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Leetcode 756. Pyramid Transition Matrix

Problem Statement

The problem is known as “756. Pyramid Transition Matrix” on LeetCode.

We are given a bottom string and a list of transition rules in the format “XYZ” where “XY” is the base and “Z” is the resulting character. Our task is to determine if we can build a pyramid to the top using these rules.

Example:

The rules mean:

We need to determine if we can build a pyramid where bottom is the base and we can place characters on top according to the allowed transition rules.

Clarifying Questions

  1. Can we assume all characters in bottom and allowed are uppercase letters?
    • Yes, all characters will be uppercase.
  2. What is the maximum length of bottom and the number of transitions in allowed?
    • bottom.length() can be up to 8.
    • allowed.length() can be up to 200.
  3. Is there any guarantee that the transitions will not contain invalid strings?
    • Yes, all transitions are valid and follow the format described.
  4. What should be the return type?
    • Return true if we can build the pyramid to the top using the given transitions, otherwise, return false.

Strategy

We can use a Depth First Search (DFS) approach with memoization to solve this problem. We will start from the bottom and recursively try to construct the top of the pyramid using the allowed transitions. Memoization will help avoid recalculating the same subproblems.

  1. Parse the allowed transitions into a more accessible structure, such as a dictionary.
  2. Define a recursive function to check if we can build the pyramid.
  3. Use memoization to store results for previously computed states to avoid redundant calculations.

Code

Here is a possible implementation of the solution in C++:

#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <string>
#include <functional>

using namespace std;

class Solution {
public:
    bool pyramidTransition(string bottom, vector<string>& allowed) {
        // Build the dictionary for transitions
        unordered_map<string, unordered_set<char>> transitions;
        for (const string& triplet : allowed) {
            transitions[triplet.substr(0, 2)].insert(triplet[2]);
        }

        // Memoization map: Cache previously computed results
        unordered_map<string, bool> memo;

        // Function for recursion and DFS
        function<bool(string)> dfs = [&](string bottom) {
            // If the bottom is of length 1, we reached the top
            if (bottom.size() == 1) return true;
            // If we have seen this bottom before return the cached result
            if (memo.find(bottom) != memo.end()) return memo[bottom];

            // Trying to build the next level
            for (int i = 0, n = bottom.size(); i < n - 1; ++i) {
                string base = bottom.substr(i, 2);
                if (transitions.find(base) == transitions.end()) return memo[bottom] = false;
            }

            // To store all combinations of possible next levels
            vector<string> nextLevelCandidates;
            generateNextLevel(bottom, "", 0, nextLevelCandidates, transitions);
            for (const string& nextLevel : nextLevelCandidates) {
                if (dfs(nextLevel)) return memo[bottom] = true;
            }

            return memo[bottom] = false;
        };

        return dfs(bottom);
    }

private:
    void generateNextLevel(const string& bottom, const string& current, int pos, vector<string>& nextLevelCandidates,
                           const unordered_map<string, unordered_set<char>>& transitions) {
        if (pos == bottom.size() - 1) {
            nextLevelCandidates.push_back(current);
            return;
        }
        string base = bottom.substr(pos, 2);
        const auto& possibleChars = transitions.at(base);
        for (char ch : possibleChars) {
            generateNextLevel(bottom, current + ch, pos + 1, nextLevelCandidates, transitions);
        }
    }
};

Time Complexity

The time complexity of this solution is difficult to bound tightly, but it can be exponential in the worst case due to the combination of states. Practical performance will depend on the branching factor of each level and the effectiveness of memoization in pruning the search space.

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