LeetCode #756 — MEDIUM

Pyramid Transition Matrix

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

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 represent the left and right bottom blocks respectively, and the third character is the top block.

For example, "ABC" represents a triangular pattern with a 'C' block stacked on top of an 'A' (left) and 'B' (right) block. Note that this is different from "BAC" where 'B' is on the left bottom and 'A' is on the right bottom.

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.

Example 1:

Input: bottom = "BCD", allowed = ["BCC","CDE","CEA","FFF"]
Output: true
Explanation: The allowed triangular patterns are shown on the right.
Starting from the bottom (level 3), we can build "CE" on level 2 and then build "A" on level 1.
There are three triangular patterns in the pyramid, which are "BCC", "CDE", and "CEA". All are allowed.

Example 2:

Input: bottom = "AAAA", allowed = ["AAB","AAC","BCD","BBE","DEF"]
Output: false
Explanation: The allowed triangular patterns are shown on the right.
Starting from the bottom (level 4), there are multiple ways to build level 3, but trying all the possibilites, you will get always stuck before building level 1.

Constraints:

2 <= bottom.length <= 6
0 <= allowed.length <= 216
allowed[i].length == 3
The letters in all input strings are from the set {'A', 'B', 'C', 'D', 'E', 'F'}.
All the values of allowed are unique.

Step 01

Brute Force Baseline

Problem summary: 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. For example, "ABC" represents a triangular pattern with a 'C' block stacked on top of an 'A' (left) and 'B' (right) block. Note that this is different from "BAC" where 'B' is on the left bottom and 'A' is on the right bottom. You start with a bottom row of blocks bottom, given as a single string, that you

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Backtracking · Bit Manipulation

Example 1

"BCD"
["BCC","CDE","CEA","FFF"]

Example 2

"AAAA"
["AAB","AAC","BCD","BBE","DEF"]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #756: Pyramid Transition Matrix
class Solution {
    private final int[][] d = new int[7][7];
    private final Map<String, Boolean> f = new HashMap<>();

    public boolean pyramidTransition(String bottom, List<String> allowed) {
        for (String s : allowed) {
            int a = s.charAt(0) - 'A', b = s.charAt(1) - 'A';
            d[a][b] |= 1 << (s.charAt(2) - 'A');
        }
        return dfs(bottom, new StringBuilder());
    }

    private boolean dfs(String s, StringBuilder t) {
        if (s.length() == 1) {
            return true;
        }
        if (t.length() + 1 == s.length()) {
            return dfs(t.toString(), new StringBuilder());
        }
        String k = s + "." + t.toString();
        Boolean res = f.get(k);
        if (res != null) {
            return res;
        }
        int a = s.charAt(t.length()) - 'A', b = s.charAt(t.length() + 1) - 'A';
        int cs = d[a][b];
        for (int i = 0; i < 7; ++i) {
            if (((cs >> i) & 1) == 1) {
                t.append((char) ('A' + i));
                if (dfs(s, t)) {
                    f.put(k, true);
                    return true;
                }
                t.setLength(t.length() - 1);
            }
        }
        f.put(k, false);
        return false;
    }
}

// Accepted solution for LeetCode #756: Pyramid Transition Matrix
func pyramidTransition(bottom string, allowed []string) bool {
	d := make([][]int, 7)
	for i := 0; i < 7; i++ {
		d[i] = make([]int, 7)
	}

	for _, s := range allowed {
		a := int(s[0] - 'A')
		b := int(s[1] - 'A')
		c := int(s[2] - 'A')
		d[a][b] |= 1 << c
	}

	f := make(map[string]bool)

	var dfs func(s string, t []byte) bool
	dfs = func(s string, t []byte) bool {
		if len(s) == 1 {
			return true
		}
		if len(t)+1 == len(s) {
			return dfs(string(t), []byte{})
		}

		key := s + "." + string(t)
		if v, ok := f[key]; ok {
			return v
		}

		i := len(t)
		a := int(s[i] - 'A')
		b := int(s[i+1] - 'A')
		cs := d[a][b]

