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Problem solution · C++

Count Fancy Numbers in a Range

Count Fancy Numbers in a Range: a C++ solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Breadth-first search
Source
Kamyu LeetCode Solutions
Length
116 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

Breadth-first search

For Count Fancy Numbers in a Range, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 116 lines of C++ from the credited upstream file count-fancy-numbers-in-a-range.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 16 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeCount Fancy Numbers in a Range · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(dlogd + g + d^2 + g * d) = O(g * d + d^2), d = logr, g = len(good)// Space: O(g + d) // bfs, dp, principle of inclusion and exclusionclass Solution {public:    long long countFancy(long long l, long long r) {        const auto& count = [&](int64_t x) {            const auto& total = [](int64_t n) { // Time: O(logn)                int result = 0;                for (; n; n /= 10) {                    result += n % 10;                }                return result;            };             const auto& length = [](int64_t n) { // Time: O(logn)                int result = 0;                for (; n; n /= 10) {                    ++result;                }                return result;            };             const auto& check = [](int n) { // Time: O(logn)                bool asc = true, desc = true;                for (; n >= 10; n /= 10) {                    if (!((n / 10) % 10 < n % 10)) {                        asc = false;                    }                    if (!((n / 10) % 10 > n % 10)) {                        desc = false;                    }                }                return asc || desc;            };             const auto& bfs = [](int64_t x) {  // Time: O(g), g = len(result)                vector<int64_t> result;                for (int i = 1; i <= min(static_cast<int64_t>(9), x); ++i) {                    result.emplace_back(i);                }                for (const auto& diff : {1, -1}) {                    vector<int64_t> q;                    for (int i = 1; i <= min(static_cast<int64_t>(9), x); ++i) {                        q.emplace_back(i);                    }                    while (!empty(q)) {                        vector<int64_t> new_q;                        for (const auto& u : q) {                            const auto& curr = u % 10;                            for (int d = curr + diff; 0 <= d && d <= 9; d += diff) {                                const auto& v = u * 10 + d;                                if (v <= x) {                                    new_q.emplace_back(v);                                    result.emplace_back(v);                                }                            }                        }                        q = move(new_q);                    }                }                return result;            };             vector<bool> lookup(length(x) * 9 + 1);            const auto& dp = [&](int64_t x) {  // Time: O(d^2), d = logr                const auto& l = length(x);                const auto& mx = l * 9;                vector<vector<int64_t>> dp(2, vector<int64_t>(mx + 1));  // dp[tight][sum]                dp[1][0] = 1;                int64_t base = pow(10LL, l - 1);                for (int i = 0; i < l; ++i) {                    vector<vector<int64_t>> new_dp(2, vector<int64_t>(mx + 1));                    const auto& v = (x / base) % 10;                    base /= 10;                    for (int t = 0; t < 2; ++t) {                        for (int s = 0; s <= mx; ++s) {                            if (dp[t][s] == 0) {                                continue;                            }                            const auto& bound = (t == 1) ? v : 9;                            for (int d = 0; d <= bound; ++d) {                                new_dp[t == 1 && d == v][s + d] += dp[t][s];                            }                        }                    }                    dp = move(new_dp);                };                 int64_t result = 0;                for (int i = 0; i <= mx; ++i) {                    if (lookup[i]) {                        result += dp[0][i] + dp[1][i];                    }                }                return result;            };                for (int i = 0; i < size(lookup); ++i) {                lookup[i] = check(i);            }            const auto& good = bfs(x);            int64_t cnt = 0;            for (const auto& x : good) {                if (lookup[total(x)]) {                    ++cnt;                }            }            return size(good) + dp(x) - cnt;        };         return count(r) - count(l - 1);    }};

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