What is the time complexity of Pyramid Transition Matrix?

The optimal solution has a time complexity of O(A^(N^2)).

What is the space complexity of Pyramid Transition Matrix?

The space complexity is O(N^2).

What data structures are used to solve Pyramid Transition Matrix?

This problem uses Bit Manipulation, Depth-First Search, Breadth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Pyramid Transition Matrix?

Insight 1: The problem is essentially a constrained search problem where each level of the pyramid depends on the level below it. Insight 2: Memoization is crucial to avoid redundant calculations as many intermediate rows might lead to the same subproblems.

Which companies ask LeetCode 756. Pyramid Transition Matrix in interviews?

This problem is frequently asked at Airbnb.

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Pyramid Transition Matrix - LeetCode 756 Solution

Pyramid Transition Matrix - Complete Solution Guide

Pyramid Transition Matrix is LeetCode problem 756, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 repre

Detailed Explanation

The problem asks whether a pyramid can be built from a given bottom row of blocks and a set of allowed triangular patterns. Each block has a color (A, B, C, D, E, F), and the allowed patterns define which color block can be placed on top of which two color blocks below. The goal is to determine if it's possible to construct a valid pyramid all the way to the top (a single block) following the allowed pattern rules.

Solution Approach

The provided solutions use a recursive Depth-First Search (DFS) approach with memoization. The `solve` function recursively checks if a given row can be built upon to form a valid pyramid. It generates the next possible row above the current row and then recursively calls itself with the new row. The `find_next_level` function generates all possible next rows based on allowed transitions from the given `bottom` level to the top level of a single block. A hashmap is used to store allowed transitions for quick lookup. Memoization prevents redundant calculations by storing the results of previously computed rows.

Step-by-Step Algorithm

Step 1: Create a data structure (hash map) to store the allowed transitions, mapping pairs of bottom blocks to possible top blocks.
Step 2: Implement a recursive function `solve(row)` that takes a row of blocks as input.
Step 3: Base Case: If the `row` has only one block, the pyramid is complete, return `true`.
Step 4: Check Memoization: If the result for the `row` is already stored in the memo, return the stored result.
Step 5: Build the next level row using a helper function `find_next_level` to generate all possible rows.
Step 6: Inside `find_next_level`, iterate through all possible combinations of top blocks allowed by the allowed transitions.
Step 7: For each generated `next_row`, recursively call `solve(next_row)`. If `solve(next_row)` returns `true`, the pyramid can be completed, so return `true`.
Step 8: If no `next_row` leads to a complete pyramid, return `false` and store this result in the memo.
Step 9: The main function calls `solve(bottom)` to start the recursion from the given bottom row.

Key Insights

Insight 1: The problem is essentially a constrained search problem where each level of the pyramid depends on the level below it.
Insight 2: Memoization is crucial to avoid redundant calculations as many intermediate rows might lead to the same subproblems.
Insight 3: Representing allowed transitions as a dictionary/hashmap (two blocks below -> possible block above) significantly speeds up pattern lookup.
Insight 4: Due to the constraints, recursion depth is limited, making a recursive approach with memoization feasible.

Complexity Analysis

Time Complexity: O(A^(N^2))

Space Complexity: O(N^2)

Topics

This problem involves: Bit Manipulation, Depth-First Search, Breadth-First Search.

Companies

Asked at: Airbnb.

Pyramid Transition Matrix | LeetCode 756

Medium

Bit Manipulation Depth-First Search Breadth-First Search

Time: O(A^(N^2))Space: O(N^2)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;

class Solution {
    public boolean pyramidTransition(String bottom, List<String> allowed) {
        Map<String, Set<Character>> transitions = new HashMap<>();
        for (String triplet : allowed) {
            String key = triplet.substring(0, 2);
            char value = triplet.charAt(2);
            if (!transitions.containsKey(key)) {
                transitions.put(key, new HashSet<>());
            }
            transitions.get(key).add(value);
        }

        Map<String, Boolean> memo = new HashMap<>();

        return solve(bottom, transitions, memo);
    }

    private boolean solve(String row, Map<String, Set<Character>> transitions, Map<String, Boolean> memo) {
        if (row.length() == 1) {
            return true;
        }
        if (memo.containsKey(row)) {
            return memo.get(row);
        }

        boolean result = findNextLevel(0, "", row, transitions, memo);
        memo.put(row, result);
        return result;
    }

    private boolean findNextLevel(int index, String nextRowBuild, String row, Map<String, Set<Character>> transitions, Map<String, Boolean> memo) {
        if (index == row.length() - 1) {
            return solve(nextRowBuild, transitions, memo);
        }

        String pair = row.substring(index, index + 2);
        if (!transitions.containsKey(pair)) {
            return false;
        }

        Set<Character> possibleChars = transitions.get(pair);
        for (char c : possibleChars) {
            if (findNextLevel(index + 1, nextRowBuild + c, row, transitions, memo)) {
                return true;
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 756: Pyramid Transition Matrix - Detailed explanation and solution approaches.

