CPMagical CP Library

Problem solution · C++

Count Ways to Build Rooms in an Ant Colony

Count Ways to Build Rooms in an Ant Colony: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first 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

Depth-first search

For Count Ways to Build Rooms in an Ant Colony, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 65 lines of C++ from the credited upstream file count-ways-to-build-rooms-in-an-ant-colony.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 Ways to Build Rooms in an Ant Colony · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n)// Space: O(n) class Solution {public:    int waysToBuildRooms(vector<int>& prevRoom) {        vector<vector<int>> adj(size(prevRoom));        for (int i = 1; i < size(prevRoom); ++i) {            adj[prevRoom[i]].emplace_back(i);        }        return dfs(adj, 0).first;    } private:    pair<int, int> dfs(const vector<vector<int>>& adj, int curr) {        static const int MOD = 1e9 + 7;        int total_ways = 1, total_cnt = 0;        for (const auto& child : adj[curr]) {            const auto& [ways, cnt] = dfs(adj, child);            total_cnt += cnt;            total_ways = mulmod(mulmod(total_ways, ways, MOD), nCr(total_cnt, cnt, MOD), MOD);        }        return {total_ways, total_cnt + 1};    }     int nCr(int n, int k, uint32_t mod) {        while (size(inv_) <= n) {  // lazy initialization            fact_.emplace_back(mulmod(fact_.back(), size(inv_), mod));            inv_.emplace_back(mulmod(inv_[mod % size(inv_)], mod - mod / size(inv_), mod));  // https://cp-algorithms.com/algebra/module-inverse.html            inv_fact_.emplace_back(mulmod(inv_fact_.back(), inv_.back(), mod));        }        return mulmod(mulmod(fact_[n], inv_fact_[n - k], mod), inv_fact_[k], mod);    }     uint32_t addmod(uint32_t a, uint32_t b, uint32_t mod) {  // avoid overflow        a %= mod, b %= mod;        if (mod - a <= b) {            b -= mod;  // relied on unsigned integer overflow in order to give the expected results        }        return a + b;    }     // reference: https://stackoverflow.com/questions/12168348/ways-to-do-modulo-multiplication-with-primitive-types    uint32_t mulmod(uint32_t a, uint32_t b, uint32_t mod)  {  // avoid overflow        a %= mod, b %= mod;        uint32_t result = 0;        if (a < b) {            swap(a, b);        }        while (b > 0)  {             if (b % 2 == 1) {                result = addmod(result, a, mod);            }            a = addmod(a, a, mod);            b /= 2;         }         return result;     }     static const int MOD = 1e9 + 7;    vector<int> fact_ = {1, 1};    vector<int> inv_ = {0, 1};    vector<int> inv_fact_ = {1, 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 ↗