This was an interview question I was asked. My answer is below, but I wonder if there is a more efficient way. Possible a combinatorics type function, or is an exhaustive generation + count the only way?

Rules: Given a character, the next character can be any one of How many N character combinations can you make? For example

  • a -> i,e

  • e -> i,o

  • i -> a,e,o

  • u -> a,e,i,o

  • o -> i,u,o

For N = 1 possible string combinations are a , e, i ,o ,u for N = 2 possible string combinations are [ai, ae], [ei ,eo] ,[ia , ie , io],[ua,ue,ui,uo],[oi,ou,oo] and so on. I solved it by constructing a tree and then counting the leaves

struct Node {
    Node(char v) : v_(v) {}
    void                     add_child(shared_ptr<Node> n) { children.emplace_back(n); }
    char                     v_ = 0;
    vector<shared_ptr<Node>> children;

void add_child(shared_ptr<Node> root, int n, int depth, const map<char, vector<char>> &combo) {
    if (depth == n) return;
    auto ch = root->v_;
    auto it = combo.find(ch);
    for (auto ch : it->second) {
        //printf("Adding to %c to %c\n", ch, root->v_);
    for (auto &child : root->children) { add_child(child, n, depth, combo); }

int count_leaf(shared_ptr<Node> root) {
    if (root->children.size() == 0) return 0;
    int sum = 0;
    for (auto &child : root->children) sum += count_leaf(child);
    return sum > 0 ? sum : root->children.size();
void password_combination() {   
    // Find combos until this depth
    int n = 3;

    map<char, vector<char>> combo;
    combo.insert(make_pair('a', vector<char>{'i', 'e'}));
    combo.insert(make_pair('e', vector<char>{'i', 'o'}));
    combo.insert(make_pair('i', vector<char>{'a', 'e', 'o'}));
    combo.insert(make_pair('u', vector<char>{'a', 'e', 'i', 'o'}));
    combo.insert(make_pair('o', vector<char>{'i', 'u', 'o'}));

    // Create a N depth tree and count leaves
    // Count all children until n = 3;
    auto root  = make_shared<Node>(0);
    int  depth = 0;
    for (auto &c : combo) { root->add_child(make_shared<Node>(c.first)); }
    for (auto child : root->children) { add_child(child, n, depth, combo); }
    printf("%d\n", count_leaf(root));

Is there a better way. Critique on the code/bugs, and if there's a c++14 idiomatic way of doing it, even better.

  • \$\begingroup\$ Do you know yet if your solution passed? \$\endgroup\$
    – yuri
    Jul 11, 2017 at 14:19
  • \$\begingroup\$ @yuri Not yet, but so far, not found any obvious bugs \$\endgroup\$
    – Delta_Fore
    Jul 11, 2017 at 14:21

1 Answer 1


TL;DR: dynamic programming.

If the alphabet has \$k\$ symbols (5 in the example), in the worst case there are \$k^N\$ possible strings and generating them is \$\Omega(k^N)\$. But it's possible to count them in \$O(k^2 N)\$ time and \$O(k^2 + N)\$ space using a simple dynamic programming approach.

The input gives the successor function: \$a\$ must be followed by one of \$\{i,e\}\$. The first step is to reverse that to get the predecessor function: \$a\$ must be preceded by one of \$\{i,u\}\$.

Then the number of strings of length \$l\$ which end in \$a\$ is \$1\$ if \$l=1\$, or the sum of the number of strings of length \$l-1\$ which end in \$i\$ or \$u\$ otherwise.

Actually, if \$N\$ is large enough relative to \$k\$ you can do even better with a more complicated approach. Treat it as a Markov chain and find a matrix power of the \$0-1\$ matrix which indicates the successor function. That takes \$O(\lg N)\$ matrix multiplications of a \$k\times k\$ matrix.

  • \$\begingroup\$ We can actually use the successor-function directly. \$\endgroup\$ Jul 11, 2017 at 17:42

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