Topics

Bit Manipulation Depth-First Search Breadth-First Search

Companies

Airbnb

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:756

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Create a data structure (hash map) to store the allowed transitions, mapping pairs of bottom blocks to possible top blocks.
2
Step 2: Implement a recursive function `solve(row)` that takes a row of blocks as input.
3
Step 3: Base Case: If the `row` has only one block, the pyramid is complete, return `true`.
4
Step 4: Check Memoization: If the result for the `row` is already stored in the memo, return the stored result.
5
Step 5: Build the next level row using a helper function `find_next_level` to generate all possible rows.
6
Step 6: Inside `find_next_level`, iterate through all possible combinations of top blocks allowed by the allowed transitions.
7
Step 7: For each generated `next_row`, recursively call `solve(next_row)`. If `solve(next_row)` returns `true`, the pyramid can be completed, so return `true`.
8
Step 8: If no `next_row` leads to a complete pyramid, return `false` and store this result in the memo.
9
Step 9: The main function calls `solve(bottom)` to start the recursion from the given bottom row.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(A^(N^2))

💾

Space Complexity

O(N^2)

📝

Why?

The time complexity is exponential because, at each level, we try multiple possibilities. The space complexity comes from storing intermediate row results in the memoization table to avoid recomputation. The memoization table stores at most N different row sizes, each of max length N, hence O(N^2).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Breadth-First Search (BFS)

Instead of recursion, BFS can be used to explore possible pyramid constructions level by level. Start with the bottom row and generate all possible rows above it, adding them to a queue. Continue until either a single-block top is reached or the queue is empty (no solution).

Time: O(A^(N^2)) - Same as DFS in the worst case.Space: O(A^(N^2)) - Can be higher than DFS due to storing all possible rows at a given level in the queue.

Trade-offs: BFS might be preferable if the solution is expected to be closer to the bottom, but it generally requires more memory compared to DFS with memoization. DFS is likely more efficient due to memoization avoiding redundant computations.

✨

Key Insights

1Insight 1: The problem is essentially a constrained search problem where each level of the pyramid depends on the level below it.
2Insight 2: Memoization is crucial to avoid redundant calculations as many intermediate rows might lead to the same subproblems.
3Insight 3: Representing allowed transitions as a dictionary/hashmap (two blocks below -> possible block above) significantly speeds up pattern lookup.
4Insight 4: Due to the constraints, recursion depth is limited, making a recursive approach with memoization feasible.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `bottom` contains characters not present in `allowed`. In such cases, no pyramid can be formed. The solutions handle this by checking if the pairs exist in the transitions map, returning `false` if not.
Edge case 2: `allowed` is empty. In this scenario, no pyramid can be built unless the base case conditions are met (single block). The solutions correctly handle this by either returning false if there are no entries in transitions HashMap or only returning true if the bottom level already contains a single block.
Edge case 3: The input `bottom` can range in size from 2 to 6. The solution correctly handles different sizes of `bottom` through the recursive structure.
Edge case 4: A single character that has no transition to the character above would immediately cause the function to return `false`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization. Without memoization, the code will have exponential time complexity and will time out for larger inputs.

✗Common mistake 2: Incorrectly building the `transitions` map. Ensure the correct mapping between bottom pairs and possible top blocks.

✗Common mistake 3: Not handling the edge cases where no transitions are possible, or `allowed` is empty, or the input characters are invalid.

✗Common mistake 4: Incorrectly copying the intermediate rows, leading to incorrect results or crashes.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This type of pattern matching and constraint satisfaction problem is relevant in compiler design, where syntax must conform to grammar rules (the `allowed` set).

🏦Real-world scenario 2: Similar algorithms can be applied in automated planning systems, where actions (transitions) are constrained by preconditions, and the goal is to reach a desired state (the top of the pyramid).

🎮Real-world scenario 3: Constraint satisfaction problems frequently occur in scheduling problems. This pyramid problem mimics finding a valid schedule when certain resources can only be used one after another.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

XOR is your friend for "find the single number" problems. Remember: n & (n-1) clears the lowest set bit.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Pyramid Transition Matrix"?

For this medium-level Bit Manipulation problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?