		for c := 0; c < 7; c++ {
			if (cs>>c)&1 == 1 {
				t = append(t, byte('A'+c))
				if dfs(s, t) {
					f[key] = true
					return true
				}
				t = t[:len(t)-1]
			}
		}

		f[key] = false
		return false
	}

	return dfs(bottom, []byte{})
}

# Accepted solution for LeetCode #756: Pyramid Transition Matrix
class Solution:
    def pyramidTransition(self, bottom: str, allowed: List[str]) -> bool:
        @cache
        def dfs(s: str) -> bool:
            if len(s) == 1:
                return True
            t = []
            for a, b in pairwise(s):
                cs = d[a, b]
                if not cs:
                    return False
                t.append(cs)
            return any(dfs("".join(nxt)) for nxt in product(*t))

        d = defaultdict(list)
        for a, b, c in allowed:
            d[a, b].append(c)
        return dfs(bottom)

// Accepted solution for LeetCode #756: Pyramid Transition Matrix
use std::collections::HashMap;

impl Solution {
    pub fn pyramid_transition(bottom: String, allowed: Vec<String>) -> bool {
        let mut d = vec![vec![0; 7]; 7];
        for s in allowed {
            let a = (s.as_bytes()[0] - b'A') as usize;
            let b = (s.as_bytes()[1] - b'A') as usize;
            let c = (s.as_bytes()[2] - b'A') as usize;
            d[a][b] |= 1 << c;
        }

        let mut f = HashMap::<String, bool>::new();

        fn dfs(s: &str, t: &mut Vec<u8>, d: &Vec<Vec<i32>>, f: &mut HashMap<String, bool>) -> bool {
            if s.len() == 1 {
                return true;
            }

            if t.len() + 1 == s.len() {
                let next = String::from_utf8_lossy(t).to_string();
                let mut nt = Vec::new();
                return dfs(&next, &mut nt, d, f);
            }

            let key = format!("{}.{}", s, String::from_utf8_lossy(t));
            if let Some(&res) = f.get(&key) {
                return res;
            }

            let i = t.len();
            let a = (s.as_bytes()[i] - b'A') as usize;
            let b = (s.as_bytes()[i + 1] - b'A') as usize;
            let mut cs = d[a][b];

            for c in 0..7 {
                if (cs >> c) & 1 == 1 {
                    t.push(b'A' + c as u8);
                    if dfs(s, t, d, f) {
                        f.insert(key.clone(), true);
                        t.pop();
                        return true;
                    }
                    t.pop();
                }
            }

            f.insert(key, false);
            false
        }

        let mut t = Vec::new();
        dfs(&bottom, &mut t, &d, &mut f)
    }
}

// Accepted solution for LeetCode #756: Pyramid Transition Matrix
function pyramidTransition(bottom: string, allowed: string[]): boolean {
    const d: number[][] = Array.from({ length: 7 }, () => Array(7).fill(0));
    for (const s of allowed) {
        const a = s.charCodeAt(0) - 65;
        const b = s.charCodeAt(1) - 65;
        const c = s.charCodeAt(2) - 65;
        d[a][b] |= 1 << c;
    }

    const f = new Map<string, boolean>();

    const dfs = (s: string, t: string[]): boolean => {
        if (s.length === 1) return true;
        if (t.length + 1 === s.length) {
            return dfs(t.join(''), []);
        }

        const key = s + '.' + t.join('');
        if (f.has(key)) return f.get(key)!;

        const i = t.length;
        const a = s.charCodeAt(i) - 65;
        const b = s.charCodeAt(i + 1) - 65;
        let cs = d[a][b];

        for (let c = 0; c < 7; c++) {
            if ((cs >> c) & 1) {
                t.push(String.fromCharCode(65 + c));
                if (dfs(s, t)) {
                    f.set(key, true);
                    return true;
                }
                t.pop();
            }
        }

        f.set(key, false);
        return false;
    };

    return dfs(bottom, []);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.