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Problem solution · C++

Sum of Perfect Square Ancestors

Sum of Perfect Square Ancestors: 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
121 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 Sum of Perfect Square Ancestors, 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

  • 121 lines of C++ from the credited upstream file sum-of-perfect-square-ancestors.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 8 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 codeSum of Perfect Square Ancestors · C++C++
Use this to learn the idea, then write your own version.
// Time:  precompute: O(r)//        runtime:    O(nlogx)// Space: O(r + n) vector<int> linear_sieve_of_eratosthenes(int n) {  // Time: O(n), Space: O(n)    vector<int> spf(n + 1, -1);    vector<int> primes;    for (int i = 2; i <= n; ++i) {        if (spf[i] == -1) {            spf[i] = i;            primes.emplace_back(i);        }        for (const auto& p : primes) {            if (i * p > n || p > spf[i]) {                break;            }            spf[i * p] = p;        }    }    return spf;} const int MAX_NUMS = 1e5;const auto& SPF = linear_sieve_of_eratosthenes(MAX_NUMS); // number theory, iterative dfs, freq tableclass Solution {public:    long long sumOfAncestors(int n, vector<vector<int>>& edges, vector<int>& nums) {        const auto& prime_factors = [](int x) {            int result = 1;            for (; x != 1; x /= SPF[x]) {                if (result % SPF[x] == 0) {                    result /= SPF[x];                } else {                    result *= SPF[x];                }            }            return result;        };         vector<vector<int>> adj(n);        for (const auto& e : edges) {            adj[e[0]].emplace_back(e[1]);            adj[e[1]].emplace_back(e[0]);        }        unordered_map<int, int> cnt;        const auto& iter_dfs = [&]() {            int64_t result = 0;            using P = tuple<int, int, int, int, int>;            vector<P> stk = {{1, 0, -1, -1, 0}};            while (!empty(stk)) {                const auto [step, u, p, i, x] = stk.back(); stk.pop_back();                if (step == 1) {                    const auto& x = prime_factors(nums[u]);                    result += cnt[x]++;                    stk.emplace_back(3, -1, -1, -1, x);                    stk.emplace_back(2, u, p, 0, 0);                } else if (step == 2) {                    if (i == size(adj[u])) {                        continue;                    }                    stk.emplace_back(2, u, p, i + 1, 0);                    const auto& v = adj[u][i];                    if (v == p) {                        continue;                    }                    stk.emplace_back(1, v, u, -1, 0);                } else if (step == 3) {                    --cnt[x];                }            }            return result;        };         return iter_dfs();    }}; // Time:  precompute: O(r)//        runtime:    O(nlogx)// Space: O(r + n)// number theory, dfs, freq tableclass Solution2 {public:    long long sumOfAncestors(int n, vector<vector<int>>& edges, vector<int>& nums) {        const auto& prime_factors = [](int x) {            int result = 1;            for (; x != 1; x /= SPF[x]) {                if (result % SPF[x] == 0) {                    result /= SPF[x];                } else {                    result *= SPF[x];                }            }            return result;        };         vector<vector<int>> adj(n);        for (const auto& e : edges) {            adj[e[0]].emplace_back(e[1]);            adj[e[1]].emplace_back(e[0]);        }        unordered_map<int, int> cnt;        const function<int64_t (int, int)> dfs = [&](int u, int p) {            const auto& x = prime_factors(nums[u]);            int64_t result = cnt[x]++;            for (const auto& v : adj[u]) {                if (v == p) {                    continue;                }                result += dfs(v, u);            }            --cnt[x];            return result;        };         return dfs(0, -1);    }};

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