Problem Description

Alice is texting Bob using a phone keypad where digits map to letters (like old phone keyboards):

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

To type a letter, Alice must press the corresponding digit key multiple times based on the letter's position. For example:

• To type 's', press '7' four times (since 's' is the 4th letter on key 7)
• To type 'k', press '5' twice (since 'k' is the 2nd letter on key 5)

When Alice sends a message, Bob doesn't receive the actual text but instead receives the sequence of key presses. For instance, if Alice sends "bob", Bob receives "2266622" (where 2 is pressed twice for 'b', 666 for 'o', and 22 for 'b').

Given a string pressedKeys representing what Bob received, determine how many different text messages Alice could have originally sent. Since the answer can be very large, return it modulo 10^9 + 7.

The challenge is that the same sequence of key presses can represent different messages. For example, "222" could mean:

• "aaa" (pressing 2 once, three times)
• "ab" (pressing 2 once for 'a', then twice for 'b')
• "ba" (pressing 2 twice for 'b', then once for 'a')
• "c" (pressing 2 three times for 'c')

The solution uses dynamic programming to count possibilities. For consecutive identical digits, it calculates how many ways they can be interpreted:

• For digits 2, 3, 4, 5, 6, 8 (which map to 3 letters each), a sequence can be broken into groups of 1, 2, or 3 presses
• For digits 7 and 9 (which map to 4 letters each), a sequence can be broken into groups of 1, 2, 3, or 4 presses

The arrays f and g precompute the number of ways to interpret sequences of length i for 3-letter keys and 4-letter keys respectively, using the recurrence relations:

• f[i] = f[i-1] + f[i-2] + f[i-3] for 3-letter keys
• g[i] = g[i-1] + g[i-2] + g[i-3] + g[i-4] for 4-letter keys

The solution groups consecutive identical digits and multiplies the number of ways for each group to get the final answer.

Intuition

The key insight is recognizing that consecutive identical digits create ambiguity in interpretation. When we see "222", we need to figure out all valid ways to split it into letters.

Let's think about this step by step. When we have a sequence of the same digit, we can choose to interpret it in different ways based on how many presses form each letter. For digit '2' (which maps to 'a', 'b', 'c'):

• 1 press → 'a'
• 2 presses → 'b'
• 3 presses → 'c'

So "222" can be split as:

• [2][2][2] → "aaa"
• [2][22] → "ab"
• [22][2] → "ba"
• [222] → "c"

This is essentially a partitioning problem - we need to count how many ways we can partition a sequence of length n into groups of size 1, 2, or 3 (or 1, 2, 3, 4 for digits 7 and 9).

This naturally leads to a dynamic programming approach. If we know the number of ways to interpret sequences of length i-1, i-2, and i-3, we can compute the ways for length i by considering:

• Taking the last digit as a single press (use solution for i-1)
• Taking the last 2 digits as two presses (use solution for i-2)
• Taking the last 3 digits as three presses (use solution for i-3)

This gives us the recurrence: f[i] = f[i-1] + f[i-2] + f[i-3]

For digits 7 and 9 (which have 4 letters), we extend this to include groups of 4: g[i] = g[i-1] + g[i-2] + g[i-3] + g[i-4]

Since different digit groups are independent (a sequence of 2's doesn't affect how we interpret a sequence of 3's), we can process each group separately and multiply the results together. This multiplication principle works because each choice in one group can be combined with any choice in another group.

The precomputation of arrays f and g is an optimization - instead of computing these values on the fly for each test case, we calculate them once up to a reasonable maximum length (100000 in this case).

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The solution implements the dynamic programming approach with precomputation and grouping.

