LeetCode #1079 — MEDIUM

Letter Tile Possibilities

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

The Problem

Problem Statement

You have n tiles, where each tile has one letter tiles[i] printed on it.

Return the number of possible non-empty sequences of letters you can make using the letters printed on those tiles.

Example 1:

Input: tiles = "AAB"
Output: 8
Explanation: The possible sequences are "A", "B", "AA", "AB", "BA", "AAB", "ABA", "BAA".

Example 2:

Input: tiles = "AAABBC"
Output: 188

Example 3:

Input: tiles = "V"
Output: 1

Constraints:

1 <= tiles.length <= 7
tiles consists of uppercase English letters.

Step 01

Brute Force Baseline

Problem summary: You have n tiles, where each tile has one letter tiles[i] printed on it. Return the number of possible non-empty sequences of letters you can make using the letters printed on those tiles.

Baseline thinking

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

Pattern signal: Hash Map · Backtracking

Example 1

"AAB"

Example 2

"AAABBC"

Example 3

"V"

Step 02

Core Insight

What unlocks the optimal approach

Try to build the string with a backtracking DFS by considering what you can put in every position.

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 #1079: Letter Tile Possibilities
class Solution {
    public int numTilePossibilities(String tiles) {
        int[] cnt = new int[26];
        for (char c : tiles.toCharArray()) {
            ++cnt[c - 'A'];
        }
        return dfs(cnt);
    }

    private int dfs(int[] cnt) {
        int res = 0;
        for (int i = 0; i < cnt.length; ++i) {
            if (cnt[i] > 0) {
                ++res;
                --cnt[i];
                res += dfs(cnt);
                ++cnt[i];
            }
        }
        return res;
    }
}

// Accepted solution for LeetCode #1079: Letter Tile Possibilities
func numTilePossibilities(tiles string) int {
	cnt := [26]int{}
	for _, c := range tiles {
		cnt[c-'A']++
	}
	var dfs func(cnt [26]int) int
	dfs = func(cnt [26]int) (res int) {
		for i, x := range cnt {
			if x > 0 {
				res++
				cnt[i]--
				res += dfs(cnt)
				cnt[i]++
			}
		}
		return
	}
	return dfs(cnt)
}

# Accepted solution for LeetCode #1079: Letter Tile Possibilities
class Solution:
    def numTilePossibilities(self, tiles: str) -> int:
        def dfs(cnt: Counter) -> int:
            ans = 0
            for i, x in cnt.items():
                if x > 0:
                    ans += 1
                    cnt[i] -= 1
                    ans += dfs(cnt)
                    cnt[i] += 1
            return ans

        cnt = Counter(tiles)
        return dfs(cnt)

// Accepted solution for LeetCode #1079: Letter Tile Possibilities
struct Solution;

impl Solution {
    fn num_tile_possibilities(tiles: String) -> i32 {
        let mut counts: Vec<usize> = vec![0; 26];
        for c in tiles.chars() {
            counts[(c as u8 - b'A') as usize] += 1;
        }
        let mut res = 0;
        Self::dfs(&mut res, &mut counts);
        res
    }

    fn dfs(sum: &mut i32, counts: &mut Vec<usize>) {
        for i in 0..26 {
            if counts[i] > 0 {
                *sum += 1;
                counts[i] -= 1;
                Self::dfs(sum, counts);
                counts[i] += 1;
            }
        }
    }
}

#[test]
fn test() {
    let tiles = "AAB".to_string();
    let res = 8;
    assert_eq!(Solution::num_tile_possibilities(tiles), res);
    let tiles = "AAABBC".to_string();
    let res = 188;
    assert_eq!(Solution::num_tile_possibilities(tiles), res);
}

// Accepted solution for LeetCode #1079: Letter Tile Possibilities
function numTilePossibilities(tiles: string): number {
    const cnt: number[] = new Array(26).fill(0);
    for (const c of tiles) {
        ++cnt[c.charCodeAt(0) - 'A'.charCodeAt(0)];
    }
    const dfs = (cnt: number[]): number => {
        let res = 0;
        for (let i = 0; i < 26; ++i) {
            if (cnt[i] > 0) {
                ++res;
                --cnt[i];
                res += dfs(cnt);
                ++cnt[i];
            }
        }
        return res;
    };
    return dfs(cnt);
}

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.