Leetcode 1079: Letter Tile Possibilities
You are given a string tiles containing uppercase English letters, where each letter represents a tile. You need to return the number of distinct non-empty sequences of letters that can be formed by selecting tiles from the string. A sequence can be formed by selecting tiles in any order and any number of times, but no tile can be used more than once in a sequence.
Problem
Approach
Steps
Complexity
Input: The input is a string tiles, consisting of uppercase English letters.
Example: Input: tiles = "ABAC"
Constraints:
• 1 <= tiles.length <= 7
• tiles consists of uppercase English letters
Output: The output should be an integer representing the number of distinct non-empty sequences that can be formed from the tiles.
Example: Output: 12
Constraints:
• The number of possible distinct sequences should be computed.
Goal: To find the number of distinct non-empty sequences that can be formed from the given string of tiles.
Steps:
• 1. Generate all possible non-empty subsequences of the input string.
• 2. Use a set to keep track of distinct sequences.
• 3. Ensure no tile is used more than once in any sequence.
• 4. Return the size of the set as the result.
Goal: The solution must handle strings of varying lengths, and ensure no tile is used more than once in any subsequence.
Steps:
• 1 <= tiles.length <= 7
• tiles consists of uppercase English letters
Assumptions:
• Each letter in the input string tiles can be used in any sequence, but no letter can be used more than once.
• Input: Input: tiles = "ABAC"
• Explanation: The distinct sequences that can be formed are: "A", "B", "C", "AB", "AC", "BA", "BC", "AA", "ABAC", "ACA", "BCA", and "AAB". So, the output is 12.
• Input: Input: tiles = "ABC"
• Explanation: The distinct sequences that can be formed are: "A", "B", "C", "AB", "AC", "BC", and "ABC", resulting in 7 distinct sequences.
Approach: To solve this problem, we can use a backtracking approach to generate all possible subsequences of the given string and count the distinct ones. We use a set to ensure all subsequences are unique.
Observations:
• We can treat this as a combination problem where each letter can either be included or excluded from the subsequence.
• A set can be used to automatically handle duplicate subsequences.
• The solution will involve recursively generating subsequences and using a set to avoid counting duplicates.
Steps:
• 1. Create a set to store unique subsequences.
• 2. Define a helper function that will generate subsequences using backtracking.
• 3. For each letter in the string, either include it in the current subsequence or exclude it, recursively.
• 4. Add each generated subsequence to the set.
• 5. Return the size of the set as the number of distinct subsequences.
Empty Inputs:
• If the input string is empty, return 0, as no sequences can be formed.
Large Inputs:
• The input string will not exceed 7 characters, so the solution should handle small inputs efficiently.
Special Values:
• If the input string consists of only one letter, the output should be 1, as there is only one possible sequence.
Constraints:
• The solution should work for inputs with repeated letters and handle small input sizes efficiently.
int numTilePossibilities(string tiles) {
set<string> ans;
set<int> seen;
gen(tiles, 0, "", ans, seen);
return ans.size();
}
void gen(string tiles, int idx, string res, set<string> &ans, set<int> &vis ) {
if(idx == tiles.size()) return;
for(int i = 0; i < tiles.size(); i++) {
if(!vis.count(i)) {
vis.insert(i);
ans.insert(res + tiles[i]);
gen(tiles, idx + 1, res + tiles[i], ans, vis);
vis.erase(i);
}
}
}
1 : Function Definition
int numTilePossibilities(string tiles) {
This line defines the function `numTilePossibilities`, which takes a string `tiles` as input and returns an integer representing the number of distinct tile sequences that can be formed.
2 : Variable Initialization
set<string> ans;
This initializes the set `ans`, which will store all distinct tile sequences generated.
3 : Set Initialization
set<int> seen;
This initializes the set `seen`, which will keep track of which tiles have been used during the sequence generation to avoid duplicates.
4 : Call to gen Function
gen(tiles, 0, "", ans, seen);
This calls the recursive function `gen`, passing the initial state of the variables: the tiles string, an index starting from 0, an empty string for the current sequence, the `ans` set to store results, and the `seen` set to track used tiles.
5 : Return Statement
return ans.size();
This returns the size of the set `ans`, which contains all distinct sequences generated.
6 : Recursive Function Definition
void gen(string tiles, int idx, string res, set<string> &ans, set<int> &vis ) {
This line defines the recursive function `gen`, which takes the current state of the tiles, the index, the current sequence, the set of results `ans`, and the set of visited tiles `vis`.
7 : Base Case
if(idx == tiles.size()) return;
This is the base case of the recursion. If the index `idx` reaches the size of `tiles`, the function returns without doing anything further.
8 : Loop Over Tiles
for(int i = 0; i < tiles.size(); i++) {
This loop iterates over each tile in the `tiles` string.
9 : Check if Tile is Used
if(!vis.count(i)) {
This checks if the tile at index `i` has already been used in the current sequence by checking the `vis` set.
10 : Mark Tile as Used
vis.insert(i);
This marks the tile at index `i` as used by inserting it into the `vis` set.
11 : Add Sequence to Result Set
ans.insert(res + tiles[i]);
This adds the new sequence, formed by appending the current tile to the current result string `res`, to the `ans` set.
12 : Recursive Call
gen(tiles, idx + 1, res + tiles[i], ans, vis);
This makes the recursive call to generate further sequences, incrementing the index `idx` and appending the current tile to the sequence.
13 : Unmark Tile as Used
vis.erase(i);
This unmarks the tile at index `i` as used by removing it from the `vis` set, allowing it to be reused in other branches of the recursion.
Best Case: O(n!)
Average Case: O(n!)
Worst Case: O(n!)
Description: The time complexity is dominated by the need to generate all possible subsequences, where n is the length of the input string.
Best Case: O(1)
Worst Case: O(2^n)
Description: The space complexity is O(2^n) due to the set storing all distinct subsequences. The worst case occurs when all subsequences are distinct.
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