Step 1: Precomputation of DP Arrays

We precompute two arrays f and g to store the number of ways to interpret sequences of different lengths:

• f[i]: Number of ways for a sequence of length i where each digit maps to 3 letters (digits 2, 3, 4, 5, 6, 8)
• g[i]: Number of ways for a sequence of length i where each digit maps to 4 letters (digits 7, 9)

Base cases:

• f[0] = f[1] = 1: Empty string or single press has 1 way
• f[2] = 2: Two presses can be "aa" or "b"
• f[3] = 4: Three presses can be "aaa", "ab", "ba", or "c"
• Similarly for g with the same initial values

Recurrence relations:

f[i] = (f[i-1] + f[i-2] + f[i-3]) % mod
g[i] = (g[i-1] + g[i-2] + g[i-3] + g[i-4]) % mod

We precompute these arrays up to length 100000 to handle any possible input efficiently.

Step 2: Grouping Consecutive Identical Digits

Using Python's groupby function, we group consecutive identical characters in pressedKeys. For example, "22233" becomes groups: ['2', '2', '2'] and ['3', '3'].

Step 3: Calculate Ways for Each Group

For each group:

1. Determine the character c and count m (length of the group)
2. If c is '7' or '9', use g[m] (4-letter mapping)
3. Otherwise, use f[m] (3-letter mapping)

Step 4: Multiply Results

Since groups are independent, we multiply the number of ways for all groups:

ans = ans * (g[m] if c in "79" else f[m]) % mod

The modulo operation is applied at each multiplication step to prevent overflow.

Example Walkthrough:

For input "222":

• Group: ['2', '2', '2'] with length 3
• Since '2' maps to 3 letters, use f[3] = 4
• Result: 4 ways ("aaa", "ab", "ba", "c")

For input "2279":

• Groups: ['2', '2'] (length 2) and ['7', '9'] (length 1 each)
• '2' group: f[2] = 2 ways
• '7' group: g[1] = 1 way
• '9' group: g[1] = 1 way
• Total: 2 * 1 * 1 = 2 ways

The algorithm runs in O(n) time where n is the length of pressedKeys, with O(1) space for the actual computation (excluding the precomputed arrays).

Example Walkthrough

Let's walk through the example "22233" step by step to understand how the solution works.

Input: pressedKeys = "22233"

Step 1: Understanding the Mapping

• Digit '2' maps to 'abc' (3 letters)
• Digit '3' maps to 'def' (3 letters)

Step 2: Group Consecutive Digits Using groupby, we get:

• Group 1: "222" (three 2's)
• Group 2: "33" (two 3's)

Step 3: Calculate Ways for Each Group

For Group 1 ("222"):

• Length = 3
• Since '2' maps to 3 letters, we use f[3]
• From our precomputed array: f[3] = 4
• The 4 ways to interpret "222" are:
  • [2][2][2] → 'a' + 'a' + 'a' = "aaa"
  • [2][22] → 'a' + 'b' = "ab"
  • [22][2] → 'b' + 'a' = "ba"
  • [222] → 'c' = "c"

For Group 2 ("33"):

• Length = 2
• Since '3' maps to 3 letters, we use f[2]
• From our precomputed array: f[2] = 2
• The 2 ways to interpret "33" are:
  • [3][3] → 'd' + 'd' = "dd"
  • [33] → 'e' = "e"

Step 4: Multiply Results

• Total ways = 4 × 2 = 8

The 8 possible messages are:

1. "aadd" (from [2][2][2] + [3][3])
2. "aade" (from [2][2][2] + [33])
3. "abdd" (from [2][22] + [3][3])
4. "abe" (from [2][22] + [33])
5. "badd" (from [22][2] + [3][3])
6. "bae" (from [22][2] + [33])
7. "cdd" (from [222] + [3][3])
8. "ce" (from [222] + [33])

Final Answer: 8

This demonstrates how the solution:

1. Groups consecutive identical digits
2. Uses precomputed DP values to find the number of interpretations for each group
3. Multiplies the results since groups are independent

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string pressedKeys. This is because:

