Goal: The goal is to find all possible letter combinations corresponding to the given digits.
Steps:
• Use a map to associate each digit with its corresponding letters.
• Iterate over the digits and for each digit, expand the existing combinations by appending each letter mapped to that digit.
Goal: The constraints ensure that the length of the input string is manageable and the digits are within the specified range.
Steps:
• 0 <= digits.length <= 4
• digits[i] is a digit in the range ['2', '9']
Assumptions:
• Each digit corresponds to a set of characters as per the old mobile keypad mappings.
• The input string could be empty, in which case the output should be an empty list.
• Input: digits = "45"
• Explanation: The digit 4 maps to 'g', 'h', 'i' and the digit 5 maps to 'j', 'k', 'l'. The output combinations are formed by combining each letter from the first digit with each letter from the second digit.
Approach: The approach uses a map to represent the letter mappings for each digit, and iteratively constructs combinations using a loop or recursive function.
Observations:
• The problem is about generating combinations, which can be done using a backtracking approach or an iterative approach.
• We can initialize the result with the letters corresponding to the first digit and progressively build the combinations for each subsequent digit.
Steps:
• Initialize a map that associates each digit (2-9) with its corresponding letters.
• Start with the combinations corresponding to the first digit.
• Iteratively expand the combinations by appending each letter of the next digit to the existing combinations.
Empty Inputs:
• An empty string should return an empty list.
Large Inputs:
• Input strings of length 4 are the largest possible, and the solution should handle that case efficiently.
Special Values:
• Digits corresponding to the same set of letters (e.g., '2' or '3') should be handled as usual.
Constraints:
• The input length is constrained to 0 to 4 digits, ensuring the problem can be solved efficiently.
Declare the `letterCombinations` function, which takes a string of digits `digits` as input and returns a vector of strings representing all possible letter combinations.
2 : Map Initialization
map<char, vector<string>> mp;
Initialize a map `mp` to store the mapping of digits to their corresponding letters.
3 : Map Initialization
mp['2'] = { "a", "b", "c" };
Map the digit '2' to the letters 'a', 'b', and 'c'.
4 : Map Initialization
mp['3'] = { "d", "e", "f" };
Map the digit '3' to the letters 'd', 'e', and 'f'.
5 : Map Initialization
mp['4'] = { "g", "h", "i" };
Map the digit '4' to the letters 'g', 'h', and 'i'.
6 : Map Initialization
mp['5'] = { "j", "k", "l" };
Map the digit '5' to the letters 'j', 'k', and 'l'.
7 : Map Initialization
mp['6'] = { "m", "n", "o" };
Map the digit '6' to the letters 'm', 'n', and 'o'.
8 : Map Initialization
mp['7'] = { "p", "q", "r", "s" };
Map the digit '7' to the letters 'p', 'q', 'r', and 's'.
9 : Map Initialization
mp['8'] = { "t", "u", "v" };
Map the digit '8' to the letters 't', 'u', and 'v'.
10 : Map Initialization
mp['9'] = { "w", "x", "y", "z" };
Map the digit '9' to the letters 'w', 'x', 'y', and 'z'.
11 : Variable Initialization
vector<string> ans, tmp;
Initialize two empty vectors `ans` and `tmp` to store the letter combinations.
12 : Array Access
ans = mp[digits[0]];
Initialize `ans` with the letter combinations for the first digit.
13 : Loop Iteration
for(int i =1; i < digits.size(); i++) {
Start a loop to iterate through the digits from the second digit to the last digit.
14 : Variable Assignment
tmp = ans;
Copy the current letter combinations in `ans` to `tmp`.
15 : Variable Initialization
ans = {};
Clear the `ans` vector to store the new combinations.
16 : Nested Loops
for(int j =0; j < mp[digits[i]].size(); j++) {
Iterate through the letters corresponding to the current digit.
17 : Nested Loops
for(int k =0; k < tmp.size(); k++)
Iterate through the existing letter combinations in `tmp`.
18 : String Manipulation
ans.push_back(tmp[k] + mp[digits[i]][j]);
Append the current letter from `mp[digits[i]]` to each combination in `tmp` and add the new combination to `ans`.
19 : Return Value
return ans;
Return the final vector `ans` containing all possible letter combinations.
Best Case: O(1)
Average Case: O(3^n)
Worst Case: O(4^n)
Description: In the worst case, for 4 digits, we need to compute up to 4^4 combinations.
Best Case: O(1)
Worst Case: O(4^n)
Description: The space complexity is determined by the number of combinations generated.
