Letter Combinations Of A Phone Number Leetcode Problem 17 [Python]

Welcome back, everyone!

Today, we're going to tackle the Letter Combinations Of A Phone Number problem.

This is a LeetCode problem with the ID 17, categorized under "Backtracking." It's an interesting problem that involves mapping digits to letters, just like we used to do on old mobile phones.

Let's dive in and explore how to solve it efficiently.

Problem Overview

In this problem, we're given a string containing digits from 2 to 9 (inclusive).

Our task is to return all possible letter combinations that these digits could represent.

The order of the output doesn't matter.

Here's the mapping of digits to letters, just like on a telephone keypad:

• 2: "abc"
• 3: "def"
• 4: "ghi"
• 5: "jkl"
• 6: "mno"
• 7: "pqrs"
• 8: "tuv"
• 9: "wxyz"

Please note that the digit 1 doesn't map to any letters.

Let's look at a few examples to better understand the problem:

Example 1:

Input: digits = "23"
Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]

Example 2:

Input: digits = ""
Output: []

Example 3:

Input: digits = "2"
Output: ["a","b","c"]

Constraints

Before we proceed to the solution, it's important to understand the constraints of the problem:

• The input string can have a length ranging from 0 to 4.
• Each character in the input string is a digit in the range ['2', '9'].

Now that we have a clear picture of the problem, let's explore the efficient Python solution for it.

Efficient Python Code Solution

Here's a Python solution to the Letter Combinations Of A Phone Number problem:

def letterCombinations(self, digits: str) -> List[str]:
    res = []
    digitToChar = {
        "2": "abc",
        "3": "def",
        "4": "ghi",
        "5": "jkl",
        "6": "mno",
        "7": "pqrs",
        "8": "tuv",
        "9": "wxyz",
    }

    def backtrack(i, curStr):
        if len(curStr) == len(digits):
            res.append(curStr)
            return
        for c in digitToChar[digits[i]]:
            backtrack(i + 1, curStr + c)

    if digits:
        backtrack(0, "")

    return res

Let's break down this solution step by step.

1. Backtracking Approach

To solve this problem efficiently, we can use a backtracking approach.

Backtracking is a recursive technique that involves exploring all possible solutions and building them step by step.

In our solution, we define a backtracking function called backtrack.

This function takes two parameters:

• i: An index representing the current digit we are processing.
• curStr: The current string we are building.

2. Base Case

The base case for our backtracking function is as follows:

if len(curStr) == len(digits):
    res.append(curStr)
    return

This base case checks whether the length of curStr is equal to the length of the input digits.

If they are equal, it means we have successfully built a valid combination, so we add curStr to our result list res.

3. Recursive Exploration

The core of the backtracking process is the loop that explores the possibilities:

for c in digitToChar[digits[i]]:
    backtrack(i + 1, curStr + c)

Here, we iterate through the characters that the current digit digits[i] maps to.

For each character c, we recursively call the backtrack function with an updated index i and an updated curStr by adding the character c to it.

This exploration continues until we reach the base case, and valid combinations are added to the result.

4. Initialize and Call Backtrack

Before we start the backtracking process, we initialize our result list res and create a dictionary digitToChar to map digits to their corresponding characters.

Finally, we check if the digits input is non-empty, and if it is, we call the backtrack function with an initial index of 0 and an empty curStr.

Time and Space Complexity

Now, let's analyze the time and space complexity of our solution.

• Time Complexity: The time complexity of this solution is O(4^N), where N is the length of the input string digits.

This is because, in the worst case, each digit can map to four characters (e.g., '7' maps to "pqrs").

As a result, we have to explore all possible combinations, leading to an exponential time complexity.

• Space Complexity: The space complexity is O(N) because the depth of the recursion is limited by the length of the input string.

The result list res and the digitToChar dictionary consume additional space, but they are not included in the space complexity analysis because their sizes do not depend on the input.

Reasoning Behind Our Approach

We have adopted a backtracking approach to solve this problem efficiently.

The reason behind using backtracking is that it allows us to systematically explore all possible combinations of letters that the given digits could represent.

By doing so, we can construct the output in a way that ensures we cover all valid combinations.

The backtracking process begins with an empty string and the first digit in the input.

We then iterate through the characters that the digit maps to and recursively explore all possible combinations.

When the length of our constructed string matches the length of the input digits, we add it to our result list.

This process continues for each digit in the input, creating a systematic way to generate all valid combinations.

While the time complexity of our solution is exponential, it's the most efficient way to solve this problem, as we need to consider all possible combinations.

Backtracking ensures that no valid combination is left unexplored.

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Conclusion

In this blog post, we've tackled the Letter Combinations Of A Phone Number problem, which is a medium difficulty problem on LeetCode.

We've provided a Python solution that uses a backtracking approach to efficiently generate all possible letter combinations for a given string of digits.

By understanding the problem's constraints and the backtracking approach, we can systematically explore and construct all valid combinations.

Although the time complexity is exponential, this is the most efficient solution for this problem.

Feel free to experiment with this solution and try it out on LeetCode.

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If you have any questions or suggestions, don't hesitate to comment below and share this content with others who might find it useful.

Question Link

Happy coding!