• The groupby function iterates through the string once, taking O(n) time
• For each group, we convert the iterator to a list to get its length, which takes time proportional to the group size
• The total time for all groups is still O(n) since the sum of all group sizes equals n
• Array lookups f[m] and g[m] are O(1) operations
• The modulo and multiplication operations for each group are O(1)

The space complexity is O(n). While the pre-computed arrays f and g have fixed size of 100004 elements (which could be considered O(1) since it's constant), the dominant factor is:

• The groupby function and converting its result to a list can potentially use O(n) space in the worst case when processing groups
• In the worst case, if all characters are the same, list(s) creates a list of size n

Therefore, the overall space complexity is O(n) due to the space needed during the grouping and list conversion operations.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Handling of Keys 0 and 1

Pitfall: The problem states that the keypad maps digits 2-9 to letters, but doesn't explicitly mention what happens with 0 and 1. A common mistake is assuming the input will never contain these digits or not handling them properly.

Issue: If '0' or '1' appears in pressedKeys, the current solution will incorrectly treat them as 3-letter keys (since they're not in "79"), leading to wrong results. In reality, these keys don't map to any letters on traditional phone keypads, making any sequence containing them invalid.

Solution:

def countTexts(self, pressedKeys: str) -> int:
    # Check for invalid keys first
    if any(c in '01' for c in pressedKeys):
        return 0  # No valid text can be formed
  
    result = 1
    for digit, group in groupby(pressedKeys):
        count = len(list(group))
        if digit in "79":
            result = (result * dp_four[count]) % MOD
        else:
            result = (result * dp_three[count]) % MOD
    return result

2. Integer Overflow in Intermediate Calculations

Pitfall: Forgetting to apply modulo operation consistently during multiplication can cause integer overflow, especially with large inputs.

Issue: Writing result = result * dp_four[count] % MOD applies modulo after multiplication, which could overflow before the modulo is applied.

Solution: Use parentheses to ensure modulo is applied correctly:

result = (result * dp_four[count]) % MOD

3. Memory Issues with Precomputation

Pitfall: Precomputing arrays up to a fixed size (100000) wastes memory if most test cases are small, and fails if input exceeds this limit.

Issue: The current solution precomputes values for sequences up to length 100000, which uses significant memory and might still be insufficient for edge cases.

Solution: Compute values on-demand with memoization:

from functools import lru_cache

class Solution:
    @lru_cache(maxsize=None)
    def count_ways(self, length: int, max_press: int) -> int:
        if length == 0:
            return 1
        if length < 0:
            return 0
      
        total = 0
        for press in range(1, min(length, max_press) + 1):
            total = (total + self.count_ways(length - press, max_press)) % MOD
        return total
  
    def countTexts(self, pressedKeys: str) -> int:
        result = 1
        for digit, group in groupby(pressedKeys):
            count = len(list(group))
            max_press = 4 if digit in "79" else 3
            result = (result * self.count_ways(count, max_press)) % MOD
        return result

4. Incorrect Base Cases for DP Arrays

Pitfall: Setting wrong initial values for the DP arrays, particularly for index 0.

Issue: Some might initialize dp_three[0] = 0 thinking an empty sequence has no ways, but it should be 1 (representing the empty string).

Solution: Ensure base cases are:

• dp[0] = 1 (empty string)
• dp[1] = 1 (single press = single letter)
• dp[2] = 2 (two presses = two possibilities)
• dp[3] = 4 (three presses = four possibilities)

5. Misunderstanding the Recurrence Relation

Pitfall: Using the wrong recurrence relation, such as thinking it's a simple Fibonacci sequence.

Issue: Writing dp[i] = dp[i-1] + dp[i-2] instead of considering all possible last letter formations.

Solution: Remember that for 3-letter keys, the last letter can be formed by 1, 2, or 3 presses, so:

dp_three[i] = dp_three[i-1] + dp_three[i-2] + dp_three[i-3]

And for 4-letter keys (7 and 9):

dp_four[i] = dp_four[i-1] + dp_four[i-2] + dp_four[i-3] + dp_four[i-4]