dp[key];                                        cnt += new_cnt;                                        if (!zero && prev2 != -1 && ((prev2 < prev && prev > d) || (prev2 > prev && prev < d))) {                                            w += new_cnt;                                        }                                        w += nw;                                    }                                }                                const auto& key = encode(prev, prev2, zero, tight);                                new_dp[key] = {cnt, w};                            }                        }                    }                }                swap(dp, new_dp);            }             return dp[encode(-1, -1, true, true)].second;        };         return count(num2) - count(num1 - 1);    }}; // Time:  O(logn * 11 * 11 * 2 * 2 * 10)// Space: O(logn * 11 * 11 * 2 * 2)// memoization by dictclass Solution3 {private:    struct TupleHash {        template <typename... T>        std::size_t operator()(const std::tuple<T...>& t) const {            return apply([](const auto&... args) {                std::size_t seed = 0;                ((seed ^= std::hash<std::decay_t<decltype(args)>>{}(args) +                        0x9e3779b9 + (seed << 6) + (seed >> 2)), ...);                return seed;            }, t);        }    }; public:    long long totalWaviness(long long num1, long long num2) {        auto count = [&](long long x) {            const auto& s = to_string(x);            using State = tuple<int, int, int, bool, bool>;            unordered_map<State, pair<long long, long long>, TupleHash> lookup;            const auto dp = [&](this auto&& dp, int i, int prev, int prev2, bool zero, bool tight) -> pair<long long, long long> {                if (i == size(s)) {                    return {1, 0};                }                State key = {i, prev, prev2, zero, tight};                if (!lookup.count(key)) {                    long long cnt = 0, w = 0;                    const auto& mx = tight ? (s[i] - '0') : 9;                    for (int d = 0; d <= mx; ++d) {                        const auto& new_tight = tight && (d == mx);                        const auto& new_zero = zero && (d == 0);                        const auto& new_prev2 = prev;                        const auto& new_prev = !new_zero ? d : -1;                        const auto& [new_cnt, nw] = dp(i + 1, new_prev, new_prev2, new_zero, new_tight);                        cnt += new_cnt;                        if (!zero && prev2 != -1 && ((prev2 < prev && prev > d) || (prev2 > prev && prev < d))) {                            w += new_cnt;                        }                        w += nw;                    }                    lookup[key] = {cnt, w};                }                return lookup[key];            };             return dp(0, -1, -1, true, true).second;        };         return count(num2) - count(num1 - 1);    }}; // Time:  O(logn * 11 * 11 * 2 * 2 * 10)// Space: O(11 * 11 * 2 * 2)// dp by dictclass Solution4 {private:    struct TupleHash {        template <typename... T>        std::size_t operator()(const std::tuple<T...>& t) const {            return apply([](const auto&... args) {                std::size_t seed = 0;                ((seed ^= std::hash<std::decay_t<decltype(args)>>{}(args) +                        0x9e3779b9 + (seed << 6) + (seed >> 2)), ...);                return seed;            }, t);        }    }; public:    long long totalWaviness(long long num1, long long num2) {        const auto& count = [&](long long x) {            const auto& s = to_string(x);            using State = tuple<int, int, int, int>;            unordered_map<State, pair<long long, long long>, TupleHash> dp, new_dp;            for (int prev = -1; prev <= 9; ++prev) {                for (int prev2 = -1; prev2 <= 9; ++prev2) {                    for (int zero = 0; zero <= 1; ++zero) {                        for (int tight = 0; tight <= 1; ++tight) {                            dp[{prev, prev2, zero, tight}] = {1, 0};                        }                    }                }            }            for (int i = size(s) - 1; i >= 0; --i) {                new_dp.clear();                for (int prev = -1; prev <= 9; ++prev) {                    for (int prev2 = -1; prev2 <= 9; ++prev2) {                        for (int zero = 0; zero <= 1; ++zero) {                            for (int tight = 0; tight <= 1; ++tight) {                                long long cnt = 0, w = 0;                                const auto& mx = tight ? s[i] - '0' : 9;                                for (int d = 0; d <= mx; ++d) {                                    const auto& new_tight = tight && (d == s[i] - '0');                                    const auto& new_zero = zero && (d == 0);                                    const auto& new_prev2 = prev;                                    const auto& new_prev = !new_zero ? d : -1;                                    State key = {new_prev, new_prev2, new_zero, new_tight};                                    if (dp.count(key)) {                                        const auto& [new_cnt, nw] = dp[key];                                        cnt += new_cnt;                                        if (!zero && prev2 != -1 && ((prev2 < prev && prev > d) || (prev2 > prev && prev < d))) {                                            w += new_cnt;                                        }                                        w += nw;                                    }                                }                                new_dp[{prev, prev2, zero, tight}] = {cnt, w};                            }                        }                    }                }                swap(dp, new_dp);            }             return dp[{-1, -1, true, true}].second;        };         return count(num2) - count(num1 - 1);    }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