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Pyramid Transition Matrix

Difficulty: Medium

Problem Description

You are stacking blocks to form a pyramid. Each block has a color, which is represented by a single letter. Each row of blocks contains one less block than the row beneath it and is centered on top. To make the pyramid aesthetically pleasing, there are only specific triangular patterns that are allowed. A triangular pattern consists of a single block stacked on top of two blocks. The patterns are given as a list of three-letter strings allowed, where the first two characters of a pattern represent the left and right bottom blocks respectively, and the third character is the top block. You start with a bottom row of blocks bottom, given as a single string, that you must use as the base of the pyramid. Given bottom and allowed, return true if you can build the pyramid all the way to the top such that every triangular pattern in the pyramid is in allowed, or false otherwise.

Key Insights

  • The pyramid is constructed from a bottom row where each subsequent row has one less block.
  • Each block can be formed by two blocks below it according to defined allowed patterns.
  • A depth-first search (DFS) or backtracking approach can be utilized to explore potential pyramid configurations.
  • The problem can be represented as a graph where each state is a possible configuration of blocks at a given level.

Space and Time Complexity

Time Complexity: O(3^N), where N is the length of the bottom string, because for each pair of blocks, there can be up to 3 possible blocks above them. Space Complexity: O(N), for the recursion stack in the DFS approach.

Solution

To solve this problem, we can use a depth-first search (DFS) approach to explore all possible configurations for building the pyramid. We start with the base layer and recursively build upwards by checking the allowed triangular patterns. We use a set to store allowed patterns for quick lookup. At each level, we generate possible combinations of blocks for the row above based on the current row, and continue until we either successfully build the pyramid to the top or exhaust all possibilities.

Code Solutions

def pyramidTransition(bottom: str, allowed: List[str]) -> bool:
    from collections import defaultdict
    
    # Create a mapping for allowed transitions
    allowed_map = defaultdict(set)
    for pattern in allowed:
        allowed_map[pattern[:2]].add(pattern[2])
    
    # Function to perform DFS
    def canBuild(current_row):
        if len(current_row) == 1:  # Reached the top
            return True
        
        next_row = []
        for i in range(len(current_row) - 1):
            block_pair = current_row[i:i+2]
            if block_pair in allowed_map:
                for top_block in allowed_map[block_pair]:
                    next_row.append(top_block)
            else:
                return False
        
        return canBuild(next_row)  # Recur with the next row
    
    return canBuild(bottom)  # Start with the base row
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